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Approximations of one singular integral on an interval by Fourier–Chebyshev rational integral operators P. G. Potseiko, E. A. Rovba Yanka Kupala State University of Grodno, Grodno, Belarus
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     § 1. IntroductionVarious problems in mathematics and its applications involve singular integrals with Cauchy-type kernel of the form
$$
\begin{equation}
\widehat{f}(x)=\int_{-1}^{1}\frac{f(t)}{t-x}\sqrt{1-t^2}\,dt, \qquad x \in [-1,1],
\end{equation}
\tag{1.1}
$$
which are understood in the sense of the Cauchy principal value. A sufficient condition for such an integral to exist is that the density $f(t)$ obeys a Hölder condition with some exponent (see [1] and [2]). The above integrals can be evaluated precisely (in closed form) only in quite rare particular cases. This calls for the development of approximate methods. At present, the technique of numerical integration of singular integrals of the form (1.1) is well developed (see, for example, [3]–[8]). For a sufficiently complete account of the results in this direction, see Gabdulkhaev [9]. In 1993 Rusak [10] proposed a method of rational approximation of singular integrals of the form (1.1) with density $f(t)$ in certain function classes on an interval. His studies were continued by his student Boksha [11] and also in the paper [12]. Pointwise approximations by algebraic polynomials of function classes defined by singular integrals of the form (1.1) were studied by Motornyi [13], who obtained asymptotically sharp estimates for approximations. Note that the above studies did not use the classical methods based on Fourier series. In 1925–1926 Takenaka [14] and Malmquist [15] introduced an orthogonal system of rational functions on the unit circle which generalizes the trigonometric system. Dzhrbashyan [16] constructed rational Fourier series in this system, evaluated the Dirichlet integral, and established analogues of the Jordan–Dirichlet and Dini–Lipschitz tests under the assumption that the poles of the rational functions have no accumulation points on the unit circle. Dzhrbashyan and Kitbalyan [17] constructed systems of rational functions which are orthogonal on $[-1,1]$ and generalize the classical Chebyshev polynomials of the first and second kind. In 1979 Rovba [18] introduced an integral operator associated with the system of Chebyshev–Markov rational functions and generalizing partial sums of the polynomial Fourier–Chebyshev series. Let $\{a_k\}_{k=1}^n$ be an arbitrary set of numbers, where either the $a_k$, $|a_k|<1$, are real, or they are complex conjugates of one another. On the set of functions $f(x)$ integrable with weight $(1-x^2)^{-1/2}$ on $[-1,1]$ consider the rational integral operator (see [18])
$$
\begin{equation}
s_n(f,x)=\frac{1}{2\pi} \int_{-\pi}^{\pi}f(\cos v) \frac{\sin\lambda_n(v,u)}{\sin\frac{v-u}{2}}\,dv, \qquad x=\cos u,
\end{equation}
\tag{1.2}
$$
where
$$
\begin{equation}
\begin{gathered} \, \notag \lambda_n(v,u)=\int_{u}^{v} \biggl(\frac{1}{2}+\lambda_n(y)\biggr)\,dy, \\ \lambda_n(y)=\sum_{k=1}^{n}\frac{1-|z_k|^2}{1+2|z_k| \cos (y-\operatorname{arg}z_k) +|z_k|^2}\quad\text{and}\quad z_k=\frac{a_k}{1+\sqrt{1-a_k^2}}, \quad |z_k|<1. \end{gathered}
\end{equation}
\tag{1.3}
$$
The operator $s_n\colon f \to \mathbb{R}_n(A_n)$ acts identically on constants. Here $\mathbb{R}_n(A_n)$ is the set of rational functions of the form
$$
\begin{equation*}
\frac{p_n(x)}{\prod_{k=1}^{n}(1+a_k x)}, \qquad p_n \in \mathbb{P}_n,
\end{equation*}
\notag
$$
and $A_n$ is the set of parameters $(a_1,\dots,a_{n})$. In particular, if $a_k=0$, $k=1,2,\dots,n$, then $s_n(f,x)$ is a partial sum of the polynomial Fourier–Chebyshev series. Along with the integral representation (1.2), upper estimates for approximations by the operator $s_n(\,\cdot\,{,}\,\cdot\,)$ of some function classes on an interval were also obtained in [18]. Rational integral operators (1.2) are widely useful in rational approximation [19]–[21]. On the basis of Fourier–Chebyshev rational integral operator (1.2), we introduce the operator
$$
\begin{equation}
\widehat{s}_{n+1}(f,x)=\int_{-1}^{1}\frac{s_{n}(f,t)}{t-x}\sqrt{1-t^2}\,dt, \qquad x \in (-1,1).
\end{equation}
\tag{1.4}
$$
It is known (see, for example, [10]) that $s_{n+1}(f,x)$ is also a rational function of order $\leqslant n+1$ and has the same poles as ${s}_{n}(f,x)$. It would be interesting to study approximations of singular integrals (1.1) on $[-1,1]$ by the Fourier–Chebyshev rational integral operator (1.2) and the above operator (1.4). In our paper we consider both approaches. As a separate problem, we study approximations of individual singular integrals (1.1) in which the density has a power-law singularity. We consider the cases where the approximating rational function has constraints on the number of geometrically different parameters and where the parameters of the approximating function are certain modifications of the ‘Newman’ parameters. § 2. Integral representations of approximationsWe set
$$
\begin{equation*}
{\varepsilon}_n(\widehat{f},x,A_n)=\widehat{f}(x) - s_n(\widehat{f},x), \qquad x \in [-1,1],
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
{\varepsilon}_n(\widehat{f},A_n)=\| \widehat{f}(x) - s_n(\widehat{f},x)\|_{C[-1,1]}, \qquad n \in \mathbb{N}.
\end{equation*}
\notag
$$
Theorem 1. The following integral representation holds: Proof. Operator (1.2) acts identically on constants, and so
$$
\begin{equation*}
\varepsilon_n(\widehat{f},x,A_n) =\int_{-\pi}^{\pi}\biggl(\int_{-1}^{1}\frac{f(t)}{t-\cos u}\sqrt{1-t^2}\,dt -\int_{-1}^{1}\frac{f(t)}{t-\cos v}\sqrt{1-t^2}\,dt\!\biggr)D_n(v,u)\,dv,
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
D_n(v,u)=\frac{1}{2\pi}\frac{\sin\lambda_n(v,u)}{\sin\frac{v-u}{2}}, \qquad x=\cos u.
\end{equation*}
\notag
$$
Using Fubini’s theorem to change the order of integration in the last expression we have
$$
\begin{equation}
\varepsilon_n(\widehat{f},x,A)=\int_{-1}^{1}\frac{f(t)\sqrt{1-t^2}}{t-x}I_n(x,t)\,dt,
\end{equation}
\tag{2.2}
$$
where
$$
\begin{equation*}
I_n(x,t)=\int_{-\pi}^{\pi}\frac{\cos u - \cos v}{\cos \tau - \cos v}D_n(v,u)\,dv, \qquad x=\cos u, \quad t=\cos \tau.
\end{equation*}
\notag
$$
Note that the integral $I_n(x,t)$ is singular and is understood in the sense of the Cauchy principal value. Let us evaluate this integral. To do this we change the integration variable to $\zeta=\mathrm{e}^{ i v}$ and set $\xi=\mathrm{e}^{ i u}$ and $z=\mathrm{e}^{ i \tau}$. Then we have
$$
\begin{equation*}
I_n(x,t) =\frac{1}{2\pi i\xi}\oint_{\Gamma}\frac{1-\xi \zeta}{(\zeta-z)(\zeta-\frac1{z})} \biggl(\zeta\frac{\omega_n(\zeta)}{\omega_n(\xi)}-\xi\frac{\omega_n(\xi)}{\omega_n(\zeta)}\biggr) \frac{d\zeta}{\zeta},
\end{equation*}
\notag
$$
where
$$
\begin{equation}
\omega_n(\zeta)=\prod_{k=1}^{n}\frac{\zeta+z_k}{1+z_k \zeta},
\end{equation}
\tag{2.3}
$$
$\Gamma=\{\zeta\colon \zeta=\mathrm{e}^{i v}, -\pi \leqslant v \leqslant \pi \}$. The integral $I_n(x,t)$ contains the points of singularity $\zeta=z$ and $\zeta=1/z$. To evaluate this integral we split it into two. We have
$$
\begin{equation}
I_n(x,t)=\frac{1}{2\pi i\xi}\bigl[\overline{\omega_n(\xi)}I^{(1)}(x,t) -\xi\omega_n(\xi)I^{(2)}(x,t)\bigr],
\end{equation}
\tag{2.4}
$$
where
$$
\begin{equation*}
I^{(1)}(x,t) =\oint_{\Gamma}\frac{1-\xi \zeta}{(\zeta-z)(\zeta-\frac1{z})}\, \omega_n(\zeta)\,d\zeta
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
I^{(2)}(x,t)=\oint_{\Gamma}\frac{1-\xi \zeta}{(\zeta-z)(\zeta-\frac1{z})\zeta}\, \overline{\omega_n(\zeta)}\,d\zeta.
\end{equation*}
\notag
$$
The integrals $I^{(1)}(x,t)$ and $I^{(2)}(x,t)$ are also understood in the sense of the Cauchy principal value. We write $I^{(1)}(x,t)$ as
$$
\begin{equation*}
I^{(1)}(x,t) =\frac{1-\xi z}{z-\frac1{z}}\oint_{\Gamma}\frac{\omega_n(\zeta)}{\zeta-z}\,d\zeta +\frac{\frac{\xi}{z}-1}{z-\frac1{z}}\oint_{\Gamma}\frac{\omega_n(\zeta)}{\zeta-\frac1{z}}\,d\zeta, \qquad z=\mathrm{e}^{ i \tau}.
\end{equation*}
\notag
$$
Applying Sokhotskii’s formulae to each integral on the right yields
$$
\begin{equation}
I^{(1)}(x,t) =\frac{i \pi }{z^2-1}[z \omega_n(z) (1-\xi z) - \overline{\omega_n(z)}(z-\xi)].
\end{equation}
\tag{2.5}
$$
A similar analysis shows that
$$
\begin{equation}
I^{(2)}(x,t) =\frac{i \pi }{z^2-1}[z \omega_n(z) (z-\xi) - \overline{\omega_n(z)}(1 -\xi z)].
\end{equation}
\tag{2.6}
$$
Plugging (2.5) and (2.6) into (2.4) we obtain
$$
\begin{equation*}
\begin{aligned} \, I_n(x,t) &=\frac{1}{2 \sin \tau} \biggl[\sin \frac{\tau +u}{2}\biggl(\sqrt{\frac{z}{\xi}}\, \frac{\omega_n(z)}{\omega_n(\xi)} +\sqrt{\frac{\xi}{z}}\, \frac{\omega_n(\xi)}{\omega_n(z)}\biggr) \\ &\qquad -\sin \frac{\tau -u}{2}\biggl(\frac{1}{\sqrt{\xi z}\, \omega_n(\xi) \omega_n(z)} +\sqrt{\xi z} \, \omega_n(\xi) \omega_n(z)\biggr)\biggr]. \end{aligned}
\end{equation*}
\notag
$$
Here we choose the branch of the root function so that $\sqrt{z} \big|_{z=1}=1$ and $\sqrt{\xi} \big|_{\xi=1}=1$. Substituting $I_n(x,t)$ into the integral representation (2.2) we obtain
$$
\begin{equation*}
\begin{aligned} \, \varepsilon_n(\widehat{f},x,A_n) &=\frac{1}{2}\int_{-1}^{1}\frac{f(t)\sqrt{1-t^2}}{t-x} \biggl[\sin \frac{\tau +u}{2}\biggl(\sqrt{\frac{z}{\xi}}\, \frac{\omega_n(z)}{\omega_n(\xi)} +\sqrt{\frac{\xi}{z}}\, \frac{\omega_n(\xi)}{\omega_n(z)}\biggr) \\ &\qquad -\sin \frac{\tau -u}{2}\biggl(\frac{1}{\sqrt{\xi z}\, \omega_n(\xi) \omega_n(z)} +\sqrt{\xi z}\, \omega_n(\xi) \omega_n(z)\biggr)\biggr]\frac{dt}{\sqrt{1-t^2}}, \end{aligned}
\end{equation*}
\notag
$$
where $t=\cos \tau$, $x=\cos u$, $\xi=\mathrm{e}^{i u}$ and $z=\mathrm{e}^{ i \tau}$.
Integration with respect to $\tau$ in the integral on the right yields
$$
\begin{equation*}
\begin{aligned} \, &\varepsilon_n(\widehat{f},x,A_n) =\frac{1}{2}\biggl[\int_{0}^{\pi}\frac{f(\cos \tau) \sin \tau}{\cos \tau-\cos u}\sin \frac{\tau +u}{2}\biggl(\sqrt{\frac{z}{\xi}}\, \frac{\omega_n(z)}{\omega_n(\xi)} +\sqrt{\frac{\xi}{z}}\, \frac{\omega_n(\xi)}{\omega_n(z)}\biggr)\,d \tau \\ &\qquad -\int_{0}^{\pi}\frac{f(\cos \tau) \sin \tau}{\cos \tau-\cos u} \sin \frac{\tau -u}{2}\biggl(\frac{1}{\sqrt{\xi z} \, \omega_n(\xi) \omega_n(z)} +\sqrt{\xi z} \, \omega_n(\xi) \omega_n(z)\biggr)\,d\tau\biggr]. \end{aligned}
\end{equation*}
\notag
$$
Making the change $\tau \mapsto -\tau$ in the second integral we obtain
$$
\begin{equation*}
\varepsilon_n(\widehat{f},x,A_n) =\frac{1}{2}\int_{-\pi}^{\pi}\frac{f(\cos \tau) \sin \tau}{\cos \tau-\cos u} \sin \frac{\tau +u}{2}\biggl(\sqrt{\frac{z}{\xi}}\, \frac{\omega_n(z)}{\omega_n(\xi)} +\sqrt{\frac{\xi}{z}}\, \frac{\omega_n(\xi)}{\omega_n(z)}\biggr)\,d \tau.
\end{equation*}
\notag
$$
Noting that
$$
\begin{equation*}
\frac{1}{2}\biggl(\sqrt{\frac{z}{\xi}}\, \frac{\omega_n(z)}{\omega_n(\xi)} +\sqrt{\frac{\xi}{z}}\, \frac{\omega_n(\xi)}{\omega_n(z)}\biggr) =\cos \lambda_n(\tau,u)
\end{equation*}
\notag
$$
(see [16]), where $\lambda_n(\tau,u)$ is defined in (1.3), we arrive at (2.1).
This proves Theorem 1. In Theorem 1 we set $z_k=0$, $k=1,2,\dots,n$. In this case
$$
\begin{equation*}
\varepsilon_n (\widehat{f},x,O)=\varepsilon_n^{(0)} (\widehat{f},x), \qquad O=(\underbrace{0,0,\dots,0}_{n}), \quad n=0,1,2,\dots,
\end{equation*}
\notag
$$
are the approximations of the singular integral (1.1) by partial sums of the polynomial Fourier–Chebyshev series. Now consider the operator (1.4). We set
$$
\begin{equation*}
\widehat{\varepsilon}_{n+1}(f,x,A_n)=\widehat{f}(x) - \widehat{s}_{n+1}(f,x), \qquad x \in [-1,1],
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\widehat{\varepsilon}_{n+1}(f,A_n)=\| \widehat{f}(x) - \widehat{s}_{n+1}(f,x)\|_{C[-1,1]}, \qquad n \in \mathbb{N} \cup \{0\}.
