Convolution of a Function I am calculating $f*f$, where $f(x) = \frac{1}{a^2 + x^2}$. So I need to calculate
$$ (f*f)(x) = \int^\infty_{-\infty} \frac{1}{(a^2+y^2)(a^2+(x-y)^2)} dy$$
I guess the answer is $\frac{2\pi}{a(4a^2+x^2)}$. Also, we can assume $a > 0$. I have tried really hard on this problem, and I am at a complete loss. I appreciate your help tremendously!
 A: One way to solve this is using partial fraction to write the integrand as

$$ {\frac {x+2\,y}{ \left( {a}^{2}+{y}^{2} \right) x \left( {x}^{2}+4\,{a
}^{2} \right) }}+{\frac {3\,x-2\,y}{ \left( {a}^{2}+(x-y)^2 \right) x \left( {x}^{2}+4\,{a}^{2} \right) }}$$

$$ = {\frac {x}{ \left( {a}^{2}+{y}^{2} \right) x \left( {x}^{2}+4\,{a
}^{2} \right) }} +{\frac {3\,x}{ \left( {a}^{2}+(x-y)^2 \right) x \left( {x}^{2}+4\,{a}^{2} \right) }}$$ 
$$+{\frac {2\,y}{ \left( {a}^{2}+{y}^{2} \right) x \left( {x}^{2}+4\,{a
}^{2} \right) }}-{\frac {2\,y}{ \left( {a}^{2}+(x-y)^2 \right) x \left( {x}^{2}+4\,{a}^{2} \right) }} $$
Now, you can use the linearity of the integral to evaluate the above. Note that for the second and fourth terms, make the change of variable $x-y=u$ and some integrals are going to cancel without the need of evaluating them and the rest is just straightforward calculations..
A: Things can get easier a lot if you can use Fourier transform. Note that:
$\mathcal{F}(\frac{1}{x^2+a^2})=\frac{1}{a}\sqrt{\frac{\pi}{2}}e^{-a|\omega|}$
$\mathcal{F}(\frac{1}{x^2+a^2}*\frac{1}{x^2+a^2})=(\frac{1}{a}\sqrt{\frac{\pi}{2}}e^{-a|\omega|})(\frac{1}{a}\sqrt{\frac{\pi}{2}}e^{-a|\omega|})=\frac{1}{a^2}\frac{\pi}{2}e^{-2a|\omega|}$
$\frac{1}{x^2+a^2}*\frac{1}{x^2+a^2} = \mathcal{F^{-1}}(\frac{1}{a^2}\frac{\pi}{2}e^{-2a|\omega|})=\frac{\sqrt{2\pi}}{a(4a^2+x^2)}$
Even if you cannot use Fourier transform in your solution may be looking at the proofs related to it help.
