Compute $\int_{0}^{\infty} \frac{dx}{(a^2+x^2)(b^2+x^2)} $ using fourier transform of $e^{-a|x|}$

Question

Compute $\int_{0}^{\infty} \frac{dx}{(a^2+x^2)(b^2+x^2)} $ using fourier transform of $e^{-a|x|}$.

I computed the fourier transform of $e^{-a|x|}$, which is $\frac{a}{\pi (a^2+w^2)}$

I'm not sure how to continue from here, I tried using this formula $f(x) = \int_{-\infty}^{\infty} \frac{dx}{(a^2+x^2)}e^{iwx}dw $. But it didn't work. Any suggestions?

I looked at this post: but I still didn't understand how parseval's theorem is used there. 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?

This wont work: Parseval's theorem

Answer 1

Actually:

$$ \mathcal{F} \left [ e^{- a |x|} \right ] (\omega) = \frac{ 2a}{a^2 + \omega^2} $$

As for the integral, you could use one of Plancherel's formulae:

$$ \int_{-\infty}^{+\infty} f(t) \overline{g (t)} \ \mathrm dt = \frac{1}{2\pi}\int_{-\infty}^{+\infty} \widehat{f} (\omega) \overline{ \widehat{g} (\omega)} \ \mathrm d \omega $$

In this case:

$$ \begin{aligned} \int_{0}^{+\infty} \frac{\mathrm dx}{ \left ( a^2 + x^2 \right ) \left ( b^2 + x^2 \right )} &= \frac{1}{2} \int_{-\infty}^{+\infty} \frac{\mathrm dx}{ \left ( a^2 + x^2 \right ) \left ( b^2 + x^2 \right )} \\ &= \frac{\pi}{4ab} \int_{-\infty}^{+\infty} e^{- (a + b) |\omega |} \ \mathrm d \omega \\ &= \frac{\pi}{4ab} \left [ \int_{-\infty}^{0} e^{ (a+b) \omega} \ \mathrm d\omega + \int_{0}^{+\infty} e^{-(a+b) \omega} \ \mathrm d\omega \right ] \\ &= \frac{\pi}{4ab} \left [ \frac{1}{a + b} + \frac{1}{a+b} \right ] \\ &= \frac{ \pi}{2ab (a+b)} \end{aligned} $$