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.
 A: Parseval's theorem is definitely how you want to go about this. Parseval states that
$$\langle g,h\rangle = \langle\mathcal{F}g,\mathcal{F}h\rangle.$$
In your case, $g = \sqrt{\frac{\pi}{2}}\frac{1}{a}f_a$ and $h = \sqrt{\frac{\pi}{2}}\frac{1}{b}f_b$ since we know that $\mathcal{F}g = \frac{1}{a^2+s^2}$ and similarly for $h$. Hence by Parseval (and noting that all of the functions are real)
$$\int_{-\infty}^{\infty} \frac{1}{(a^2+s^2)(b^2+s^2)}\,ds = \int_{-\infty}^{\infty} \left(\sqrt{\frac{\pi}{2}}\frac{1}{a}f_a(t)\right)\left(\sqrt{\frac{\pi}{2}}\frac{1}{b}f_b(t)\right)\,dt.$$
Or equivalently,
$$\int_{-\infty}^{\infty} \frac{1}{(a^2+s^2)(b^2+s^2)}\,ds = \frac{\pi}{2ab} \int_{-\infty}^{\infty} e^{-(a+b)|t|}\,dt.$$
Can you take it from here? (Hint: use evenness of $e^{-c|t|}$.)
A: 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:
$$ \int_{-\infty}^{+\infty}\frac{ds}{(s^2+a^2)(s^2+b^2)}=\frac{1}{b^2-a^2}\left(\frac{\pi}{a}-\frac{\pi}{b}\right)=\frac{\pi}{ab(a+b)}.\tag{3}$$
A: Using partial fractions we can assume there are numbers $A$ and $B$ where
$$
\frac{1}{(a^2 + s^2)(b^2 + s^2)} \;\; =\;\; \frac{A}{a^2 + s^2} + \frac{B}{b^2 + s^2}.
$$
This is equivalent to writing 
$$
1 \;\; =\;\; A(b^2 + s^2) + B(a^2 + s^2).
$$
If you pick the values of $s = ia$ and $s = ib$ you'll find that $A = -B = \frac{1}{b^2 - a^2}$.  