\end{equation*}
\notag
$$
Theorem 2. The following integral representation holds: Proof. It is known (see [18]) that the Fourier–Chebyshev rational integral operator (1.2) can be written in the form where
$$
\begin{equation*}
D_n(v,u)=\frac{1}{2\pi} \biggl(\frac{\zeta\frac{\omega_n(\zeta)}{\omega_n(\xi)} -\xi\frac{\omega_n(\xi)}{\omega_n(\zeta)}}{\zeta - \xi} +\frac{\frac{1}{\omega_n(\zeta)\omega_n(\xi)}-\xi \zeta\omega_n(\xi)\omega_n(\zeta)} {1-\zeta \xi}\biggr),
\end{equation*}
\notag
$$
$\xi=\mathrm{e}^{ i u}$, $\zeta=\mathrm{e}^{i v}$ and $\omega_n(y)$ is defined in (2.3).
Substituting the last expression into (1.4) and using Fubini’s theorem to change the order of integration we obtain
$$
\begin{equation}
\widehat{s}_{n+1}(f,x)=\int_{0}^{\pi}f(\cos v)I_{n+1}(x,v)\,dv, \qquad x=\cos u,
\end{equation}
\tag{2.9}
$$
where
$$
\begin{equation*}
I_{n+1}(x,v)=\int_{-1}^{1}D_n(v,\tau)\frac{\sqrt{1-t^2}}{t-x}\,dt, \qquad t=\cos \tau.
\end{equation*}
\notag
$$
Note that integral $I_{n+1}(x,v)$ is singular and is understood in the sense of the Cauchy principal value. Putting $t=\cos \tau$ and $x=\cos u$ we can write this integral as
$$
\begin{equation*}
\begin{aligned} \, I_{n+1}(x,v) &=\frac{1}{2\pi}\int_{0}^{\pi}\biggl[z\frac{\omega_n(z)}{\omega_n(\zeta)} -\zeta\frac{\omega_n(\zeta)}{\omega_n(z)}\biggr] \frac{\sin^2 \tau\, d\tau}{(z-\zeta)(\cos \tau-\cos u)} \\ &\qquad +\frac{1}{2\pi}\int_{0}^{\pi} \biggl[\frac{1}{\omega_n(\zeta)\omega_n(z)}-z \zeta\omega_n(z)\omega_n(\zeta)\biggr] \frac{\sin^2 \tau \,d\tau}{(1 - z \zeta)(\cos \tau-\cos u)}, \\ &\qquad\qquad\qquad \zeta=\mathrm{e}^{iv},\qquad z=\mathrm{e}^{i\tau}. \end{aligned}
\end{equation*}
\notag
$$
After the change of the variable $\tau \mapsto -\tau$ in the second integral we have
$$
\begin{equation*}
I_{n+1}(x,v) =\frac{1}{2\pi}\int_{-\pi}^{\pi} \biggl[z\frac{\omega_n(z)}{\omega_n(\zeta)}-\zeta\frac{\omega_n(\zeta)}{\omega_n(z)}\biggr] \frac{\sin^2 \tau\, d\tau}{(z-\zeta)(\cos \tau-\cos u)},
\end{equation*}
\notag
$$
where $x=\cos u$, $\zeta=\mathrm{e}^{iv}$ and $z=\mathrm{e}^{i\tau}$.
The further substitution $z=\mathrm{e}^{i \tau}$, $\zeta=\mathrm{e}^{iv}$ gives
$$
\begin{equation}
I_{n+1}(x,v)=-\frac{1}{4\pi i}\oint_{\Gamma} \biggl[\frac{z^2}{\zeta}\frac{\omega_{n-1}(z)}{\omega_{n-1}(\zeta)}- \frac{\zeta^2}{z}\frac{\omega_{n-1}(\zeta)}{\omega_{n-1}(z)}\biggr] \frac{(z^2-1)^2\,dz}{z^2 (z-\zeta)(z-\xi)(z-\frac1{\xi})},
\end{equation}
\tag{2.10}
$$
where $\Gamma=\{z\colon z=\mathrm{e}^{i \tau}, -\pi \leqslant v \leqslant \pi \}$.
The integral $I_{n+1}(x,v)$ contains the points of singularity $z=\xi$ and $z=1/\xi$. Our further analysis depends on the method proposed in [22], p. 115. From (1.2) it follows that $I_{n+1}(x,v)$ is a rational function of $\xi=\mathrm{e}^{i u}$ and $\zeta=\mathrm{e}^{i v}$ with first-order poles at the points $z_k$ and $\overline{z}_k$, $k=1,2,\dots,n-1$ (see (1.3)). So it suffices to evaluate the integral $I_{n+1}(x,v)$ for $\zeta=\delta \mathrm{e}^{i v}$, $\delta \in (0,1)$, $v \in (0,\pi)$, and then let $\delta \to 1$. In view of the above we write the integral $I_{n+1}(x,v)$ as
$$
\begin{equation}
I_{n+1}(x,v)=-\frac{1}{4\pi i} \bigl[\overline{\zeta \omega_{n-1}(\zeta)}J^{(1)}-\zeta^2 \omega_{n-1}(\zeta)J^{(2)}\bigr],
\end{equation}
\tag{2.11}
$$
where
$$
\begin{equation*}
J^{(1)}(x,v) =\oint_{\Gamma} \frac{ {\omega_{n-1}(z) (z^2-1)^2}\,dz}{(z-\zeta)(z-\xi)(z-\frac{1}{\xi})}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
J^{(2)}(x,v) =\oint_{\Gamma}\frac{\overline{\omega_{n-1}(z)}(z^2-1)^2\,dz} {z^3(z-\zeta)(z-\xi)(z-\frac{1}{\xi})}.
\end{equation*}
\notag
$$
The integrals $J^{(1)}(x,v)$ and $J^{(2)}(x,v)$ contain the points of singularity $z=\xi$ and $z=1/\xi$ on the boundary of the unit disc. To evaluate these integrals we use Sokhotskii’s formulae. Proceeding as with the expression (2.4), we find that
$$
\begin{equation}
\begin{aligned} \, \notag J^{(1)}(x,v) &=2 \pi i\frac{\xi\omega_{n-1}(\zeta)(\zeta^2-1)^2}{(\zeta-\xi)(\zeta \xi-1)} \\ &\qquad +\pi i\biggl(\frac{\xi \omega_{n-1}(\xi)(\xi^2-1)}{\xi-\zeta} +\frac{\overline{\omega_{n-1}(\xi)} (1-\xi^2)}{\xi^2(1 - \xi \zeta)}\biggr) \end{aligned}
\end{equation}
\tag{2.12}
$$
and
$$
\begin{equation}
J^{(2)}(x,v) =- \pi i\biggl(\frac{\xi\omega_{n-1}(\xi)(1-\xi^2)}{1-\xi \zeta} +\frac{\overline{\omega_{n-1}(\xi)}(\xi^2-1)}{\xi^2(\xi-\zeta)}\biggr),
\end{equation}
\tag{2.13}
$$
where $x=\cos u$, $\zeta=\mathrm{e}^{iv}$ and $\xi=\mathrm{e}^{i u}$.
Substituting (2.12) and (2.13) into (2.11) we have
$$
\begin{equation*}
\begin{aligned} \, I_{n+1}(x,v) &=\frac{ \sin^2 v}{\cos v-\cos u} +\frac{1}{2}\sin u \frac{\cos \lambda_{n}(v,u)}{\sin \frac{v-u}{2}} \\ &\qquad -\frac{1}{4}\frac{\sin u}{\sin \frac{v+u}{2}} \bigl((\xi \zeta)^{3/2}\omega_n(\xi)\omega_n(\zeta) +(\xi \zeta)^{-3/2}\overline{\omega_n(\xi)\omega_n(\zeta)}\bigr), \end{aligned}
\end{equation*}
\notag
$$
where $\lambda_{n}(v,u)$ was defined in (1.3) and $z_n=0$.
Inserting the last equality into (2.9), splitting the resulting expression into three integrals and changing the variable $v$ to $ -v$ in the integral corresponding to the third term, we obtain
$$
\begin{equation*}
\begin{aligned} \, \widehat{s}_{n+1}(f,x) &=\int_{0}^{\pi}f(\cos v)\frac{\sin^2 v}{\cos v - \cos u}\,dv \\ &\qquad +\frac{\sqrt{1-x^2}}{2}\int_{-\pi}^{\pi}f(\cos v)\frac{\cos \lambda_{n}(v,u)}{\sin \frac{v-u}{2}}\,dv, \qquad z_n=0. \end{aligned}
\end{equation*}
\notag
$$
Now (2.8) follows on noticing that
$$
\begin{equation*}
\widehat{f}(x)=\int_{0}^{\pi}f(\cos v)\frac{\sin^2 v}{\cos v - \cos u}\,dv, \qquad x=\cos u.
\end{equation*}
\notag
$$
This proves Theorem 2. In Theorem 2, we set $z_k=0$, $k=1,2,\dots,n-1$. In this case
$$
\begin{equation*}
\widehat{\varepsilon}_{n+1}(f,x,O)=\widehat{\varepsilon}_{n+1}^{(0)}(f,x),\quad\text{where } O=(\underbrace{0,0,\dots,0}_{n-1}), \quad n=0,1,2,\dots,
\end{equation*}
\notag
$$
are the approximations of the singular integral (1.1) by the polynomial integral operator that is the image under the transformation (1.1) of partial sums of the Fourier–Chebyshev series. As a result, we have the following statement. § 3. Approximations of a singular integral with density satisfying a Hölder condition (the polynomial case)Consider the classes $H^{(\alpha)}[-1,1]$, $\alpha \in (0,1]$, of functions $f(x)$ satisfying the Hölder condition of exponent $\alpha$ with constant 1, that is,
$$
\begin{equation*}
|f(x_1)- f(x_2)| \leqslant | x_1 - x_2 |^\alpha, \qquad x_1,x_2 \in [-1,1].
\end{equation*}
\notag
$$
Let us study the approximations (2.7) and (2.14) in the case where $f \in H^{(\alpha)}[-1,1]$, $\alpha \in (0,1]$. Theorem 3. The approximations on $[-1,1]$ of the singular integral (1.1) with density $f \in H^{(\alpha)} [-1,1]$, $\alpha \in (0,1]$, by partial sums of the Fourier–Chebyshev series satisfy the upper estimate Proof. From (2.7) it is easily seen that
$$
\begin{equation*}
\begin{aligned} \, \varepsilon_n^{(0)}(\widehat{f},x) &=-\frac{1}{4}\int_{0}^{2\pi}\bigl[f(\cos(u+\tau))\sin(u+\tau)- f(\cos(u-\tau))\sin(u-\tau)\bigr] \\ &\qquad\qquad\times \frac{\cos(n+\frac12) \tau}{\sin \frac{\tau}{2}}\,d\tau. \end{aligned}
\end{equation*}
\notag
$$
This can be rewritten as
$$
\begin{equation}
\varepsilon_n^{(0)}(\widehat{f},x) =-\frac{1}{4}\bigl(\sin u I_1 +2 \cos u I_2\bigr), \qquad x=\cos u, \quad x \in [-1,1],
\end{equation}
\tag{3.2}
$$
where and
We consider each integral separately. The corresponding results are formulated as two lemmas. Lemma 1. For $\alpha \in (0,1)$, and for $\alpha=1$, where $n_0(\alpha)$ is a natural number depending only on $\alpha$. Proof. We write the integral $I_1$ as where and
Consider the integral $I_{11}$ (the arguments for the integral $I_{12}$ are the same). We use the method proposed by Besov in [23] and [24]. The integrand is $2\pi$-periodic, and we can write the integral $I_{11}$ as
$$
\begin{equation*}
I_{11}=\frac{1}{2}\int_{0}^{2\pi}\biggl[\varphi_u(\tau) -\varphi_u\biggl(\tau+\frac{\pi}{\lambda_1}\biggr)\biggr]\cos \lambda_1 \tau\,d\tau, \qquad \lambda_1=n+\frac{1}{2}.
\end{equation*}
\notag
$$
Splitting the integral on the right into three we have
$$
\begin{equation}
I_{11}=\frac{1}{2}[I_{111} +I_{112}+I_{113}], \qquad x=\cos u, \quad x \in [-1,1],
\end{equation}
\tag{3.5}
$$
where
$$
\begin{equation*}
\begin{gathered} \, I_{111} =\int_{0}^{{\pi}/{\lambda_1}} \biggl[\varphi_u(\tau)-\varphi_u\biggl(\tau+\frac{\pi}{\lambda_1}\biggr)\biggr] \cos \lambda_1 \tau\,d\tau, \\ I_{112} =\int_{{\pi}/{\lambda_1}}^{2\pi-{\pi}/{\lambda_1}} \biggl[\varphi_u(\tau)-\varphi_u\biggl(\tau+\frac{\pi}{\lambda_1}\biggr)\biggr] \cos \lambda_1 \tau\,d\tau \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
I_{113} =\int_{2\pi-{\pi}/{\lambda_1}}^{2\pi} \biggl[\varphi_u(\tau)-\varphi_u\biggl(\tau+\frac{\pi}{\lambda_1}\biggr)\biggr] \cos \lambda_1 \tau\,d\tau.
\end{equation*}
\notag
$$
Since $(2/\pi) \tau < \sin \tau < \tau$, $\tau \in (0,\pi/2)$, for the function $\varphi_u(\tau)$ we have
$$
\begin{equation*}
|\varphi_u(\tau)|\leqslant\frac{2^\alpha \pi |{\sin u}|^\alpha |{\sin \tau}|^{\alpha}}{\tau}, \qquad u \in [0,\pi], \quad \alpha \in (0,1].
\end{equation*}
\notag
$$
For the integral $I_{111}$ we have
$$
\begin{equation}
\begin{aligned} \, \notag |I_{111}| &\leqslant 2^\alpha \pi |{\sin u}|^\alpha \biggl(\int_{0}^{{\pi}/{\lambda_1}}\tau^{\alpha-1}\,d\tau +\int_{0}^{{\pi}/{\lambda_1}}\biggl(\tau+\frac{\pi}{\lambda_1}\biggr)^{\alpha-1}\,d\tau\biggr) \\ \notag &=\frac{2^\alpha \pi|{\sin u}|^\alpha}{\alpha} \biggl(\biggl(\frac{\pi}{\lambda_1}\biggr)^{\alpha} +\biggl(\frac{2\pi}{\lambda_1}\biggr)^{\alpha} -\biggl(\frac{\pi}{\lambda_1}\biggr)^{\alpha} \biggr) \\ &=\frac{2^{2\alpha}\pi^{1+\alpha}|{\sin u}|^\alpha}{\alpha \lambda_1^\alpha}, \qquad \alpha \in (0,1]. \end{aligned}
\end{equation}
\tag{3.6}
$$
Changing the variable to $\tau \mapsto 2\pi-\tau$ in the integral $I_{113}$ and proceeding similarly, we obtain
$$
\begin{equation}
|I_{113}| \leqslant\frac{2^{2\alpha}\pi^{1+\alpha}|{\sin u}|^\alpha}{\alpha \lambda_1^\alpha}, \qquad u \in [0,\pi], \quad \alpha \in (0,1], \quad \lambda_1=n+\frac{1}{2}.
