Example 7. Solve the integral equation

$$
\varphi(x)=x e^{-x}+\lambda \int_{0}^{\infty} J_{0}(2 \sqrt{x t}) \varphi(t) d t \quad(|\lambda| \neq 1)
$$
Solution. Let $\varphi(x) \risingdotseq \Phi(p)$. Applying the Laplace transform to both sides of (20) and taking into account the Efros theorem, we find

$$
\Phi(p)=\frac{1}{(p+1)^{2}}+\lambda \frac{1}{p} \Phi\left(\frac{1}{p}\right)
$$

Replacing $p$ with $\frac{1}{p}$, we get

$$
\Phi\left(\frac{1}{p}\right)=\frac{p^{2}}{(p+1)^{2}}+\lambda p \Phi(p)
$$

From (21) and (22), we find

$$
\Phi(p)=\frac{1}{(p+1)^{2}}+\frac{\lambda}{p}\left[\frac{p^{2}}{(p+1)^{2}}+\lambda p \Phi(p)\right],
$$

or

$$
\Phi(p)=\frac{1}{1-\lambda^{2}}+\left[\frac{1}{(p+1)^{2}}+\frac{\lambda p}{(p+1)^{2}}\right]
$$

From this,

$$
\varphi(x)=e^{-x}\left(\frac{x}{1+\lambda}+\frac{\lambda}{1-\lambda^{2}}\right)
$$

## Problems for Independent Solution

Solve the following integral equations ( $\lambda \neq \pm 1$ ):

270. $\varphi(x)=e^{2}+\lambda \int_{0}^{\infty} \sqrt{\frac{x}{t}} J_{1}(2 \sqrt{x t}) \varphi(t) d t$.
271. $\varphi(x)=\cos x+\lambda \int_{0}^{\infty} J_{0}(2 \sqrt{x t}) \varphi(t) d t$.
272. $\varphi(x)=\cos x+\lambda \int_{0}^{\infty} \frac{x}{t} J_{2}(2 \sqrt{x t}) \varphi(t) d t$.
273. $\varphi(x)=\sin x+\lambda \int_{0}^{\infty} \sqrt{\frac{x}{t}} J_{1}(2 \sqrt{x t}) \varphi(t) d t$.

## § 19. Application of the Mellin Transform to the Solution of Some Integral Equations

Let the function $f(t)$ be defined for positive $t$ and satisfy the conditions

$$
\int_{0}^{1}|f(t)| t^{\sigma_{1}-1} d t<\infty, \quad \int_{1}^{\infty}|f(t)| t^{\sigma_{2}-1} d t<\infty \quad\left(\sigma_{1}<\sigma<\sigma_{2}\right)
$$

The Mellin transform of the function $f(t)$ is defined by the formula

$$
F(s)=M\{f(t)\}=\int_{0}^{\infty} f(t) t^{s-1} d t \quad\left(\sigma_{1}<\sigma<\sigma_{2}\right)
$$

where the integral is taken along the line $l: \operatorname{Re} s=\sigma$, parallel to the imaginary axis of the $s$-plane, and is understood in the sense of the principal value. If the behavior of the function $f(t)$ as $t \rightarrow 0$ and $t \rightarrow \infty$ is known, for example, from physical considerations, the boundaries of the strip $\left(\sigma_{1}, \sigma_{2}\right)$ can be determined from the conditions of absolute convergence of the integral (2). If the behavior of $f(t)$ is known only at one end of the interval $(0,+\infty)$, for example, as $t \rightarrow 0$, then only $\sigma_{1}$ is determined, the line of integration $l$ in (3) should be chosen to the right of the line $\sigma=\sigma_{1}$ and to the left of the nearest singular point of the function $F(s)$.

The Mellin transform is closely related to the Fourier and Laplace transforms, and many theorems concerning the Mellin transform can be obtained from the corresponding theorems for the Fourier and Laplace transforms by a change of variables.

The convolution theorem for the Mellin transform has the following form:

$$
M\left\{\int_{0}^{\infty} f(t) \varphi\left(\frac{x}{t}\right) \frac{d t}{t}\right\}=F(s) \cdot \Phi(s)
$$

From this, it can be concluded that the Mellin transform is convenient to apply when solving integral equations of the form

$$
\varphi(x)=f(x)+\int_{0}^{\infty} K\left(\frac{x}{t}\right) \varphi(t) \frac{d t}{t}
$$

Indeed, let the functions $\varphi(x), f(x)$, and $K(x)$ admit the Mellin transform and let $\varphi(x) \rightarrow \Phi(s), f(x) \rightarrow F(s), K(x) \rightarrow \widetilde{K}(s)$, and the regions of analyticity of $F(s)$ and $\widetilde{K}(s)$ have a common strip $\sigma_{1}<\operatorname{Re} s=\sigma<\sigma_{2}$. Applying the Mellin transform to both sides of equation (5) and using the convolution theorem (4), we get

$$
\Phi(s)=F(s)+\tilde{K}(s) \cdot \Phi(s)
$$

from which

$$
\Phi(s)=\frac{F(s)}{1-\tilde{K}(s)} \quad(\widetilde{K}(s) \neq 1)
$$

This is the operator solution of the integral equation (5). By the inversion formula (3), we find the solution $\varphi(x)$ of this equation:

$$
\varphi(x)=\frac{1}{2 \pi i} \int_{\sigma-i \infty}^{\sigma+i \infty} \frac{F(s)}{1-\widetilde{K}(s)} x^{-s} d s
$$

Consider an integral equation of the form

$$
\varphi(x)=f(x)+\int_{0}^{\infty} K(x \cdot t) \varphi(t) d t
$$

(Fox's equation). Multiplying both sides of (6) by $x^{s-1}$ and integrating over $x$ from 0 to $\infty$, we get

$$
\int_{0}^{\infty} \varphi(x) x^{s-1} d x=\int_{0}^{\infty} f(x) x^{s-1} d x+\int_{0}^{\infty} \varphi(t) d t \int_{0}^{\infty} K(x \cdot t) x^{s-1} d x
$$

Denoting the Mellin transforms of the functions $\varphi(x), f(x), K(x)$ by $\Phi(s), F(s), \widetilde{K}(s)$ respectively, after some transformations, we get

$$
\Phi(s)=F(s)+\widetilde{K}(s) \int_{0}^{\infty} \varphi(t) t^{-s} d t
$$

It is easy to see that $\int_{0}^{\infty} \varphi(t) t^{-s} d t=\Phi(1-s)$, so (7) can be written as

$$
\Phi(s)=F(s)+\Phi(1-s) \tilde{K}(s)
$$

Replacing $s$ with $1-s$ in equation (8), we get

$$
\Phi(1-s)=F(1-s)+\Phi(s) \widetilde{K}(1-s)
$$

From equations (8) and (9), we find

$$
\Phi(s)=F(s)+F(1-s) \tilde{K}(s)+\Phi(s) \tilde{K}(s) \widetilde{K}(1-s)
$$

from which

$$
\Phi(s)=\frac{F(s)+F(1-s) \widetilde{K}(s)}{1-\widetilde{K}(s) \cdot \widetilde{K}(1-s)}
$$

This is the operator solution of equation (6).

By the inversion formula of the Mellin transform, we find

$$
\varphi(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} \frac{F(s)+F(1-s) \tilde{K}(s)}{1-\tilde{K}(s) \cdot \tilde{K}(1-s)} x^{-s} d s
$$

- the solution of the integral equation (6).