Example 4. Solve the integral equation

$$
\int_{0}^{\infty} \frac{e^{-x t /(1-x)}}{1-x} e^{-t} \varphi(t) d t=1-x, \quad|x|<1
$$
Solution. The function

$$
G(x, t)=\frac{e^{-x t /(1-x)}}{1-x}
$$

is the generating function for the Chebyshev-Laguerre polynomials $L_{n}(t)$:

$$
G(x, t)=\sum_{n=0}^{\infty} L_{n}(t) x^{n}
$$

We seek the solution of equation (19) in the form

$$
\varphi(t)=\sum_{k=0}^{\infty} a_{k} L_{k}(t)
$$

Substituting (20) and (21) into (19), we get

$$
\int_{0}^{\infty} \sum_{n=0}^{\infty} e^{-t} L_{n}(t) x^{n} \cdot \sum_{k=0}^{\infty} a_{k} L_{k}(t) d t=1-x
$$

Due to the orthogonality of the Chebyshev-Laguerre polynomials with weight $\rho(t)=e^{-t}$ on the interval $(0, \infty)$, we have

$$
\sum_{n=0}^{\infty}(n!)^{2} x^{n} a_{n}=1-x
$$

Here, we used the fact that

$$
\left\|L_{n}\right\|^{2}=\int_{0}^{\infty} e^{-t} L_{n}^{2}(t) d t=(n!)^{2}
$$

By equating the coefficients of the same powers of $x$ on both sides of (22), we find $a_{0}=1, a_{1}=-1, a_{k}=0(k=2,3, \ldots)$. Therefore,

$$
\varphi(t)=L_{0}(t)-L_{1}(t)=1-(1-t)=t
$$

Thus, $\varphi(t)=t$.

## Exercises for Independent Solution

Solve the integral equations using the method of generating functions:

306. $\int_{-1}^{1} \frac{\varphi(t) d t}{\sqrt{1+x^{2}-2 x t}}=2 x^{3}-2 x$.
307. $\int_{-1}^{1} \frac{\varphi(t) d t}{\sqrt{1+x^{2}-2 x t}}=\frac{1}{1-x}$.
308. $\int_{0}^{\infty} \frac{e^{-x t /(1-x)}}{1-x} e^{-t} \varphi(t) d t=2-x^{2}$.

Integral Equation of Schlömilch

The integral equation of Schlömilch is of the form

$$
f(x)=\frac{2}{\pi} \int_{0}^{\pi / 2} \varphi(x \sin \theta) d \theta
$$

where the function $f(x)$ has a continuous derivative $f^{\prime}(x)$ for $-\pi \leqslant x \leqslant \pi$, and $\varphi(x)$ is the unknown function.

We show that equation (23) has a unique continuously differentiable solution on the interval $-\pi \leqslant x \leqslant \pi$, namely:

$$
\varphi(x)=f(0)+x \int_{0}^{\pi / 2} f^{\prime}(x \sin \theta) d \theta
$$

From equation (23), we find

$$
f(0)=\frac{2}{\pi} \int_{0}^{\pi / 2} \varphi(0) d \theta=\varphi(0), \quad \text { or } \varphi(0)=f(0)
$$

Differentiating both sides of equation (23) with respect to $x$, we get

$$
f^{\prime}(x)=\frac{2}{\pi} \int_{0}^{\pi / 2} \varphi^{\prime}(x \sin \theta) \sin \theta d \theta
$$

Substituting $x$ with $x \sin \psi$ in (24):

$$
f^{\prime}(x \sin \psi)=\frac{2}{\pi} \int_{0}^{\pi / 2} \varphi^{\prime}(x \sin \psi \sin \theta) \sin \theta d \theta
$$

Multiplying both sides of (25) by $x$ and integrating over $\psi$ from 0 to $\frac{\pi}{2}$:

$$
x \int_{0}^{\pi / 2} f^{\prime}(x \sin \psi) d \psi=\frac{2 x}{\pi} \int_{0}^{\pi / 2}\left\{\int_{0}^{\pi / 2} \varphi^{\prime}(x \sin \psi \sin \theta) \sin \theta d \theta\right\} d \psi
$$