\end{equation}
\tag{3.7}
$$
Consider the integral $I_{112}$. As before, the integrand is $2\pi$-periodic, and so it suffices to estimate the absolute value of the integral over $[\pi/\lambda_1,\pi]$ (the estimate for the rest of the integral is the same). We have
$$
\begin{equation}
\int_{{\pi}/{\lambda_1}}^{\pi}\biggl[\varphi_u(\tau)-\varphi_u \biggl(\tau+\frac{\pi}{\lambda_1}\biggr)\biggr] \cos \lambda_1 \tau \,d\tau =I_3+I_4,
\end{equation}
\tag{3.8}
$$
where
$$
\begin{equation*}
\begin{gathered} \, I_3=\int_{{\pi}/{\lambda_1}}^{\pi} \frac{g(u,\tau)}{\sin \frac{\tau+{\pi}/{\lambda_1}}{2}}\cos \lambda_1 \tau\,d\tau, \\ \begin{aligned} \, g(u,\tau) &=f(\cos(u+\tau))-f(\cos(u-\tau)) \\ &\qquad-f\biggl(\cos\biggl(u+\tau+\frac{\pi}{\lambda_1}\biggr)\biggr) +f\biggl(\cos\biggl(u-\tau-\frac{\pi}{\lambda_1}\biggr)\biggr), \end{aligned} \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
I_4=\int_{{\pi}/{\lambda_1}}^{\pi} [f(\cos(u+\tau))-f(\cos(u-\tau))] \biggl[\frac{1}{\sin \frac{\tau}{2}}-\frac{1}{\sin\frac{\tau+\pi/\lambda_1}{2}}\biggr] \cos \lambda_1 \tau\,d\tau.
\end{equation*}
\notag
$$
Next, $f \in H^{(\alpha)} [-1,1]$, $\alpha \in (0,1]$, and so
$$
\begin{equation}
\begin{aligned} \, \notag &\biggl|f(\cos(u \pm \tau))-f\biggl(\cos\biggl(u \pm \tau \pm \frac{\pi}{\lambda_1}\biggr)\biggr)\biggr| \\ \notag &\qquad\leqslant\biggl|2 \sin \biggl(u \pm \tau \pm \frac{\pi}{2\lambda_1}\biggr) \sin \frac{\pi}{2\lambda_1}\biggr|^\alpha \\ \notag &\qquad=\biggl|\sin(u \pm \tau )\sin \frac{\pi}{\lambda_1}\pm2\cos(u \pm \tau ) \sin^2 \frac{\pi}{2\lambda_1}\biggr|^\alpha \\ \notag &\qquad=\biggl|\sin u \cos \tau \sin \frac{\pi}{\lambda_1} \pm\cos u \sin \tau \sin \frac{\pi}{\lambda_1} +2\cos(u \pm \tau )\sin^2 \frac{\pi}{2\lambda_1}\biggr|^\alpha \\ &\qquad\leqslant |{\sin u}|^\alpha\sin^\alpha \frac{\pi}{\lambda_1} +|{\cos u}|^\alpha\tau^\alpha \sin^\alpha \frac{\pi}{\lambda_1} + 2^\alpha\sin^{2\alpha} \frac{\pi}{2 \lambda_1}, \notag \\ &\qquad\qquad\qquad\qquad u \in [0,\pi], \qquad \tau \in \biggl[\frac{\pi}{\lambda_1},\pi\biggr]. \end{aligned}
\end{equation}
\tag{3.9}
$$
Therefore, for $I_3$ we have the estimate
$$
\begin{equation}
\begin{aligned} \, \notag |I_3| &\leqslant2\pi\biggl(|{\sin u}|^\alpha\biggl(\frac{\pi}{\lambda_1}\biggr)^\alpha +2^\alpha\biggl(\frac{\pi}{2\lambda_1}\biggr)^{2\alpha}\biggr) \int_{\pi/\lambda_1}^{\pi}\frac{d\tau}{\tau} \\ &\qquad +2\pi|{\cos u}|^\alpha\biggl(\frac{\pi}{\lambda_1}\biggr)^\alpha \int_{{\pi}/{\lambda_1}}^{\pi}\tau^{\alpha-1}\,d\tau \notag \\ &\leqslant2\pi^{1+\alpha}|{\sin u}|^\alpha \frac{\log \lambda_1}{\lambda_1^\alpha} +\frac{2\pi^{2\alpha+1}|{\cos u}|^\alpha}{\alpha \lambda_1^\alpha} +2^{1-\alpha}\pi^{1+2\alpha}\frac{\log \lambda_1}{\lambda_1^{2\alpha}}, \qquad u \in [0,\pi] . \end{aligned}
\end{equation}
\tag{3.10}
$$
Let us now consider the integral $I_4$. We have
$$
\begin{equation*}
f(\cos(u+\tau))-f(\cos(u-\tau))| \leqslant2^\alpha |{\sin u}|^\alpha \tau^\alpha, \qquad u \in [0,\pi], \quad \tau \in \biggl[\frac{\pi}{\lambda_1},\pi\biggr],
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\biggl|\frac{1}{\sin\frac{\tau}{2}}-\frac{1}{\sin \frac{\tau+{\pi}/{\lambda_1}}{2}}\biggr| \leqslant\frac{\pi^3}{2\lambda_1 \tau^2}, \qquad \tau \in \biggl[\frac{\pi}{\lambda_1},\pi\biggr].
\end{equation*}
\notag
$$
As a result,
$$
\begin{equation*}
|I_4|\leqslant\frac{2^{\alpha-1}\pi^3}{\lambda_1}|{\sin u}|^\alpha \int_{\pi/\lambda_1}^{\pi}\tau^{\alpha-2}\,d\tau.
\end{equation*}
\notag
$$
Hence
$$
\begin{equation}
|I_4|\leqslant\begin{cases} \displaystyle \frac{\pi^{2+\alpha}|{\sin u}|^\alpha}{2^{1-\alpha}(1-\alpha)\lambda_1^\alpha}, &\alpha \in (0,1), \\ \displaystyle \pi^3 |{\sin u}|\frac{\log \lambda_1}{\lambda_1}, &\alpha=1. \end{cases}
\end{equation}
\tag{3.11}
$$
From (3.8), using estimates (3.10) and (3.11), for $I_{112}$ we obtain
$$
\begin{equation}
|I_{112}|\leqslant\begin{cases} \displaystyle 8\pi^{1+\alpha}|{\sin u}|^\alpha\frac{\log \lambda_1}{\lambda_1^\alpha} +\frac{4\pi^{2\alpha+1}|{\cos u}|^\alpha}{\alpha \lambda_1^\alpha} +2^{2-\alpha}\pi^{1+2\alpha}\frac{\log \lambda_1}{\lambda_1^{2\alpha}}, &\alpha \in (0,1), \\ \displaystyle 4\pi^3|{\sin u}|\frac{\log \lambda_1}{\lambda_1} +\frac{4\pi^3 |{\cos u}|}{\lambda_1}+2\pi^3\frac{\log \lambda_1}{\lambda_1^2}, &\alpha=1. \end{cases}
\end{equation}
\tag{3.12}
$$
Now, in view of (3.6), (3.7) and (3.12), by (3.5) we have
$$
\begin{equation}
|I_{11}|\leqslant\begin{cases} \displaystyle 8\pi^{1+\alpha}|{\sin u}|^\alpha\frac{\log \lambda_1}{\lambda_1^\alpha} +\frac{2\pi^{2\alpha+1}|{\cos u}|^\alpha}{\alpha \lambda_1^\alpha} +2^{1-\alpha}\pi^{1+2\alpha}\frac{\log \lambda_1}{\lambda_1^{2\alpha}}, &\alpha \in (0,1), \\ \displaystyle 4\pi^3|{\sin u}|\frac{\log \lambda_1}{\lambda_1}+\frac{2\pi^3 |{\cos u}|}{\lambda_1} +\pi^3\frac{\log \lambda_1}{\lambda_1^2}, &\alpha=1. \end{cases}
\end{equation}
\tag{3.13}
$$
Let us now consider the integral $I_{12}$ (see (3.4)). We write it as a sum: where
$$
\begin{equation*}
\begin{aligned} \, I_5 &=\int_{0}^{2\pi}[f(\cos(u+\tau))-f(\cos(u-\tau))]\sin n \tau\,d\tau, \\ I_6 &=\int_{0}^{2\pi}[f(\cos(u+\tau))-f(\cos(u-\tau))]\sin (n+1) \tau\,d\tau. \end{aligned}
\end{equation*}
\notag
$$
Let us estimate $I_5$. The integrand is $2\pi$-periodic, and so the integral can be written as
$$
\begin{equation*}
I_5=\frac{1}{2}\int_{0}^{2\pi}g(u,\tau)\sin n \tau\,d\tau,
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
\begin{aligned} \, g(u,\tau) &=f(\cos(u+\tau))-f(\cos(u-\tau)) \\ &\qquad-f\biggl(\cos\biggl(u+\tau+\frac{\pi}{n}\biggr)\biggr) +f\biggl(\cos\biggl(u-\tau-\frac{\pi}{n}\biggr)\biggr). \end{aligned}
\end{equation*}
\notag
$$
Using (3.9), this establishes
$$
\begin{equation*}
|I_5|\leqslant\frac{2\pi^{1+\alpha}|{\sin u}|^\alpha}{n^\alpha} +\frac{2^{1+\alpha}\pi^{1+2\alpha}|{\cos u}|^\alpha}{(1+\alpha)n^\alpha} +\frac{2^{1-\alpha}\pi^{1+2\alpha}}{n^{2\alpha}}, \qquad n >n_0(\alpha),
\end{equation*}
\notag
$$
where $n_0(\alpha)$ is a natural number depending only on $\alpha$.
Proceeding similarly with the integral $I_6$, we obtain the estimate
$$
\begin{equation}
|I_{12}|\leqslant\frac{4\pi^{1+\alpha}|{\sin u}|^\alpha}{n^\alpha} +\frac{2^{2+\alpha}\pi^{1+2\alpha}|{\cos u}|^\alpha}{(1+\alpha)n^\alpha} +\frac{2^{2-\alpha}\pi^{1+2\alpha}}{n^{2\alpha}}, \qquad \alpha \in (0,1].
\end{equation}
\tag{3.14}
$$
Now a substitution of (3.13) and (3.14) into (3.4) produces (3.3). This completes the proof of Lemma 1. Let us now consider the integral $I_2$ (see (3.2)). Lemma 2. For each $\alpha \in (0,1]$, where $n_0(\alpha)$ is a natural number depending only on $\alpha$. The proof of the above lemma is similar to the proof of estimate (3.14), and we omit it. Now we return to the proof of Theorem 3. To derive estimates (3.1) it suffices to substitute (3.3) and (3.15) into (3.2), collect the resulting estimates, and perform some algebraic transformations. This proves Theorem 3. The following theorem gives an estimate for the approximations (2.14) under the condition that $f \in H^{(\alpha)}[-1,1]$, $\alpha \in (0,1]$. We formulate this result without proof because it is similar to the proof of Theorem 3. Theorem 4. The approximations (2.14) on $[-1,1]$ of the singular integral (1.1) with density $f \in H^{(\alpha)} [-1,1]$, $\alpha \in (0,1]$, satisfy the upper estimate Remark 1. Estimates (3.1) and (3.16) depend substantially on the position of the point $x$ on $[-1,1]$ (the approximations at the endpoints of the interval have a greater rate of decay than the ones on the whole interval). § 4. Approximation of singular integrals with density $|x|^s$ by the Fourier–Chebyshev rational integral operator4.1. An integral representation and estimates of approximationsSetting $f_s(t)=|t|^s$, $s \in (0,+\infty)$, in representation (1.1) we obtain
$$
\begin{equation}
\widehat{f}_{s}(x)=2x\int_{0}^{1}\frac{t^s }{t^2-x^2}\sqrt{1-t^2}\,dt, \qquad x \in [-1,1].
\end{equation}
\tag{4.1}
$$
We look at the properties of the approximations (2.1) in this case. Since $f_s(t)=|t|^s$, we have
$$
\begin{equation*}
\varepsilon_n(\widehat{f}_{s},x,A_n) =-\frac{1}{2}\int_{-\pi}^{\pi}|{\cos \tau}|^s \sin \tau \frac{\cos \lambda_n(\tau,u)}{\sin \frac{\tau-u}{2}}\,d\tau, \qquad x=\cos u, \quad x \in [-1,1].
\end{equation*}
\notag
$$
Splitting the integral into the three integrals over the intervals $[-\pi,-\pi/2]$, $[-\pi/2,\pi/2]$ and $[\pi/2,\pi]$ and making the changes $\tau+\pi \mapsto \tau$ in the first integral and ${\tau-\pi \mapsto \tau}$ in the third we have
$$
\begin{equation*}
\begin{aligned} \, &\varepsilon_n(\widehat{f}_{s},x,A_n) \\ &=-\frac{i}{2}\int_{-\pi/2}^{\pi/2}\cos^s \tau \sin \tau \biggl[\frac{z\frac{\omega_n(z)}{\omega_n(\xi)}+\xi\frac{\omega_n(\xi)}{\omega_n(z)}}{z-\xi} -(-1)^n\frac{z\frac{\omega_n^{-}(z)}{\omega_n(\xi)}-\xi\frac{\omega_n(\xi)}{\omega_n^{-}(z)}}{z+\xi} \biggr]\,d\tau, \qquad n \,{\in}\, \mathbb{N}, \end{aligned}
\end{equation*}
\notag
$$
where $\omega_n(y)$ was defined in (2.3) and
$$
\begin{equation*}
\omega_n^{-}(y)=\prod_{k=1}^{n}\frac{y-z_k}{1-\overline{z}_k y}.
