Solution of inhomogeneous Fredholm integral equation of the first kind with symmetric rational kernel

Question

Please explain how to solve this inhomogeneous Fredholm integral equation of the first kind: $$f(x)=\frac{1}{\pi}\int_{0}^{\infty}\frac{g(y)}{x+y}dy$$

Answer 1

The equation

$$ f(x)=\frac{1}{\pi}\int_{0}^{\infty}\frac{g(y)}{x+y}\mathrm{d}y $$

has solution

$$\begin{align} y(x) &= \frac{1}{2 i} \lim_{\epsilon \to 0^+} \left\{f(-x-i\epsilon)-f(-x+i\epsilon)\right\} \\ &= \frac{1}{\sqrt{x}} \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} \left(\frac{\pi}{x} \frac{\mathrm{d}}{\mathrm{d}x}\right)^{2k} \left\{\sqrt{x}f(x)\right\}. \end{align}$$

Source: Polyanin and Manzhirov, Handbook of Integral Equations, section 3.1-3, #17.

Numerous other sources are cited below the entry there.

Answer 2

You could try a Mellin transform. Since $\int _{0}^{\infty }\!{\frac {{x}^{s-1}}{x+y}}{dx}={y}^{s-1}\pi \,\csc \left( \pi \,s \right) $ for $y > 0$ and $0 < \Re s < 1$, the Mellin transforms of $f$ and $g$ satisfy $Mf(s) = \csc(\pi s) Mg(s)$ for $0 < \Re s < 1$. You might then try inverting $Mg(s)$ using the inversion formula

$$g(y) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} Mf(s) \sin(\pi s)\ ds$$

where $0 < c < 1$, under appropriate convergence assumptions.

Answer 3

Assuming that $x>0$, and that $f$ and $g$ are sufficiently well-behaved, we can write \begin{align} f(x)&=\frac{1}{\pi}\int_0^{\infty}g(y)\int_0^{\infty}e^{-s(x+y)}\,ds\,dy \\ &=\frac{1}{\pi}\int_0^{\infty}e^{-sx}\int_0^{\infty}e^{-sy}g(y)\,dy\,ds \\ &=\frac{1}{\pi}\int_0^{\infty}e^{-sx}\,\cal{L}[g](s)\,ds \\ &=\frac{1}{\pi}\cal{L}[\cal{L}[g]](x), \tag{1} \end{align} where $\cal{L}[g]$ denotes the Laplace transform of $g$. Therefore, formally, $$ g(x)=\pi\cal{L}^{-1}[\cal{L}^{-1}[f]](x). \tag{2} $$