Example 1. Solve the integral equation

$$
\varphi(x)=f(x)+\lambda \int_{-\infty}^{+\infty} e^{-|x-t|} \varphi(t) d t \quad\left(\lambda<\frac{1}{2}\right)
$$
Solution. Let $F(\omega)$ be the Fourier transform of the function $f(x)$, and $\tilde{K}(\omega)$ be the Fourier transform of the kernel $K(x)=e^{-|x|}$. Here,

$$
\begin{aligned}
\tilde{K}(\omega)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} e^{-|x|} e^{-i z \omega} d x & =\frac{1}{\sqrt{2 \pi}}\left[\int_{-\infty}^{0} e^{x(1-i \omega)} d x+\int_{0}^{+\infty} e^{-x(1+i \omega)} d x\right]= \\
& =\frac{1}{\sqrt{2 \pi}}\left(\frac{1}{1-i \omega}+\frac{1}{1+i \omega}\right)=\sqrt{\frac{2}{\pi}} \frac{1}{1+\omega^{2}}
\end{aligned}
$$

Applying the Fourier transform to both sides of the given equation, we get

$$
\Phi(\omega)=F(\omega)+\lambda \sqrt{\frac{2}{\pi}} \frac{1}{1+\omega^{2}} \Phi(\omega) \sqrt{2 \pi}
$$

from which

$$
\Phi(\omega)=\frac{1+\omega^{2}}{1-2 \lambda+\omega^{2}} F(\omega)
$$

Therefore, the solution to the original equation is the function

$$
\varphi(x)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \frac{1+\omega^{2}}{1-2 \lambda+\omega^{2}} F(\omega) e^{i z \omega} d \omega
$$

Note that in this case

$$
1-\sqrt{2 \pi} \tilde{K}(\omega) \lambda=1-\frac{2 \lambda}{1+\omega^{2}}=\frac{1-2 \lambda+\omega^{2}}{1+\omega^{2}}
$$

does not vanish for any real value of $\omega$ when $\lambda<\frac{1}{2}$. Assuming, for example, $f(x)=e^{-|x|}$, we have

$$
F(\omega)=\sqrt{\frac{2}{\pi}} \frac{1}{1+\omega^{2}} .
$$

So formula (3) gives

$$
\varphi(x)=\frac{1}{\pi} \int_{-\infty}^{+\infty} \frac{e^{i z \omega}}{1-2 \lambda+\omega^{2}} d \omega
$$

To compute the last integral, we can use the method of contour integration. This yields

$$
\begin{aligned}
& \varphi(x)=\frac{e^{-x \sqrt{1-2 \lambda}}}{\sqrt{1-2 \lambda}} \quad \text { for } \quad x \geqslant 0 \\
& \varphi(x)=\frac{e^{x \sqrt{1-2 \lambda}}}{\sqrt{1-2 \lambda}} \quad \text { for } \quad x<0
\end{aligned}
$$

more concisely,

$$
\varphi(x)=\frac{e^{-\sqrt{1-2 \lambda}|x|}}{\sqrt{1-2 \lambda}}
$$