\end{equation*}
\notag
$$
Note that the integral on the right is singular and is understood in the sense of the Cauchy principal value. For further purposes, we need to choose the parameters of the approximating rational function in a special way. We set $n \mapsto 2n-1$. Assume that the $2n-1$ parameters $\{z_k\}_{k=1}^{2n-1}$ have the form
$$
\begin{equation}
\begin{gathered} \, z_{k}=- z_{n+k-1}, \quad z_{k}=i \alpha_{k}, \quad k=1,2,\dots, n-1, \qquad z_{2n-1}=0, \nonumber \\ z_{1}=z_{2}=\dots=z_{p}=0, \qquad p=\biggl[\frac s2\biggr], \quad n>p, \end{gathered}
\end{equation}
\tag{4.2}
$$
where $[\,{\cdot}\,]$ is the integral part of a number. Then we have
$$
\begin{equation}
\begin{gathered} \, \varepsilon_{2n-1}(\widehat{f}_{s},x,A_n)=-i\int_{-\pi/2}^{\pi/2}\cos^s \tau \sin \tau \biggl(\frac{z^3\omega_{2n-2}(z)}{\xi\omega_{2n-2}(\xi)} +\frac{\xi^3\omega_{2n-2}(\xi)}{z\omega_{2n-2}(z)}\biggr) \frac{d\tau}{z^2-\xi^2}, \\ \notag \omega_{2n-2}(y)= y^{2p}\prod_{k=p+1}^{n-1}\frac{y^2+\alpha_k^2}{1+\alpha_k^2 y^2}, \qquad n \in \mathbb{N}. \end{gathered}
\end{equation}
\tag{4.3}
$$
The following result holds. Theorem 5. Under conditions (4.2), for the approximations of the function $\widehat{f}_s(x)$ (see (4.1)) on the interval $[-1,1]$ by the Fourier–Chebyshev rational integral operator (1.2) we have: (1) the integral representation
$$
\begin{equation}
\begin{aligned} \, \notag &\varepsilon_{2n-1}(\widehat{f}_{s},x,A_n) \\ &\qquad =\frac{(-1)^n }{2^{s-1}}\sin \frac{\pi s}{2} \int_{0}^{1}\frac{(1-t^2)^s (1+t^2) t^{1-s}}{\sqrt{1+2t^2 \cos 2u +t^4}} \cos \psi_{2n}(x,t,A_n)\chi_{2n-2}(t)\,dt, \end{aligned}
\end{equation}
\tag{4.4}
$$
where
$$
\begin{equation}
\psi_{2n}(x,t,A_n)=\operatorname{arg}\frac{\xi^3 \omega_{2n-2}(\xi)}{1+t^2 \xi^2}, \quad \xi=\mathrm{e}^{iu}\quad\textit{and} \quad \chi_{2n-2}(t)=t^{2p} \prod_{k=p+1}^{n-1} \frac{t^2-\alpha_k^2}{1- \alpha_k^2 t^2};
\end{equation}
\tag{4.5}
$$
(2) the pointwise estimate
$$
\begin{equation}
\begin{aligned} \, \notag &|\varepsilon_{2n-1}(\widehat{f}_{s},x,A_n)| \\ &\qquad \leqslant\frac{1}{2^{s-1}}\biggl|\sin \frac{\pi s}{2}\biggr| \int_{0}^{1}\frac{(1-t^2)^s (1+t^2) t^{1-s}}{\sqrt{1+2t^2 \cos 2u +t^4}}| \chi_{2n-2}(t)|\,dt, \qquad x=\cos u; \end{aligned}
\end{equation}
\tag{4.6}
$$
(3) the estimate for uniform approximations where Proof. In representation (4.3) we make the change $z=\mathrm{e}^{ i \tau}$ and set $\xi=\mathrm{e}^{ i u}$. As a result, we have
$$
\begin{equation}
\varepsilon_{2n-1}(\widehat{f}_{s},x,A_n) =\frac{i}{2^{s+1}}\int_{C}\frac{(z^2+1)^s(z^2-1)}{z^{s+2}(z^2-\xi^2)} \biggl(\frac{z^3\omega_{2n-2}(z)}{\xi\omega_{2n-2}(\xi)} +\frac{\xi^3\omega_{2n-2}(\xi)}{z\omega_{2n-2}(z)}\biggr)\,dz,
\end{equation}
\tag{4.9}
$$
where $C=\{z\colon z=\mathrm{e}^{i \tau},\,-{\pi}/{2} \leqslant \tau \leqslant {\pi}/{2}\}$ is the right semicircle of the unit circle which is traversed anticlockwise. Note that in the last integral the integrand has branch points at $z=0$, $z=\infty$ and $z=\pm i$. It it also worth noting that $z=\pm \xi$ are simple poles of the integrand, but only one of these poles lies on the contour of integration. We can assume without loss of generality that this is the pole $z=\xi$. We split the integral (4.9) into two so that
$$
\begin{equation}
\varepsilon_{2n-1}(\widehat{f}_{s},x,A_n) =\frac{i}{2^{s+1}}\bigl[\overline{\xi \omega_{2n-2}(\xi)}J_1 +\xi^3 \omega_{2n-2}(\xi) J_2\bigr],
\end{equation}
\tag{4.10}
$$
where $\xi=\mathrm{e}^{ i u}$, $x=\cos u$, and
Let us study each integral separately. Consider the domain described by the contour $\Gamma=C\cup C_{\delta_1}^{-}\cup l_1^{-}\cup C_\delta^{-}\cup l_2^{-}\cup C_{\delta_2}^{-}$ (see Figure 1), where
$$
\begin{equation*}
\begin{gathered} \, C_\delta=\biggl\{z\colon z=\delta\mathrm{e}^{i \tau}, -\frac{\pi}{2} \leqslant \tau \leqslant \frac{\pi}{2} \biggr\}, \\ C_{\delta_1}=\biggl\{z\colon z - i=\delta_1\mathrm{e}^{i \tau}, -\frac{\pi}{2} \leqslant \tau \leqslant v_1 \biggr\}, \quad C_{\delta_2}=\biggl\{z\colon z + i=\delta_2\mathrm{e}^{i \tau}, v_2 \leqslant \tau \leqslant \frac{\pi}{2} \biggr\}, \\ l_1=\biggl\{ z\colon z=i t, \delta \leqslant t \leqslant 1-\delta_1\biggr\}\quad\text{and} \quad l_2=\biggl\{ z\colon z=- i t, 1-\delta_2 \leqslant t \leqslant \delta \biggr\}. \end{gathered}
\end{equation*}
\notag
$$
It is clear that $v_k \to 0$ as $\delta_k \to 0$, $k=1,2$. Consider the integrand of the first integral:
$$
\begin{equation*}
\varphi_{1}(z,\xi)=\frac{2^s g(z,s) z (z^2-1)}{ z^2 - \xi^2}\omega_{2n-2}(z), \quad\text{where } g(z,s)=\frac{1}{2^s}\biggl(z+\frac{1}{z} \biggr)^s.
\end{equation*}
\notag
$$
In the above domain $g(z,s)$ splits into the regular branches specified by the conditions $g(1,s)=\mathrm{e}^{2\pi k s i}$, $k\in \mathbb{Z}$. Let $g_0(z,s)$ be the branch satisfying $g_0(1,s)=1$. Then the function
$$
\begin{equation*}
{\varphi }_{1}(z,\xi)=\frac{2^s g_0(z,s) z (z^2-1)}{ z^2 - \xi^2}\omega_{2n-2}(z)
\end{equation*}
\notag
$$
is regular inside this domain and has a singular point on the boundary, which is a simple pole at $z=\xi$. Let us employ Sokhotskii’s formulae. First assume that $\xi \in D$. By Cauchy’s residue theorem we have
$$
\begin{equation*}
J_1^{+}+\int_{\Gamma_1}\varphi_1(z,\xi)\,dz =2\pi i\operatorname*{Res}_{z=\xi}\varphi_1(z,\xi),
\end{equation*}
\notag
$$
where $\Gamma_1=C_{\delta_1}^{-}\cup l_1^{-}\cup C_\delta^{-} \cup l_2^{-} \cup C_{\delta_2}^{-}$. Now assume that $\xi \mathbin{\overline{\in}} \overline{D}$. Using Cauchy’s residue theorem we find that As a result,
$$
\begin{equation}
J_1=2^{s+1} \pi\xi\omega_{2n-2}(\xi)\cos^s u\sin u-\int_{\Gamma_1}\varphi_1(z,\xi)\,dz.
\end{equation}
\tag{4.11}
$$
Now consider the integral on the right. Setting $z=\delta\mathrm{e}^{i v}$ in the integral over the semicircle $C_\delta$, we have
$$
\begin{equation*}
\int_{C_\delta}\varphi_1(z,\xi)\,dz =\int_{{\pi}/{2}}^{-{\pi}/{2}} \frac{((\delta \mathrm{e}^{i v})^2+1)^s((\delta \mathrm{e}^{i v})^2-1) (\delta \mathrm{e}^{i v})^{1-s}}{(\delta \mathrm{e}^{i v})^2-\xi^2 } \omega_{2n-2}((\delta \mathrm{e}^{i v}))\delta i \mathrm{e}^{i v}\,dv.
\end{equation*}
\notag
$$
Letting $\delta \to 0$ we obtain the asymptotic equality
$$
\begin{equation*}
\int_{C_\delta}\varphi_1(\zeta,\xi)\,d\zeta \sim(-1)^p2 i \sin \frac{\pi s}{2}\frac{\delta^{2+2p-s}}{\xi^{2}(2-s+2p)} \prod_{k=p+1}^{n-1}\alpha_k^{2}, \qquad \delta \to 0.
\end{equation*}
\notag
$$
Since $2+2p-s>0$ (see (4.2)), the integral over the semicircle $C_\delta$ tends to zero as $C_\delta$ shrinks to a point. Proceeding similarly with the integrals over the arcs $C_{\delta_1}$ and $C_{\delta_2}$ we obtain the asymptotic equalities
$$
\begin{equation*}
\int_{C_{\delta_1}}\varphi_1(z,\xi)\,dz \sim(-1)^{n+1}\frac{\delta_1^{s+1}(1-\mathrm{e}^{-i{\pi(s+1)}/2})}{(1+\xi^2)(s+1)i}, \qquad \delta_1 \to 0,
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\int_{C_{\delta_2}}\varphi_1(z,\xi)\,dz \sim(-1)^{n+1}\frac{\delta_2^{s+1}(1-\mathrm{e}^{i\pi(s+1)/2})}{(1+\xi^2)(s+1)i}, \qquad \delta_2 \to 0.
\end{equation*}
\notag
$$
Hence the corresponding integrals tend to zero as the arcs $C_{\delta_1}$ and $C_{\delta_2}$ shrink to points. Thus, as $\delta, \delta_1, \delta_2 \to 0$, (4.11) assumes the form
$$
\begin{equation*}
J_1=2^{s+1} \pi\xi\omega_{2n-2}(\xi)\cos^s u\sin u+\biggl(\int_{-i}^{0} +\int_{0}^{i}\biggr)\varphi_1(z,\xi)\,dz.
\end{equation*}
\notag
$$
Setting $z=it$ in the integrals on the right, this establishes
$$
\begin{equation*}
\begin{aligned} \, J_1 &=2^{s+1} \pi\xi\omega_{2n-2}(\xi)\cos^s u\sin u \\ &\qquad +(-1)^{n-1} i\biggl(\int_{-1}^{0} +\int_{0}^{1}\biggr) \frac{(1-t^2)^s (1+t^2)(i t)^{1-s}}{t^2 + \xi^2}\chi_{2n-2}(t)\,dt, \end{aligned}
\end{equation*}
\notag
$$
where ${{\chi }_{2n}}(t)$ is defined in (5.3). Changing the variable $t$ to $ -t$ in the first integral on the right we obtain eventually
$$
\begin{equation}
\begin{aligned} \, \notag J_1 &=2^{s+1} \pi\xi\omega_{2n-2}(\xi)\cos^s u\sin u \\ &\qquad +2 i (-1)^{n-1}\sin \frac{\pi s}{2}\int_{0}^{1}\frac{(1-t^2)^s (1+t^2) t^{1-s}}{t^2+\xi^2} \chi_{2n-2}(t)\,dt, \qquad \xi=\mathrm{e}^{i u}. \end{aligned}
\end{equation}
\tag{4.12}
$$
Now we deal with the integral $J_2$. Consider the domain bounded by the contour $\Gamma= l_3^{-}\cup C_{\delta_3}^{-}\cup C^{-}\cup C_{\delta_4}^{-}\cup l_4^{-}\cup C_R$ (see Figure 2), where
$$
\begin{equation*}
C_R=\biggl\{z\colon z=R\mathrm{e}^{i \tau}, -\frac{\pi}{2} \leqslant \tau \leqslant \frac{\pi}{2} \biggr\}
\end{equation*}
\notag
$$
is the semicircle of sufficiently large radius $R$, which encircles the point $z= \infty$ clockwise, and
$$
\begin{equation*}
\begin{gathered} \, C_{\delta_3}=\biggl\{z\colon z - i=\delta_3 \mathrm{e}^{i \tau},\, v_3 \leqslant \tau \leqslant \frac{\pi}{2} \biggr\}, \\ C_{\delta_4}=\biggl\{z\colon z + i=\delta_4\mathrm{e}^{i \tau},\, -\frac{\pi}{2} \leqslant \tau \leqslant v_4 \biggr\}, \\ l_3=\biggl\{ z\colon z=i t, 1+\delta_3 \leqslant t \leqslant R \biggr\} \quad\text{and} \quad l_4=\biggl\{ z\colon z=- i t, 1+\delta_4 \leqslant t \leqslant R \biggr\}. \end{gathered}
\end{equation*}
\notag
$$
It is clear that $v_k \to 0$ as $\delta_k \to 0$, $k=3,4$. Consider the integrand of $J_2$:
$$
\begin{equation*}
\varphi_2(z,\xi)=\frac{2^sg(z,s)(z^2-1)}{z^{3}(z^2-\xi^2)\omega_{2n-2}(z)}\,, \quad\text{where } g(z,s)= \frac{1}{2^s}\biggl(z+\frac{1}{z} \biggr)^s.
\end{equation*}
\notag
$$
In the above domain the function $g(z,s)$ splits into the regular branches specified by the conditions $g(1,s)=\mathrm{e}^{2\pi k s i}$, $k\in \mathbb{Z}$. Let $g_0(z,s)$ be the branch for which $g_0(1,s)=1$. Inside the domain the function
$$
\begin{equation*}
\varphi_2(z,\xi)= \frac{2^s g_0(z,s)(z^2-1)}{z^{3}(z^2-\xi^2)\omega_{2n-2}(z)}\,,
\end{equation*}
\notag
$$
is regular, and on its boundary it has a singular point, which is a simple pole at ${z=\xi}$. We use Sokhotskii’s formulae again. We first set $\xi \in D$, where $D=\{z\colon |z|<1, \mathrm{Re}\,z >0\}$. By Cauchy’s integral formula we have where $\Gamma_2=\Gamma \setminus C$. Now let $\xi \mathbin{\overline{\in}} \overline{D}$. In this case,
$$
\begin{equation*}
\int_{\Gamma_2}\varphi_{2}(z,\xi )\,dz- J_2^{+} =2 \pi i\operatorname*{Res}_{z=\xi}\varphi_{2}(z,\xi ).
\end{equation*}
\notag
$$
In view of Sokhotskii’s formulae, from the two last representations we obtain
$$
\begin{equation}
J_2=-2^{s+1} \pi \frac{\cos^s u \sin u}{\xi^3 \omega_{2n-2}(\xi)} +\int_{\Gamma_2}\varphi_{2}(z,\xi )\,dz.
\end{equation}
\tag{4.13}
$$
Proceeding as with the integral $J_1$, we conclude that the integrals over the arcs $C_{\delta_3}$ and $C_{\delta_4}$ tend to zero as $\delta_k \to 0$, $k=3,4$. Now consider the integral over $C_R$. Setting $z=R \mathrm{e}^{i \tau}$ yields
$$
\begin{equation*}
\begin{aligned} \, &\int_{C_R}\varphi_2(z,\xi)\,dz \\ &\quad=\int_{-{\pi}/{2}}^{{\pi}/{2}} \frac{((R \mathrm{e}^{ i \tau})^2+1)^s((R \mathrm{e}^{ i \tau})^2-1)} {(R \mathrm{e}^{ i \tau})^{s+3}((R \mathrm{e}^{ i \tau})^2 - \xi^2)} (R \mathrm{e}^{ i \tau})^{-2p}\prod_{k=p+1}^{n-1} \frac{1 + \alpha_k^2 (R \mathrm{e}^{ i \tau})^2}{(R \mathrm{e}^{ i \tau})^2 + \alpha_k^2} Ri \mathrm{e}^{ i \tau}\,d\tau. \end{aligned}
\end{equation*}
\notag
$$
As a result, we obtain the asymptotic equality
$$
\begin{equation*}
\int_{C_R}\varphi_2(z,\xi)\,dz \sim\frac{(-1)^{p+1} 2^{1-s} i}{\xi^{2}(2-s+2p)R^{2-s+2p}}\sin \frac{\pi s}{2}\prod_{k=p+1}^{n}\alpha_k^{2}, \qquad R \to \infty.
