My question seems elementary, yet I could not find the solution after working on and searching for several days... I'd like to find the eigenfunctions of a simple integral kernel: \begin{equation} \lambda f_\lambda(x) = \frac{1}{2\pi}\int_0^\infty \frac{d x'}{x^2 +x'^2} f_\lambda(x') \end{equation} There is an obvious scaling property \begin{equation} f_{\lambda'}(x)=f_{\lambda}(x \lambda/\lambda'), \end{equation} so it suffice to find an eigenfunction at a certain $\lambda$.
It looks like the theory of complex functions may be of help here, and $f(x)$ can be analytically continued to the half-plane ${\rm Re} z >0$, from the line ${\rm Im}z =0$, ${\rm Re}z >0$. One can thus derive an equation relating the values of $f(z)$ at this line and the line ${\rm Im}(z)=0$, \begin{equation} \lambda \left(f(-i x+ \delta) - f(i x + \delta)\right) = \frac{i}{2 x} f(x); \end{equation} $x>0$, $\delta>0, \delta \to 0$. However, I am not able to guess the solution of this equation.
Update:
With some sweat, I've obtained the asymptotic expression at small $x$ ($\lambda = 1/2)$, \begin{equation} f(x) \propto x^{-1/2} \cos\left(\frac{\ln^2(x)}{\pi} \right). \end{equation} No further progress so far.
Update 2:
The asymptotic expression at big $x$: \begin{equation} f(x) \propto \frac{1}{x^2} \left( 1 - \frac{1}{4\lambda x} + \frac{\ln(\lambda x)}{8 \pi (\lambda x)^2}\right) + O(x^{-4}) \end{equation}