Calculation of the convolution of Cauchy density function $\int_{-\infty}^{\infty}\frac{ab}{\pi^2}\frac{1}{y^2+a^2}\frac{1}{(x-y)^2+b^2}dy$

Question

I tried to calculate the following integral, which is the convolution of Cauchy density function:

$$\int_{-\infty}^{\infty}\frac{ab}{\pi^2}\frac{1}{y^2+a^2}\frac{1}{(x-y)^2+b^2}dy$$

I tried to use substitution, let $x-y=t$, then $y=x-t$

So $$\int_{-\infty}^{\infty}\frac{ab}{\pi^2}\frac{1}{(x-t)^2+a^2}\frac{1}{t^2+b^2}dt$$

But it is still hard to solve.

Could someone kindly provide some help? Thanks!

Answer 1

Assuming WLOG $a,b>0$, then using partial fraction decomposition and/or the residue theorem it follows that:

$$ \int_{-\infty}^{+\infty}\frac{dy}{(y^2+a^2)((y-x)^2+b^2)} = \frac{\pi(a+b)}{ab\left(x^2+(a+b)^2\right)}.\tag{1}$$

That can be achieved also by convolving two Laplace distributions, as pointed in the comments.