\end{equation*}
\notag
$$
We have $2-s+2p>0$ (see (4.2)), and so the integral over the semicircle $C_R$ tends to zero as $R \to \infty$. From representation (4.13), as $\delta_k \to 0$, $k=3,4$, and $R \to \infty$, we obtain
$$
\begin{equation*}
J_2=-2^{s+1} \pi \frac{\cos^s u \sin u}{\xi^3 \omega_{2n-2}(\xi)} +\biggl(\int_{i \infty}^{i}+\int_{-i}^{-i \infty}\biggr)\varphi_{2}(z,\xi )\,dz,
\end{equation*}
\notag
$$
where the integrals on the right are taken along the corresponding rays on the imaginary axis. The further change of the variable $z \mapsto {z^{-1}}$ gives
$$
\begin{equation*}
J_2=-2^{s+1} \pi \frac{\cos^s u \sin u}{\xi^3 \omega_{2n-2}(\xi)} +\biggl(\int_{-i}^{0}+\int_{0}^{i}\biggr)\frac{(z^2+1)^s (1-z^2)z^{1-s}}{1- \xi^2 z^2} \, \omega_{2n-2}(z)\,dz.
\end{equation*}
\notag
$$
Setting $z \mapsto it$ yields
$$
\begin{equation*}
\begin{aligned} \, J_2 &=-2^{s+1} \pi \frac{\cos^s u \sin u}{\xi^3 \omega_{2n-2}(\xi)} \\ &\qquad+(-1)^{n-1} i\biggl(\int_{-1}^{0}+\int_{0}^{1}\biggr) \frac{(1-t^2)^s (1+t^2)(i t)^{1-s}}{1+ \xi^2 t^2 }\chi_{2n-2}(t) \,dt. \end{aligned}
\end{equation*}
\notag
$$
Changing the variable of integration $t$ in the first integral to $ -t$ we obtain eventually
$$
\begin{equation}
\begin{aligned} \, \notag J_2 &=-2^{s+1} \pi \frac{\cos^s u \sin u}{\xi^3 \omega_{2n-2}(\xi)} \\ &\qquad +2(-1)^{n-1} i\sin \frac{\pi s}{2} \int_{0}^{1}\frac{(1-t^2)^s (1+t^2) t^{1-s}}{1+ \xi^2 t^2 }\chi_{2n-2}(t) \,dt. \end{aligned}
\end{equation}
\tag{4.14}
$$
From representation (4.10) and in view of (4.12) and (4.14) we see that
$$
\begin{equation*}
\begin{aligned} \, & \varepsilon_{2n-1}(\widehat{f}_{s},x,A_n) =\frac{(-1)^n }{2^{s}}\sin \frac{\pi s}{2} \\ &\qquad\qquad \times\int_{0}^{1}(1-t^2)^s (1+t^2) t^{1-s} \biggl[\frac{\xi^3 \omega_{2n-2}(\xi)}{1+ t^2 \xi^2} +\frac{\overline{\xi \omega_{2n-2}(\xi)}}{ t^2 + \xi^2}\biggr]\chi_{2n-2}(t)\,dt. \end{aligned}
\end{equation*}
\notag
$$
Now, to deduce the integral representation (4.5) it suffices to note that the terms in square brackets in the integrand are complex conjugates of each other and then make the corresponding transformations. From the integral representation (4.5) we obtain estimates (4.6) and (4.7). This completes the proof of Theorem 5. Setting $\alpha_k=0$, $k=1,2,\dots,n$, in Theorem 5 we see that the quantities
$$
\begin{equation*}
\varepsilon_{2n-1}(\widehat{f}_{s},x,O)=\varepsilon_{2n-1}^{(0)}(\widehat{f}_{s},x) \ \ \text{and}\ \ \varepsilon_{2n-1}(\widehat{f}_{s},O)=\varepsilon_{2n-1}^{(0)}(\widehat{f}_{s}), \quad O=(\underbrace{0,0,\dots,0}_{n}),
\end{equation*}
\notag
$$
are pointwise and uniform approximations, respectively, of the function $\widehat{f}_s(x)$, ${s \in (0,+\infty)}$, by the polynomial Fourier–Chebyshev sums. Corollary 3. The approximations of the function $\widehat{f}_s(x)$ on $[-1,1]$ by the Fourier–Chebyshev sums of order $n$, $n>s/2$, satisfy: (1) the integral representation
$$
\begin{equation*}
\varepsilon_{2n-1}^{(0)}(\widehat{f}_{s},x) =\frac{(-1)^n }{2^{s-1}}\sin \frac{\pi s}{2} \int_{0}^{1}\frac{ (1-t^2)^s (1+t^2) t^{2n-1-s}}{\sqrt{1+2t^2 \cos 2u +t^4}} \cos \psi_n^{(0)}(x,t)\,dt,
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
\psi_n^{(0)}(x,t)=(2n+1)u-\operatorname{arg}(1+t^2 \xi^2);
\end{equation*}
\notag
$$
(2) the pointwise estimate
$$
\begin{equation*}
|\varepsilon_{2n-1}^{(0)}(\widehat{f}_{s},x)| \leqslant \frac{1}{2^{s-1}} \biggl|\sin \frac{\pi s}{2}\biggr| \int_{0}^{1}\frac{ (1-t^2)^s (1+t^2) t^{2n-1-s}}{\sqrt{1+2t^2 \cos 2u +t^4}}\,dt, \qquad x=\cos u;
\end{equation*}
\notag
$$
(3) the estimate for uniform approximations We consider the asymptotic behaviour of (4.8) as $n \to \infty$. To do this we change the variable by the formula $t=\sqrt{(1-y)/(1+y)}$, $dt=-dy/((1+y)^{3/2} (1-y)^{1/2})$, on the right-hand side of the integral representation (4.8). As a result,
$$
\begin{equation}
\varepsilon_{2n-1}^{*}(\widehat{f}_{s},A_n) =2\biggl|\sin\frac{\pi s}{2}\biggr|\int_{0}^{1}\mu_{s}(y) \biggl|\prod_{k=1}^{n_1}\frac{\beta_k-y}{\beta_k+y}\biggr|\,dy, \qquad n_1=n-p-1, \quad n \in \mathbb{N},
\end{equation}
\tag{4.15}
$$
where
$$
\begin{equation}
\beta_k=\frac{1-\alpha_{p+k}^2}{1+\alpha_{p+k}^2}, \quad k=1,2,\dots,n_1,\quad\text{and} \quad \mu_{s}(y)=\frac{y^{s-1}}{(1+y)^2(1-y^2)^{s/2}}\biggl(\frac{1-y}{1+y}\biggr)^p.
\end{equation}
\tag{4.16}
$$
4.2. An asymptotic expression for the majorant when the approximating function has a fixed number of polesLet $n>p$, $p=[s/2]$, $n_1=n-p-1$, let $q$ be an arbitrary integer such that $0 < q < n_1$, and let $A_q$ be a set of parameters $(\beta_1,\beta_2,\dots,\beta_{n_1})$ precisely $q$ among which are different, with the multiplicity of each parameter equal to $m$: $n_1=m q$. So we are concerned with approximation by rational functions with a pole of order $2p+2$ at infinity and $2q$ geometrically different poles of multiplicity $m$ each in the extended complex plane. Note that Lungu [25], [26] was the first to consider the problem of the approximation of continuous functions with typical singularities by rational functions with a fixed number of geometrically different poles. Under the above assumptions the integral representation (4.15) can be written as
$$
\begin{equation}
\varepsilon_{2n-1,2q}^{*}(\widehat{f}_{s},A_q)= 2\biggl|\sin\frac{\pi s}{2}\biggr|\int_{0}^{1}\mu_{s}(y)|\pi_q(y)|^m\,dy, \qquad s \in (0,+\infty),
\end{equation}
\tag{4.17}
$$
where Note that in this case, for each $n \in \mathbb{N}$ the corresponding set of parameters $(\beta_1,\beta_2,\dots,\beta_q)$ can depend on $n$, that is, $\beta_k=\beta_k(n)$, $k=1,2,\dots,q$. We also assume that the following conditions are satisfied:
$$
\begin{equation}
\varliminf_{m\to \infty}\biggl(m\sum_{k=1}^q \beta_k\biggr)=\infty, \qquad n_1=mq.
\end{equation}
\tag{4.18}
$$
We order the parameters $\beta_k$, $k=1,2,\dots,q$, as follows:
$$
\begin{equation*}
0<\beta_q \leqslant \beta_{q-1} \leqslant \dots \leqslant \beta_1 \leqslant 1.
\end{equation*}
\notag
$$
Theorem 6. For each integer $q $ satisfying $0 < q<n_1$, $n_1=n-p-1$, and each $s \in (0,+\infty)$, the following asymptotic equality holds: Proof. We write the integral representation of the approximations (4.17) in the form
$$
\begin{equation}
\varepsilon_{2n-1,2q}^{*}(\widehat{f}_{s},A_q) =2\biggl|\sin\frac{\pi s}{2}\biggr| \bigl[I_n^{(1)}(A_q)+I_n^{(2)}(A_q)+I_n^{(3)}(A_q)\bigr], \qquad n\in \mathbb{N},
\end{equation}
\tag{4.20}
$$
where and
Let us study the asymptotic behaviour of each of the above three expressions as $m\to \infty $. To do this we employ the Laplace method [27], [28]. We require the following three lemmas (for a proof, see [19]). Lemma 3. The following asymptotic equality holds: where $\Gamma(\,\cdot\,)$ is the Euler gamma function. Lemma 4. The following asymptotic equality holds: Lemma 5. The following asymptotic equality holds: where $\Gamma (\,\cdot\,)$ is the Euler gamma function. Substituting the asymptotic estimates from Lemmas 3, 4 and 5 into (4.20) we obtain (4.19), completing the proof of Theorem 6. Putting $\beta_j=1$ in Theorem 6, $j=1,2,\dots,q$, we see that
$$
\begin{equation*}
\varepsilon_{2n-1,2q}^{*}(\widehat{f}_{s},O) =\varepsilon_{2n-1}^{(0)}(\widehat{f}_{s}), \qquad O=(\underbrace{0,0,\dots,0}_{n}),
\end{equation*}
\notag
$$
is an asymptotic estimate for the majorant of uniform approximations of the function $\widehat{f}_s(x)$ by partial sums of the polynomial Fourier–Chebyshev series. Corollary 4. The uniform approximations of the function $\widehat{f}_s(x)$ on the interval $[-1,1]$ by the Fourier–Chebyshev sums of order $n$, $ n>s/2$, satisfy the upper estimate 4.3. The best majorant of uniform approximations in the case of a fixed number of poles in the approximating functionIt would be interesting to minimize the right-hand side of the asymptotic equality (4.19) by choosing a set of parameters $(\beta_1^{*},\beta_2^{*},\dots,\beta_{q}^{*})$ optimal for this problem; that is, it would be interesting to find the best majorant of uniform approximations of the function $\widehat{f}_s(x)$ by Fourier–Chebyshev rational integral operators (1.2) in the case of an arbitrary fixed number of geometrically different poles. We set
$$
\begin{equation*}
\varepsilon_{2n-1,2q}(\widehat{f}_{s}) =\operatorname*{inf}_{A_q}\varepsilon_{2n-1,2q}(\widehat{f}_{s},A_q)\quad\text{and} \quad \varepsilon_{2n-1,2q}^{*}(\widehat{f}_{s}) =\operatorname*{inf}_{A_q}\varepsilon_{2n-1,2q}^{*}(\widehat{f}_{s},A_q).
\end{equation*}
\notag
$$
From (4.7) we have
$$
\begin{equation*}
\varepsilon_{2n-1,2q}(\widehat{f}_{s}) \leqslant\varepsilon_{2n-1,2q}^{*}(\widehat{f}_{s}), \qquad n \in \mathbb{N}.
\end{equation*}
\notag
$$
Theorem 7. The uniform approximations of the function $\widehat{f}_s(x)$ by the rational integral operator (1.2) with $2q$ geometrically different poles on $[-1,1]$ satisfy the upper estimate where and $n_0(s)$ is some natural number independent of $n$ but depending on $s$. Proof. Let the family $A_q$ be defined by
$$
\begin{equation*}
\beta_k=c_k \biggl( \frac{\log m}{m}\biggr)^{2k-1}, \qquad k=1,2,\dots,q,
\end{equation*}
\notag
$$
where the coefficients $c_k$, $k=1,2,\dots,q$, are independent of $m$ (the precise values of the $c_k$ are specified below). Let us find the asymptotic behaviour of the terms $S_n^{(1)}(A_q)$, $S_n^{(2)}(A_q)$ and $S_n^{(3)}(A_q)$ from (4.19) in this case. It is easily seen that $S_n^{(1)}(A_q)$ satisfies the asymptotic equality
$$
\begin{equation*}
S_n^{(1)}(A_q)= \frac{\Gamma(s)} {\bigl(2m\sum_{k=1}^{q}\frac{1}{c_k}\bigl(\frac{m}{\log m}\bigr)^{2k-1}\bigr)^{s}} \sim\frac{\Gamma(s) c_q^{s}(\log m)^{(2q-1)s}}{2^{s} m^{2qs}}, \qquad m \to \infty.
\end{equation*}
\notag
$$
Now we estimate $S_n^{(2)}(A_q)$. We need the well-known asymptotic equality
$$
\begin{equation*}
b_j \sim \sqrt{c_j c_{j+1}} \biggl(\frac{\log m}{m}\biggr)^{2j}, \qquad b_j \in (\beta_{j+1},\beta_j), \quad j=1,2,\dots,q-1, \quad m \to \infty
\end{equation*}
\notag
$$
(see [19], Lemma 6), which holds for the above parameters $\beta_k$, $k=1,2,\dots,q$. From the last asymptotic equality and Theorem 6 in [19] we find that
$$
\begin{equation*}
S_n^{(2)}(A_q) \sim \sqrt{\frac{\pi}{2}}\, \sum_{j=1}^{q-1}(c_j c_{j+1})^{s/2}\sqrt[4]{\frac{c_j}{c_{j+1}}} \, \frac{(\log m)^{2js-1/2}}{m^{2js+4\sqrt{c_{j+1}/c_j}}}, \qquad m \to \infty.
\end{equation*}
\notag
$$
For the above parameters $\beta_k$, $k=1,2,\dots,q$, we have
$$
\begin{equation*}
\biggl(\prod_{k=1}^{q}\frac{1-\beta_k}{1+\beta_k}\biggr)^m\sim \frac{1}{m^{2c_1}}, \qquad m \to \infty,
\end{equation*}
\notag
$$
and so, for the third term in (4.19),
$$
\begin{equation*}
S_n^{(3)}(A_q) \sim\frac{\Gamma(1+p-\frac{s}2)}{2^{2p+3}c_1^{1+p-s/2} m^{2c_1}(\log m)^{1+p-s/2}}, \qquad m \to \infty.
