Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.