I have tried (hard) to solve the following integral equation, but without success:
$$\frac{1}{2{\pi}i}\int_{-\infty}^\infty{\frac{\sigma(s)}{s-(2hi+x)}ds}=\frac{1}{(x+hi)^2}$$ where $h$ and $x$ are real, $h>0$ and $x\in[-\infty,\infty]$.
At first sight, and because $h>0$, $\sigma(z)$ must be an analytic function in the upper half-plane, as it is $\frac{1}{(z+hi)^2}$, where $z=x+yi$, and whose limiting value ($z\to{x}$) are $\sigma(z)$ and the right hand side of the integral equation respectively. But if the following parameter change is done, $z'=2hi+x$, the follwing integral equation arise: $$\frac{1}{2{\pi}i}\int_{-\infty}^\infty{\frac{\sigma(s)}{s-z'}ds}=\frac{1}{(z'-hi)^2}$$ and the right side function is not analytic in the upper $z'$ half-plane. Please, can someone give some hint to deal with this IE?
Thanks in advance friends!