\end{equation*}
\notag
$$
Let us choose $c_k$, $k=1,2,\dots,q$, so as to equate the exponents of $m$ in each of the asymptotic equalities for $S_n^{(1)}(A_q)$, $S_n^{(2)}(A_q)$ and $S_n^{(3)}(A_q)$. That is, the $c_k$, $k=1,2,\dots,q$, must satisfy
$$
\begin{equation*}
\begin{cases} qs=c_1, \\ \displaystyle qs=j s + 2 \sqrt{\frac{c_{j+1}}{c_j}}, &j=1,2,\dots,q-1. \end{cases}
\end{equation*}
\notag
$$
As a result, Now the asymptotic equality (4.19) can be written as
$$
\begin{equation*}
\varepsilon_{2n-1,2q}^{*}(\widehat{f}_{s},A_q^{*}) \sim2^{1-s}\biggl|\sin\frac{\pi s}{2}\biggr|\Gamma(s) \biggl(\frac{s^{2q-1}(q!)^2}{q 2^{2q-2}}\biggr)^{s} \biggl(\frac{\log^{2q-1}m}{m^{2q}}\biggr)^{s}, \qquad m \to \infty.
\end{equation*}
\notag
$$
To show that for the $c_k$ as above, $k=1,2,\dots,q$, the family of parameters $\beta_k$, $k=1,2,\dots,q$, is optimal in the sense that $\varepsilon_{2n-1,2q}^{*}(\widehat{f}_{s},A_q)$ attains its asymptotically minimum value for these $\beta_k$, it suffices to follow the arguments of [29]. Using now the formula $m=(n-1-p)/q$ to get back to the parameter $n$, we have
$$
\begin{equation*}
\varepsilon_{2n-1,2q}^{*}(\widehat{f}_{s}) \sim c(q,s)\biggl(\frac{\log^{2q-1}n}{n^{2q}}\biggr)^{s}, \qquad s \in (0,+\infty), \quad n \to \infty,
\end{equation*}
\notag
$$
where $c(q,s)$ was defined in Theorem 7. Now estimate (4.21) is secured by (4.7). This proves Theorem 7. It would be interesting to compare the uniform estimate from Theorem 7 with the one for approximations of the function $\widehat{f}_s(x)$ by partial sums of the Fourier–Chebyshev series from Corollary 4. The decay rate of uniform approximations in the polynomial case is $O(1/n^{s})$, whereas the corresponding rational approximations with $2q$ geometrically different poles in the approximating function in the open complex plane decay as $O(({\log^{2q-1}n}/n^{2q})^s)$. 4.4. Asymptotic expression for the majorant in the general caseLet us study the asymptotic behaviour of (4.8) as $n \to \infty$ without any constraints on the number of poles. To do this we use representation (4.15) again. Let the parameters $\beta_k$, $k=1,2,\dots,n_1$, $n_1=n - 1 - p$, be of the form $\beta_{k}=\xi^k$, $\xi \in (0,1)$, $k=1,2,\dots,n_1$. In addition, for each $n$, we can choose its own family of parameters $\beta_k$, that is, $\beta_k=\beta_k(n)$. So we assume that
$$
\begin{equation}
\beta_{1} \to 1, \qquad \beta_{n_1} \to 0, \qquad n \to \infty,
\end{equation}
\tag{4.22}
$$
$$
\begin{equation}
\sum_{k=1}^{n_1}\frac{1}{\beta_k}\xrightarrow{n \to \infty }\infty
\end{equation}
\tag{4.23}
$$
and
$$
\begin{equation}
\sum_{k=1}^{n_1}\frac{\beta_k}{1-\beta _{k}^{2}} \xrightarrow{n \to \infty}\infty .
\end{equation}
\tag{4.24}
$$
Conditions (4.22)–(4.24) are consistent. Below we consider a sequence of parameters $\{\beta_{k}\}_{k=1}^{n_1}$ of the approximating function that obey these conditions. Theorem 8. Under conditions (4.22)–(4.24) the uniform approximations of the function $\widehat{f}_s(x)$ on $[-1,1]$ by the Fourier–Chebyshev rational integral operator satisfy the upper estimate Proof. Again, consider the integral (4.15). We split it into three parts, so that
$$
\begin{equation}
\varepsilon_{2n-1}^{*}(\widehat{f}_{s},A_n) =2\biggl|\sin\frac{\pi s}{2}\biggr|\bigl[I_n^{(4)}+I_n^{(5)} + I_n^{(6)}\bigr], \qquad n \in \mathbb{N},
\end{equation}
\tag{4.26}
$$
where
$$
\begin{equation*}
\begin{gathered} \, I_n^{(4)}=\int_{0}^{\beta_{n_1}}\mu_{s}(y)\pi_{n_1}(y)\,dy, \qquad I_n^{(5)}=\int_{\beta_{n_1}}^{\beta_{1}}\mu_{s}(y)|\pi_{n_1}(y)|\,dy \\ \text{and}\quad I_n^{(6)}=\int_{\beta_{1}}^{1}\mu_{s}(y)|\pi_{n_1}(y)|\,dy. \end{gathered}
\end{equation*}
\notag
$$
We consider each of these three integrals separately. The integral $I_n^{(4)}$ can be written as
$$
\begin{equation*}
I_n^{(4)}=\int_{0}^{\beta_{n_1}}\mu_{s}(y)\mathrm{e}^{S(y)}\,dy, \quad\text{where } S(y)=\sum_{k=1}^{n_1}\log \frac{\beta_k-y}{\beta_k+y}.
\end{equation*}
\notag
$$
Let us use the Laplace method [27], [28]. The function $S(y)$ decreases on the interval $[0, \beta_{n_1}]$ since ${S}'(y)<0$, and therefore it attains its maximum value at $y=0$. Using the expansion
$$
\begin{equation*}
S(y)=-2y\sum_{k=1}^{n_1}\frac{1}{\beta_k}-\frac{2}{3}y^3\sum_{k=1}^{n_1}\frac{1}{\beta_k^3}+O(y^5), \qquad y\to 0,
\end{equation*}
\notag
$$
and the obvious asymptotic equality for some small $\varepsilon >0$, as $n \to \infty $, we obtain
$$
\begin{equation*}
I_n^{(4)}\sim\int_{0}^{\varepsilon}y^{s-1}\exp \biggl[ -2 y \sum_{k=1}^{n_1}{\frac{1}{\beta_k}}\biggr]\,dy.
\end{equation*}
\notag
$$
Changing the variable of integration on the right by the formula $2u\sum_{k=1}^{n_1}1/\beta_k\mapsto t$ we have
$$
\begin{equation*}
I_n^{(4)} \sim\frac{1}{\bigl(2\sum_{k=1}^{n_1}\frac1{\beta_k}\bigr)^{s}} \int_{0}^{\varphi (n,\varepsilon )}{t^{s-1}}\mathrm{e}^{-t}\,dt, \qquad n \to \infty,
\end{equation*}
\notag
$$
where $\varphi (n,\varepsilon)=2\varepsilon \sum_{k=1}^{n_1}{1/\beta_k} \to \infty $ as $n \to \infty $ by condition (4.23). Since from the last asymptotic equality we obtain where $\Gamma(\,\cdot\,)$ is the Euler gamma function.
Let us estimate the integral $I_n^{(5)}$. We have
$$
\begin{equation}
\int_{\beta_{n_1}}^{\beta_{1}}\mu_{s}(y)|\pi_{n_1}(y)|\,dy \leqslant \operatorname*{max}_{y \in [\beta_{n_1},\beta_{1}]} |\pi_{n_1}(y)|\int_{\beta_{n_1}}^{\beta_{1}}\frac{y^{s-1}}{(1-y^2)^{s/2-p}}\,dy.
\end{equation}
\tag{4.28}
$$
In view of conditions (4.22) we find that
$$
\begin{equation*}
\int_{\beta_{n_1}}^{\beta_{1}}\frac{y^{s-1}}{(1-y^2)^{s/2-p}}\,dy \xrightarrow{n \to \infty}\int_{0}^{1}\frac{y^{s-1}}{(1-y^2)^{s/2-p}}\,dy =c_1(s),
\end{equation*}
\notag
$$
where $c_1(s)$ was defined in Theorem 8. It is known (see, for example, [30]) that if $\beta_{k}=\xi^k$, $\xi \in (0,1)$, $k=1,2,\dots,n_1$, then the products on the right-hand side of (4.28) satisfy the estimate
$$
\begin{equation*}
\operatorname*{max}_{y \in [\beta_{n_1},\beta_{1}]}|\pi_{n_1}(y)| \leqslant\prod_{k=1}^{n_1}\frac{1-\beta_k}{1+\beta_k}.
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation}
I_n^{(5)} \leqslant c_1(s)\prod_{k=1}^{n_1}\frac{1-\beta_k}{1+\beta_k}.
\end{equation}
\tag{4.29}
$$
Now consider the integral $I_n^{(6)}$. We write it as
$$
\begin{equation*}
I_n^{(6)}=\int_{\beta_1}^{1}\mu_s(y)\mathrm{e}^{S(y)}\,dy, \qquad S(y)=\sum_{k=1}^{n_1}\log \frac{y-\beta_k}{y+\beta_k}.
\end{equation*}
\notag
$$
Again, we use the Laplace method [27], [28]. The function $S(y)$ is increasing for $y \in (\beta_{1},1)$ since ${S}'(y)=2\sum_{k=1}^{n_1}\beta_k/(y^2-\beta_{k}^{2})>0$. Hence it assumes its maximum value at $y=1$. From the asymptotic equalities
$$
\begin{equation*}
\begin{gathered} \, S(y)=\sum_{k=1}^{n_1}\log \frac{1-\beta_k}{1+\beta_k} +\sum_{k=1}^{n_1}\frac{2\beta_k}{1-\beta_k^2}(y-1)+o(y-1), \\ \mu_s(y)\sim \frac{(1-y)^{p-s/2}}{2^{p+s/2+2}}, \end{gathered}
\end{equation*}
\notag
$$
which hold as $y \to 1$ for some small $\varepsilon >0$ and $n \to \infty $ we find that
$$
\begin{equation*}
I_n^{(6)} \sim\frac{1}{2^{p+s/2+2}} \prod_{k=1}^{n_1}\frac{1-\beta_k}{1+\beta_k} \int_{1-\varepsilon}^{1}(1-y)^{p-s/2}\exp \biggl[ 2\sum_{k=1}^{n_1}\frac{\beta_k}{1-\beta_k^2}(y-1) \biggr]\,dy.
\end{equation*}
\notag
$$
Making the change of the variable
$$
\begin{equation*}
2(1-y) \sum_{k=1}^{n_1}\frac{\beta_k}{1-\beta_k^2} \mapsto u
\end{equation*}
\notag
$$
in this integral, we obtain the asymptotic equality
$$
\begin{equation*}
I_n^{(6)} \sim\frac{1}{2^{2p+3}\bigl(\sum_{k=1}^{n_1}\frac{\beta_k}{1-\beta_k^2}\bigr)^{1+p-s/2}} \prod_{k=1}^{n_1}\frac{1-\beta_k}{1+\beta_k} \int_{0}^{\varphi (n,\varepsilon )} u^{p-s/2}\mathrm{e}^{-u}\,du, \qquad n \to \infty,
\end{equation*}
\notag
$$
where $\varphi(n,\varepsilon)=\varepsilon \sqrt{\sum_{k=1}^{n_1}\beta_k/(1-\beta_k^2)} \to \infty $ as $n \to \infty $ in view of condition (4.24). Next,
$$
\begin{equation*}
\int_{0}^{+\infty }u^{p-s/2}\mathrm{e}^{-u}\,du =\Gamma \biggl(1+p-\frac{s}{2} \biggr),
\end{equation*}
\notag
$$
and so
$$
\begin{equation}
I_n^{(6)}=\frac{\Gamma(1+p-\frac{s}2)} {2^{2p+3}\bigl(\sum_{k=1}^{n_1}\frac{\beta_k}{1-\beta_k^2}\bigr)^{1+p-s/2}} \prod_{k=1}^{n_1}\frac{1-\beta_k}{1+\beta_k}(1+o(1)), \qquad n \to \infty.
\end{equation}
\tag{4.30}
$$
Substituting the asymptotic estimates (4.27), (4.29) and (4.30) for the integrals $I_n^{(4)}$, $I_n^{(5)}$ and $I_n^{(6)}$ into representation (4.26) we arrive at (4.25). This proves Theorem 8. 4.5. The case of ‘Newman’ parametersLet us consider the right-hand side of estimate (4.25) in the case where the parameters $\beta_{k}$, $k=1,2,\dots,n_1$, are modifications of those introduced by Newman in [31] in his famous paper on rational approximation of the function $|x|$. This problem is investigated below. Let $A_N$ be a family of parameters $\beta_k$, $k=1,2,\dots,n_1$, such that
$$
\begin{equation}
\beta_{k}=\mathrm{e}^{-ck/\sqrt{n_1}}, \qquad k=1,2,\dots,n_1, \quad n_1=n-1-p,
\end{equation}
\tag{4.31}
$$
where $n_1\in \mathbb{N}$ is fixed and $c$ is some positive constant independent of $n$. Note that this family of parameters satisfies conditions (4.22). Theorem 9. The uniform approximations of the function $\widehat{f}_s(x)$ on $[-1,1]$ by the Fourier–Chebyshev rational integral operator (1.2) with parameters (4.31) satisfy the upper estimate Proof. We examine the asymptotic behaviour of the right-hand side of (4.25) when the parameters $\beta_{k}$, $k=1,2,\dots,n_1$, are given by (4.31). The proof of the following two lemmas can be found in [32].
Lemma 6. Under conditions (4.31) the following asymptotic equality holds: Lemma 7. Under conditions (4.31) the following asymptotic equalities hold: and Note that conditions (4.23) and (4.24) are secured by Lemma 7. Let us return to the proof of Theorem 9. Substituting the asymptotic equalities (4.33)–(4.35) into (4.25), this gives
$$
\begin{equation*}
\begin{aligned} \, \varepsilon_{2n-1}^{*}(\widehat{f}_{s},A_N) &\leqslant2\biggl|\sin \frac{\pi s}{2}\biggr| \biggl(\frac{2^{1-s}\Gamma (s)c^s \mathrm{e}^{-cs\sqrt{n_1}}}{(\sqrt{n_1}\,)^s} +c_1(s)\sqrt{n_1}\exp \biggl(-\frac{\pi^2}{4c}\sqrt{n_1}\biggr) \\ &\qquad +\frac{\Gamma(1+p-\frac{s}2)(2c)^{1+p-s/2}}{2^{2p+2}(\sqrt{n_1}\, \log \sqrt{n_1}\, )^{1+p-s/2}} \sqrt{n_1}\exp \biggl(-\frac{\pi^2}{4c}\sqrt{n_1}\biggr)\biggr), \\ &\qquad\qquad s \in (0,+\infty), \qquad n > n_0(s). \end{aligned}
\end{equation*}
\notag
$$
Now estimate (4.32) is a consequence of the last relation and (4.7). This proves Theorem 9. It would be interesting to minimize the majorant in the upper estimate (4.32) by choosing an optimal parameter $c$ for this problem; in other words we are interested in a sharp upper estimate for the uniform approximations with parameters (4.31). We set
$$
\begin{equation*}
\varepsilon_{2n-1}(\widehat{f}_{s}) =\operatorname*{inf}_{A_N}\varepsilon_{2n-1}(\widehat{f}_{s},A_N)\quad\text{and} \quad \varepsilon_{2n-1}^{*}(\widehat{f}_{s}) =\operatorname*{inf}_{A_N}\varepsilon_{2n-1}^{*}(\widehat{f}_{s},A_N).
