<s>[INST] How do I show $\int_{-\infty}^\infty \frac 1{(a^2+s^2)(b^2+s^2)} ds=\frac {\pi}{ab(a+b)}$ using the solution to the following Fourier transform? For a function  $f_a(x)=e^{-a|x|}$ , where $a>0$ I have found that the fourier transform of it is as follows, i know this is correct.
$\def\F{\mathcal F}$
\begin{align*}
  \F(f_a)(s) &=  \sqrt{\frac 2\pi} \frac a{a^2 + s^2}
\end{align*}
How do I use this to show
$\begin{align*}
\int_{-\infty}^\infty \frac 1{(a^2+s^2)(b^2+s^2)} ds=\frac {\pi}{ab(a+b)}
\end{align*}$
My attempts have been useless, am I supposed to use Parceval's relation or maybe the inversion formula for fourier transforms? I really have no idea.
 [/INST] We don't even need Fourier transforms. Since:
$$ \int_{-\infty}^{+\infty}\frac{ds}{s^2+a^2}=\frac{\pi}{a}\tag{1}$$
and:
$$ \frac{1}{(s^2+a^2)(s^2+b^2)} = \frac{1}{b^2-a^2}\left(\frac{1}{s^2+a^2}-\frac{1}{s^2+b^2}\right)\tag{2}$$
it follows that:
$$ \int_{-\infty}^{+\infty}\frac{ds}{(s^2+a^2)(s^2+b^2)}=\frac{1}{b^2-a^2}\left(\frac{\pi}{a}-\frac{\pi}{b}\right)=\frac{\pi}{ab(a+b)}.\tag{3}$$
</s>