By changing the order of integration in the right-hand side of the last equation, we get

$$
x \int_{0}^{\pi / 2} f^{\prime}(x \sin \psi) d \psi=\frac{2 x}{\pi} \int_{0}^{\pi / 2}\left\{\int_{0}^{\pi / 2} \varphi^{\prime}(x \sin \psi \sin \theta) \sin \theta d \psi\right\} d \theta
$$

## Introduce a new variable by the formula

$$
\sin \chi=\sin \theta \sin \psi
$$

from which

$$
\cos \chi d \chi=\sin \theta \cos \psi d \psi
$$

so that

$$
\frac{\cos \chi}{\cos \psi} d \chi=\sin \theta d \psi
$$

When $\psi=0$, we have $\sin \chi=0$, i.e., $\chi=0$, and when $\psi=\frac{\pi}{2}$, we get $\chi=\theta$. Thus, equation (26) becomes

$$
x \int_{0}^{\pi / 2} f^{\prime}(x \sin \psi) d \psi=\frac{2 x}{\pi} \int_{0}^{\pi / 2}\left\{\int_{0}^{\theta} \frac{\varphi^{\prime}(x \sin \chi) \cos \chi}{\cos \psi} d \chi\right\} d \theta
$$

By the relation (27), we have

$$
\sin \psi=\frac{\sin \chi}{\sin \theta}, \quad \cos \psi=\frac{\sqrt{\sin ^{2} \theta-\sin ^{2} \chi}}{\sin \theta}=\frac{\sqrt{\cos ^{2} \chi-\cos ^{2} \theta}}{\sin \theta}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_d528bb4cd8a42024c50cg-144.jpg?height=505&width=538&top_left_y=968&top_left_x=405)

Figure 7

Substituting (29) into the right-hand side of (28) and changing the order of integration (Figure 7), we get

$$
\begin{aligned}
x \int_{0}^{\pi / 2} f^{\prime}(x \sin \psi) d \psi & =\frac{2 x}{\pi} \int_{0}^{\pi / 2}\left\{\int_{0}^{\theta} \frac{\varphi^{\prime}(x \sin \chi) \cos \chi \sin \theta}{\sqrt{\cos ^{2} \chi-\cos ^{2} \theta}} d \chi\right\} d \theta= \\
& =\frac{2 x}{\pi} \int_{0}^{\pi / 2}\left\{\int_{\chi}^{\pi / 2} \frac{\varphi^{\prime}(x \sin \chi) \cos \chi \sin \theta}{\sqrt{\cos ^{2} \chi-\cos ^{2} \theta}} d \theta\right\} d \chi
\end{aligned}
$$

## We have

$$
\int_{\chi}^{\pi / 2} \frac{\sin \theta d \theta}{\sqrt{\cos ^{2} \chi-\cos ^{2} \theta}}=-\left.\arcsin \frac{\cos \theta}{\cos \chi}\right|_{\theta=x} ^{\theta=\pi / 2}=\arcsin 1=\frac{\pi}{2}
$$

Thus, relation (30) becomes

$$
x \int_{0}^{\pi / 2} f^{\prime}(x \sin \psi) d \psi=x \int_{0}^{\pi / 2} \varphi^{\prime}(x \sin \chi) \cos \chi d \chi
$$

But

$$
\begin{aligned}
& x \int_{0}^{\pi / 2} \varphi^{\prime}(x \sin \chi) \cos \chi d \chi= \\
& \quad=\int_{0}^{\pi / 2} d \varphi(x \sin \chi)=\left.\varphi(x \sin \chi)\right|_{x=0} ^{x=\pi / 2}=\varphi(x)-\varphi(0)
\end{aligned}
$$

Noting that $\varphi(0)=f(0)$, from (31) and (32) we finally get

$$
\varphi(x)=f(0)+x \int_{0}^{\pi / 2} f^{\prime}(x \sin \psi) d \psi
$$