\end{equation*}
\notag
$$
Note that from (4.7) we obtain
$$
\begin{equation*}
\varepsilon_{2n-1}(\widehat{f}_{s}) \leqslant\varepsilon_{2n-1}^{*}(\widehat{f}_{s}), \qquad n \in \mathbb{N}.
\end{equation*}
\notag
$$
Theorem 10. For the uniform approximations of the function $\widehat{f}_s(x)$ on $[-1,1]$ by the Fourier–Chebyshev rational integral operator (1.2) there exists a family of parameters $A_N^{*}$ of the form (4.31) such that Proof. The constant $c$ on the right-hand side (4.32) is chosen from the condition of the equality of the arguments of the exponential in each term. Therefore, which gives As a result, (4.32) can be written as
$$
\begin{equation*}
\begin{gathered} \, \begin{aligned} \, \varepsilon_{2n-1}(\widehat{f}_{s},A_N^{*}) &\leqslant2\biggl|\sin \frac{\pi s}{2}\biggr| \biggl(\frac{2^{1-2s}\Gamma (s)}{(\sqrt{n_1}\,)^{s+1}} \biggl(\frac{\pi}{\sqrt{s}}\biggr)^s+c_1(s) \\ &\qquad+\frac{\Gamma(1+p-\frac{s}2)\pi^{1+p-s/2}}{2^{2+2p}(\sqrt{n_1 s}\, \log \sqrt{n_1}\,)^{1+p-s/2}}\biggr) \sqrt{n_1}\exp \biggl(-\frac{\pi}{2}\sqrt{n_1 s}\biggr), \end{aligned} \\ s \in (0,+\infty), \qquad p=\biggl[\frac{s}{2}\biggr], \qquad n_1>n_0(s). \end{gathered}
\end{equation*}
\notag
$$
The first and third terms in brackets tend to zero as $n_1 \to \infty$. Therefore, there exists $n_0(s)$ such that, for all $n_1 >n_0(s)$,
$$
\begin{equation*}
\varepsilon_{2n-1}(\widehat{f}_{s},A_N^{*}) \leqslant6\biggl|\sin \frac{\pi s}{2}\biggr|c_1(s) \sqrt{n_1}\exp \biggl(-\frac{\pi}{2}\sqrt{n_1 s}\biggr), \qquad n_1>n_0(s).
\end{equation*}
\notag
$$
To verify that for this $c^{*}$ the parameters of the approximating rational function are indeed optimal, namely, they deliver the best uniform approximations by the rational integral operator (1.2) with parameters satisfying condition (4.31), it suffices to employ the method proposed in [29]. Since $n_1=n - 1 - p$ and $n - 1 - p \sim n$ as ${n \to \infty}$, we have (4.36). This completes the proof of Theorem 10. Corollary 5. For the uniform approximations of functions (1.1) with density $|x|^s$, $s \in (0,2)$, on $[-1,1]$ by the Fourier–Chebyshev rational integral operator (1.2) there exists a family of parameters of the approximating function such that For the proof of the last estimate it suffices to put $p=0$ in (4.36). In relation to the above result, it is worth recalling the asymptotic estimate
$$
\begin{equation*}
R_{n,n}(|x|^s,[-1,1])\sim4^{1+s/2}\biggl|\sin \frac{\pi s}{2}\biggr| \mathrm{e}^{-\pi \sqrt{ns}}, \qquad s>0, \quad n \to \infty
\end{equation*}
\notag
$$
(see Stahl [33]), where the $R_{n,n}(|x|^s,[-1,1])$ are best uniform approximations of the function $|x|^s$ on $[-1,1]$ by rational functions of degree $\leqslant n$.
§ 5. Approximation of singular integrals with density $|x|^s$ by the rational integral operator $\widehat{s}_{n+1}$5.1. Integral representations and estimates for approximationsWe consider approximations (2.8) of the function $\widehat{f}_s(x)$ (see (4.1)) on $[-1,1]$ by the rational integral operator (1.4). For further purposes, we need to choose in a special way the parameters of the approximating rational function. Let the family of parameters $\{ z_k\}_{k=1}^{2n}$ (see (1.3)) be defined by
$$
\begin{equation}
\begin{gathered} \, \notag z_{k}=- z_{n+k}, \quad z_{k} \mapsto i \alpha_{k}, \qquad k=1,2,\dots, n, \\ z_{1}=z_{2}=\dots=z_{p}=0, \qquad p=\biggl[\frac{s}{2}\biggr], \quad n>p. \end{gathered}
\end{equation}
\tag{5.1}
$$
Then the following result holds. Theorem 11. The approximations of the function $\widehat{f}_s(x)$ on $[-1,1]$ by the rational integral operator (1.4) satisfy: (1) the integral representation
$$
\begin{equation}
\begin{aligned} \, \notag &\widehat{\varepsilon}_{2n+1}(f_s,x,A_n)={(-1)^{n}2^{2-s}\sqrt{1-x^2}}\sin\frac{\pi s}{2} \\ &\qquad\times \int_{0}^{1}\frac{(1-t^2)^s t^{1-s}} {\sqrt{1+2t^2 \cos 2 u +t^4}} \chi_{2n}(t)\sin\biggl(\operatorname{arg}\frac{\xi^2 \omega_{2n}(\xi)}{1+t^2\xi^2}\biggr)\,dt, \qquad x=\cos u, \end{aligned}
\end{equation}
\tag{5.2}
$$
where
$$
\begin{equation}
\omega_{2n}(\xi) =\prod_{k=1}^{n}\frac{\xi^2+\alpha_k^2}{1+ \alpha_k^2 \xi^2}, \qquad \chi_{2n}(t)=\prod_{k=1}^{n}\frac{t^2-\alpha_k^2}{1- \alpha_k^2 t^2}, \qquad \xi=\mathrm{e}^{iu};
\end{equation}
\tag{5.3}
$$
(2) the pointwise estimate
$$
\begin{equation}
\begin{aligned} \, \notag &|\widehat{\varepsilon}_{2n+1}(f_s,x,A_n)| \leqslant 2^{2-s}\sqrt{1-x^2}\, \biggl|\sin\frac{\pi s}{2}\biggr| \\ &\qquad\qquad \times \int_{0}^{1}\frac{(1-t^2)^s t^{1-s}}{\sqrt{1+2t^2 \cos 2 u +t^4}}|\chi_{2n}(t)|\,dt, \qquad x=\cos u; \end{aligned}
\end{equation}
\tag{5.4}
$$
(3) the estimate for uniform approximations where Proof. It is clear that
$$
\begin{equation}
\widehat{\varepsilon}_{2n+1}(f_s,x,A_n) =\int_{-1}^{1}\frac{{\varepsilon}_{2n}(f_s,t,A_n)}{t-x}\sqrt{1-t^2}\,dt, \qquad x \in [-1,1],
\end{equation}
\tag{5.7}
$$
where the ${\varepsilon}_{2n}(f_s,t,A_n)$ are approximations of the function $|x|^s$, $s \in (0,+\infty)$, on the interval $[-1,1]$ by the rational integral operator (1.2).
The integral representation
$$
\begin{equation*}
\begin{aligned} \, &\varepsilon_{2n}(f_s,t,A_n)= \frac{(-1)^n 2^{1-s}}{\pi}\sin \frac{\pi s}{2} \\ &\qquad\times\int_{0}^{1}(1-y^2)^s y^{1-s}\biggl[\frac{z^2 \omega_{2n}(z)}{1+y^2 z^2} +\frac{\overline{\omega_{2n}(z)}}{z^2+y^2} \biggr]\chi_{2n}(y)\,dy, \qquad z=\mathrm{e}^{i \tau}, \quad t=\cos \tau, \end{aligned}
\end{equation*}
\notag
$$
holds for the parameters (5.1) (see [20]), where $\omega_{2n}(\,\cdot\,)$ and $\chi_{2n}(\,\cdot\,)$ are defined in (5.3).
Substituting the integral representation into (5.7) and using Fubini’s theorem to change the order of integration we obtain
$$
\begin{equation}
\widehat{\varepsilon}_{2n+1}(f_s,x,A_n) =\frac{(-1)^n 2^{1-s}}{\pi}\sin \frac{\pi s}{2}\int_{0}^{1}(1-y^2)^s y^{1-s}I_{2n}(y,x)\chi_{2n}(y)\,dy,
\end{equation}
\tag{5.8}
$$
where
$$
\begin{equation*}
I_{2n}(y,x)=\int_{-1}^{1}\biggl[\frac{z^2 \omega_{2n}(z)}{1+y^2 z^2} + \frac{\overline{\omega_{2n}(z)}}{z^2+y^2}\biggr]\frac{\sqrt{1-t^2}}{t-x}\,dt, \qquad z=\mathrm{e}^{i \tau}, \quad t=\cos \tau.
\end{equation*}
\notag
$$
Now changing the variable of integration to $t=\cos \tau$ and setting $x=\cos u$ in the integral $I_{2n}(y,x)$ we easily obtain
$$
\begin{equation*}
I_{2n}(y,x) =\frac{1}{2}\int_{-\pi}^{\pi}\biggl[\frac{z^2 \omega_{2n}(z)}{1+y^2 z^2} +\frac{\overline{\omega_{2n}(z)}}{z^2+y^2}\biggr]\frac{\sin^2 \tau}{\cos \tau - \cos u}\,d\tau, \qquad z=\mathrm{e}^{i \tau}, \quad t=\cos \tau.
\end{equation*}
\notag
$$
Going over to integration with respect to $z=\mathrm{e}^{i \tau}$ and setting $\xi=\mathrm{e}^ {i u}$ on the right-hand side we obtain
$$
\begin{equation}
I_{2n}(y,x)=-\frac{1}{4i}\bigl[J^{(1)}(y,x) + J^{(2)}(y,x)\bigr],
\end{equation}
\tag{5.9}
$$
where
$$
\begin{equation*}
J^{(1)}(y,x) =\oint_{|z|=1}\frac{\omega_{2n}(z)(z^2-1)^2}{(1+y^2 z^2)(z-\xi)(z-\frac1{\xi})}\,dz
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
J^{(2)}(y,x) =\oint_{|z|=1}\frac{\overline{\omega_{2n}(z)}(z^2-1)^2}{(z^2+y^2) z^2(z-\xi)(z-\frac1{\xi})}\,dz.
\end{equation*}
\notag
$$
Note that the integrals $J^{(1)}(y,x)$ and $J^{(2)}(y,x)$, which contain points of singularity $z=\xi$ and $z=1/\xi$ on the boundary of the unit disc, are understood in the sense of the Cauchy principal value. To evaluate these integrals, we use the Sokhotskii’s formulae. After some calculations we obtain
$$
\begin{equation*}
J^{(1)}(y,x) =\pi i\biggl[\frac{\xi \omega_{2n}(\xi)(\xi^2-1)}{1+y^2 \xi^2} +\frac{\overline{\omega_{2n}(\xi)}(1-\xi^2)}{\xi(\xi^2+y^2)}\biggr].
\end{equation*}
\notag
$$
A similar analysis shows that
$$
\begin{equation*}
J^{(2)}(y,x) =-\pi i\biggl[\frac{\xi \omega_{2n}(\xi)(1-\xi^2)}{1+y^2 \xi^2} +\frac{\overline{\omega_{2n}(\xi)}(\xi^2-1)}{\xi(\xi^2+y^2)}\biggr].
\end{equation*}
\notag
$$
Substituting the last two expressions into (5.9) we arrive at
$$
\begin{equation*}
I_{2n}(y,x) =-\pi i \sin u\biggl[\frac{\xi^2 \omega_{2n}(\xi)}{1+\xi^2 y^2} -\frac{\overline{\omega_{2n}(\xi)}}{\xi^2+y^2}\biggr], \qquad \xi=\mathrm{e}^{ i u}, \quad x=\cos u.
\end{equation*}
\notag
$$
Plugging this into (5.8) we find that
$$
\begin{equation*}
\begin{aligned} \, \widehat{\varepsilon}_{2n+1}(f_s,x,A_n) &=(-1)^{n+1} 2^{1-s}\sqrt{1-x^2}\, i\sin \frac{\pi s}{2} \\ &\qquad\times\int_{0}^{1}(1-y^2)^s y^{1-s} \biggl[\frac{\xi^2 \omega_{2n}(\xi)}{1+\xi^2 y^2} -\frac{\overline{\omega_{2n}(\xi)}}{\xi^2+y^2}\biggr] \chi_{2n}(y)\,dy. \end{aligned}
\end{equation*}
\notag
$$
Note that the expressions in square brackets are the complex conjugates of each other. Now (5.2) follows after some simple algebra.
Estimates (5.4) and (5.5) are clear from the integral representation (5.2). This completes the proof of Theorem 11. Setting $\alpha_k=0$ in Theorem 11, $ k=1,2,\dots,n$, we see that
$$
\begin{equation*}
\widehat{\varepsilon}_{2n+1}(f_s,x,O) =\widehat{\varepsilon}_{2n+1}^{\,(0)}(f_s,x)\ \ \text{and} \ \ \widehat{\varepsilon}_{2n+1}(f_s,O) =\widehat{\varepsilon}_{2n+1}^{\,(0)}(f_s), \quad O=(\underbrace{0,0,\dots,0}_{n}),
\end{equation*}
\notag
$$
are, respectively, the pointwise and uniform approximations of the function $\widehat{f}_s(x)$ on the interval $[-1,1]$ by the polynomial analogue of operator (1.4). Corollary 6. For the approximations of the function $\widehat{f}_s(x)$ on $[-1,1]$ by the operator which is the image under the transformation (1.1) of partial sums of the polynomial Fourier–Chebyshev series of order $n$, $n>s/2$, we have: (1) the integral representation
$$
\begin{equation*}
\begin{gathered} \, \begin{aligned} \, \widehat{\varepsilon}_{2n+1}^{\,(0)}(f_s,x) &={(-1)^{n}2^{2-s}\sqrt{1-x^2}}\sin\frac{\pi s}{2} \\ &\qquad \times \int_{0}^{1}\frac{(1-t^2)^st^{2n+1-s}}{\sqrt{1+2t^2 \cos 2 u +t^4}}\sin \psi_n^{(0)}(x,t)\,dt, \qquad x=\cos u, \end{aligned} \\ \psi_n^{(0)}(x,t)=(2n+2)u-\operatorname{arg}(1+t^2\xi^2), \qquad \xi=\mathrm{e}^{ i u}; \end{gathered}
\end{equation*}
\notag
$$
(2) the pointwise estimate
$$
\begin{equation*}
|\widehat{\varepsilon}_{2n+1}^{\,(0)}(f_s,x)| \leqslant{2^{2-s}\sqrt{1-x^2}} \biggl|\sin\frac{\pi s}{2}\biggr| \int_{0}^{1}\frac{(1-t^2)^s t^{2n+1-s}}{\sqrt{1+2t^2 \cos 2 u +t^4}}\,dt, \qquad x=\cos u;
\end{equation*}
\notag
$$
(3) the estimate for uniform approximations Let us study the asymptotic behaviour of (5.6) as $n \to \infty$. To do this we change the variable to $t=\sqrt{(1-y)/(1+y)}$, $dt=-dy/((1+y)^{3/2} (1-y)^{1/2})$, on the right-hand side of (5.6). Then we have
$$
\begin{equation}
\widehat{\varepsilon}_{2n+1}^{\,*}(f_s,A_n)= 2\biggl|\sin\frac{\pi s}{2}\biggr| \int_{0}^{1}\nu_s(y)\biggl|\prod_{k=1}^{n-p}\frac{\beta_k-y}{\beta_k+y}\biggr|\,dy, \quad n \in \mathbb{N},
\end{equation}
\tag{5.10}
$$
where $\beta_k=\frac{1-\alpha_k^2}{1+\alpha_k^2}$ and
$$
\begin{equation*}
\nu_s(y)=\frac{y^{s-1}}{(1+y)(1-y^2)^{s/2}} \biggl(\frac{1-y}{1+y}\biggr)^p.
