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?

Question

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.

Answer 1

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 $e^{-c|t|}$.)

Answer 2

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 $A = -B = \frac{1}{b^2 - a^2}$.

Answer 3

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: $$ \int_{-\infty}^{+\infty}\frac{ds}{(s^2+a^2)(s^2+b^2)}=\frac{1}{b^2-a^2}\left(\frac{\pi}{a}-\frac{\pi}{b}\right)=\frac{\pi}{ab(a+b)}.\tag{3}$$