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

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