\end{equation*}
\notag
$$
5.2. An asymptotic expression for the majorant in the case of an approximating function with a fixed number of polesLet $n>p$, $p=[s/2]$, $n_1=n-p$, let $q$ be an arbitrary fixed natural number, and let $A_q$ be a set of parameters $(\beta_{1},\beta_2, \dots, \beta_{n_1})$ that contains precisely $q$ different parameters and the multiplicity of each parameter is $m$, $n_1=m q$. So we are concentred here with approximations by rational functions with a pole of order $2p$ at infinity and with $2q$ geometrically different poles (each of multiplicity $m$) in the extended complex plane. In view of the above assumptions, (5.10) can be written as
$$
\begin{equation*}
\begin{gathered} \, \widehat{\varepsilon}_{2n+1,2q}^{\,*}(f_s,A_q)= 2\biggl|\sin\frac{\pi s}{2}\biggr| \int_{0}^{1}\nu_s(y)|\pi_q(y)|^m\,dy, \\ \text{where}\quad\pi_q(y)=\prod_{k=1}^{q}\frac{\beta_k-y}{\beta_k+y}\quad\text{and}\quad s \in (0,+\infty). \end{gathered}
\end{equation*}
\notag
$$
Let the parameters $\beta_k$, $k=1,2,\dots,q$, satisfy condition (4.18) and be ordered as follows:
$$
\begin{equation*}
0<\beta_q \leqslant \beta_{q-1} \leqslant \dots \leqslant \beta_1 \leqslant 1.
\end{equation*}
\notag
$$
In the next theorem we find an asymptotic expression for the majorant of uniform approximations of the function $\widehat{f}(x)$ on the interval $[-1,1]$ by the rational integral operator (1.4) with an arbitrary fixed number of geometrically different poles. The theorem is stated without a proof, because it is based on the method described in detail in [19] and used already for the verification of Theorem 6. Theorem 12. For each natural number $q$, $0 < q<n_1$, $n_1=n-p$, $n_1=mq$, and any $s \in (0,+\infty)$, the following asymptotic equality holds: Putting $\beta_j=1$ in Theorem 12, $j=1,2,\dots,q$, we find that the quantity
$$
\begin{equation*}
\widehat{\varepsilon}_{2n+1,2q}^{\,*}(f_s,O) =\widehat{\varepsilon}_{2n+1}^{\,(0)}(f_s), \qquad O=(\underbrace{0,0,\dots,0}_{n}),
\end{equation*}
\notag
$$
is an upper bound for uniform approximations of the function $\widehat{f}_s(x)$ on the interval $[-1,1]$ by the polynomial operator equal to the image of the partial sums of the Fourier–Chebyshev series under the transformation (1.1). Corollary 7. The uniform approximations of the function $\widehat{f}_s(x)$ on $[-1,1]$ by the polynomial operator that is the image of the partial sums of the Fourier–Chebyshev series of order $n$, $n>s/2$, under the transformation (1.1), satisfy the estimate It would be interesting to compare the estimate in Corollary 7 with the one for uniform polynomial approximations from Corollary 4. 5.3. The best majorant of uniform approximations for an approximating function with a fixed number of polesOur aim here is to minimize the right-hand side of the asymptotic equality (5.11) by choosing a family of parameters $(\beta_1^{*},\beta_2^{*},\dots,\beta_q^{*})$ optimal for this problem, that is, we find a best estimate for approximations of the singular integrals under consideration by the rational integral operator (1.4) with a fixed number of poles. We set
$$
\begin{equation*}
\widehat{\varepsilon}_{2n+1,2q}(f_s)=\operatorname*{inf}_{A_q} \widehat{\varepsilon}_{2n+1,2q}(f_s,A_q)\quad\text{and} \quad \widehat{\varepsilon}_{2n+1,2q}^{\,*}(f_s) =\operatorname*{inf}_{A_q}\widehat{\varepsilon}_{2n+1,2q}^{\,*}(f_s,A_q).
\end{equation*}
\notag
$$
It is clear from (5.5) that
$$
\begin{equation*}
\widehat{\varepsilon}_{2n+1,2q}(f_s) \leqslant\widehat{\varepsilon}_{2n+1,2q}^{\,*}(f_s), \qquad n \in \mathbb{N}.
\end{equation*}
\notag
$$
Theorem 13. The uniform approximations of the function $\widehat{f}_s(x)$ on $[-1,1]$ by the rational integral operator (1.4) with $q$ geometrically different poles satisfy the upper estimate where $n_0(s)$ is a natural number independent of $n$ but depending on $s$ and $\Gamma(\,\cdot\,)$ is the Euler gamma function. Proof. Let the family $A_q$ be defined by where the $c_k$, $k=1,2,\dots,q$, are some numbers to be defined below.
Let us study the asymptotic behaviour of each of the three terms in (5.11). Considering the first term we obtain the asymptotic equality
$$
\begin{equation*}
S_n^{(1)}(A_{q})=\frac{\Gamma(s)}{\bigl(2m\sum_{k=1}^{q}\frac{1}{c_k}(\frac{m}{\log m})^{2k-1}\bigr)^{s}} \sim \frac{\Gamma(s) c_q^{s}(\log m)^{(2q-1)s}}{2^{s} m^{2qs}}, \qquad m \to \infty.
\end{equation*}
\notag
$$
For the second term an appeal to Theorem 6 in [19] shows that
$$
\begin{equation*}
S_n^{(2)}(A_{q}) \sim\sqrt{\frac{\pi}{2}}\, \sum_{j=1}^{k-1}(c_j c_{j+1})^{s/2} \sqrt[4]{\frac{c_j}{c_{j+1}}} \, \frac{(\log m)^{2js-1/2}}{m^{2js+4\sqrt{c_{j+1}/c_j}}}, \qquad m \to \infty.
\end{equation*}
\notag
$$
For the third term in (5.11) we have
$$
\begin{equation*}
S_n^{(3)}(A_q) \sim \frac{\Gamma(1+p-\frac{s}2)}{2^{2p+1}c_1^{1+p-s/2} m^{2c_1}(\log m)^{1+p-s/2}}, \qquad m \to \infty.
\end{equation*}
\notag
$$
Next we choose numbers $c_k$, $k=1,2,\dots,q$, to equate the exponents of $m$ in the expressions for $S_n^{(1)}(A_{q})$, $S_n^{(2)}(A_{q})$ and $S_n^{(3)}(A_{q})$. It is clear that the $c_k$, $k=1,2,\dots,q$, must satisfy
$$
\begin{equation*}
\begin{cases} qs=c_1, \\ qs=js+2\sqrt{\dfrac{c_j+1}{c_j}}. \end{cases}
\end{equation*}
\notag
$$
As a result, and the asymptotic equality (5.11) assumes the form
$$
\begin{equation*}
\widehat{\varepsilon}_{2n+1,2q}^{\,*}(f_s,A_q^{*}) \sim2^{1-s}\biggl|\sin\frac{\pi s}{2}\biggr|\Gamma(s) \biggl(\frac{s^{2q-1}(q!)^2}{q 2^{2q-2}}\biggr)^{s} \biggl(\frac{\log^{2q-1}m}{m^{2q}}\biggr)^{s}, \qquad m \to \infty.
\end{equation*}
\notag
$$
Now, to prove that for the above values of the $c_k$, $k=1,2,\dots,q$, the family of parameters $\beta_k$, $k=1,\dots,q$, is optimal, that is, the quantity $\varepsilon_{n}^{*}(|\cdot|^s,A_{q})$ assumes its asymptotically minimum values for these $c_k$, it suffices to follow the arguments in [29]. Consequently, we have
$$
\begin{equation*}
\widehat{\varepsilon}_{2n+1,2q}^{\,*}(f_s) =\widehat{\varepsilon}_{2n+1,2q}^{\,*}(f_s,A_q^{*}), \qquad n \in \mathbb{N}.
\end{equation*}
\notag
$$
In addition, by (5.5), uniformly with respect to $x \in [-1,1]$ we have
$$
\begin{equation*}
\widehat{\varepsilon}_{2n+1,2q}(f_s) \leqslant\widehat{\varepsilon}_{2n+1,2q}^{\,*}(f_s).
\end{equation*}
\notag
$$
It remains to note that $n_1=mq$ and $n_1=n-p$, and derive the asymptotic equality (5.12) from the last estimate.
This completes the proof of Theorem 13. Let us compare the estimate from Theorem 13 with the one from Corollary 7 for uniform polynomial approximations of the function $\widehat{f}_s(x)$. The rate of decay of polynomial analogues of the above method of rational approximation is $O(1/n^{s})$ uniformly with respect to $x \in [-1,1]$, whereas in the rational case, when the approximating function has $2q$ geometrically different poles in the open complex plane the rate of decay can be increased to $O(({\log^{2q-1}n}/{n^{2q}})^{s})$. Remark 2. From Theorems 7 and 13 it follows that, for the asymptotic expressions for best majorants of uniform approximations of the function $\widehat{f}_s(x)$ by the above rational integral operators, in the case of an arbitrary fixed number of geometrically different poles we have the equality 5.4. The majorant of uniform approximations in the general caseLet us investigate (5.10) as $n \to \infty$ without any constraints on the number of parameters of the approximating function. As above, we assume that the parameters $\beta_k$, $k=1,2,\dots,n_1$, $n_1=n-p$, are ordered as follows:
$$
\begin{equation*}
0<\beta_{n_1}<\beta_{n_1-1}<\dots <\beta_{1}\leqslant 1,
\end{equation*}
\notag
$$
and, moreover, they have the form $\beta_{k}=\xi^k$, $\xi \in (0,1)$, $k=1,2,\dots,n_1$. Theorem 14. The uniform rational approximations of the function $\widehat{f}_s(x)$ on $[-1,1]$ by the integral operator (1.4) with parameters (4.22)–(4.24) satisfy the upper estimate For a proof it suffices to find the asymptotic behaviour of the integral on the right-hand side of (5.10) in the case where the parameters of the approximating function satisfy (4.22)–(4.24). The proof repeats verbatim the argument in Theorem 8 and is omitted. 5.5. The case of ‘Newman’ parametersLet us investigate the right-hand side of estimate (5.13) in the case where the parameters $\beta_{k}$, $k=1,2,\dots,n_1$, $n_1=n-p$, are modifications of those introduced by Newman in [31]. For each fixed $n\in \mathbb{N}$ let $A_N$ be a family of parameters $\beta_k$ of the form
$$
\begin{equation}
\beta_{k}=\exp \biggl(-\frac{ck}{\sqrt{n_1}}\biggr), \qquad k=1,2,\dots,n_1,
\end{equation}
\tag{5.14}
$$
where $c$ is some positive quantity independent of $n$. Note that some modifications of the parameters $\beta_k$, $k=1,2,\dots,n_1$, were used by Bulanov (see, for example, [34]) for the solution of some approximation problems. The theorems establishing the corresponding estimates for uniform approximations with parameters (5.14) of the approximating functions are presented given without proofs, because the corresponding arguments are quite similar to the ones used for Theorems 9 and 10. Theorem 15. The uniform approximations of the function $\widehat{f}_s(x)$ on $[-1,1]$ by the rational integral operator (1.4) with parameters (5.14) satisfy the upper estimate Next we minimize (5.15) by choosing an optimal parameter $c$ for this problem, that is, we find a best estimate for uniform approximations with parameters (5.14). We set
$$
\begin{equation*}
\widehat{\varepsilon}_{2n+1}(f_{s}) =\inf_{A_N}\widehat{\varepsilon}_{2n+1}(f_{s},A_N)\quad\text{and} \quad \widehat{\varepsilon}_{2n+1}^{\,*}(f_{s}) =\inf_{A_N}\widehat{\varepsilon}_{2n+1}^{*}(f_{s},A_N).
\end{equation*}
\notag
$$
From (5.5) it is clear that
$$
\begin{equation*}
\widehat{\varepsilon}_{2n+1}(f_{s}) \leqslant\widehat{\varepsilon}_{2n+1}^{\,*}(f_{s}), \qquad n \in \mathbb{N}.
\end{equation*}
\notag
$$
Theorem 16. For the uniform approximations of the function $\widehat{f}_s(x)$ on $[-1,1]$ by the rational integral operator (1.4) there exists a family of parameters $A_N^{*}$ of the form (5.14) such that the following upper estimate holds: Corollary 8. For the uniform approximations of functions (1.1) with density $|x|^s$, $s \in (0,2)$, on $[-1,1]$ by the rational integral operator (1.4) there exists a family of parameters of the approximating function such that the upper estimate holds uniformly with respect to $x \in [-1,1]$. Remark 3. It follows from Theorems 10 and 16 that for sharp estimates of uniform approximations of the function $\widehat{f}_s(x)$ by the above rational integral operators in the case of ‘Newman’ poles we have § 6. ConclusionsIn the present paper we have studied approximations on the interval $[-1,1]$ of singular integrals of the form (1.1) by two rational integral operators. The first operator here is the Fourier–Chebyshev rational integral operator [18] associated with the system of Chebyshev–Markov rational functions and which is a natural generalization of partial sums of the polynomial Fourier–Chebyshev series. The second operator is the image of the above integral operator under the transformation (1.1). For each of these operators an integral representation of approximations has been presented. In the polynomial case, for both operators we have established estimates for approximations on $[-1,1]$ of singular integrals of the form (1.1) with density satisfying a Hölder condition of exponent $\alpha \in (0,1]$. As a separate problem for both methods we have studied rational approximations in the case where the density of the singular integrals (1.1) has a power-law singularity on the interval $[-1,1]$. Two radically different approaches have been considered. The first approach involves constraints on the number of geometrically different poles in the approximating functions. In the second approach the parameters of the approximating functions are modifications of the ‘Newman’ parameters. In both cases we have established estimates for pointwise approximations and for uniform approximations with a certain majorant, found an asymptotic expression for this estimate and provided the optimal values of the parameters under which the majorant has the greatest decay rate. As a corollary, we have obtained estimates for approximations by polynomial analogues of both integral operators of singular integrals of the form (1.1) with density having a power-law singularity on $[-1,1]$. The above findings show that the classes of singular integrals of the form (1.1) with density having a power-law singularity on the interval $[-1,1]$ reflect the peculiarities of the rational approximation by the integral operators under consideration in the following sense: for certain values of the parameters of the approximating functions these function provide approximations of higher order in comparison to their polynomial analogues. It is worth pointing out that the estimates of best uniform approximations of the function $\widehat{f}_s(x)$ on $[-1,1]$ by the above methods are asymptotically equal. 
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\Bibitem{PotRov24} |
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