I need to use the Plancherel theorem to show that :
$ \int_{-\infty}^{\infty} \frac{ab}{(x^2+a^2)(x^2+b^2)} dx = \frac{\pi}{(a+b)} $ with a,b > 0
From the Plancherel theorem I know that: $ \int_{-\infty}^{\infty} |f(x)|^2 dx = \int_{-\infty}^{\infty}|\mathfrak{F}(\xi)|^2 d\xi $
So, I tried to just apply the theorem using $ f(x) = \sqrt \frac{ab}{(x^2+a^2)(x^2+b^2)} $
$ \int_{-\infty}^{\infty} |\sqrt \frac{ab}{(x^2+a^2)(x^2+b^2)}|^2 dx = \int_{-\infty}^{\infty}|\mathfrak{F}(\xi)|^2 d\xi $
But I don't know if I'm on the right way, I tried to calculate the Fourier transform of $ \sqrt \frac{ab}{(x^2+a^2)(x^2+b^2)} $ , but it's looks more complicated than it should be.
Is there another way of doing that using the Plancherel theorem?
You may consider the Fourier transform of $\frac{1}{x^2+a^2}$ and the inverse Fourier transform of $\frac{1}{x^2+b^2}$ or... just a partial fraction decomposition. Assuming $a\neq b$ we have $$ \frac{1}{(x^2+a^2)(x^2+b^2)}=\frac{1}{b^2-a^2}\left(\frac{1}{x^2+a^2}-\frac{1}{x^2+b^2}\right) $$ and by integrating both sides over $\mathbb{R}$ we get $$ \int_{-\infty}^{+\infty}\frac{dx}{(x^2+a^2)(x^2+b^2)}=\frac{1}{b^2-a^2}\left(\frac{\pi}{a}-\frac{\pi}{b}\right) = \frac{\pi}{ab(a+b)}.$$ Considering the limit as $b\to a$ we also have $\int_{\mathbb{R}}\frac{dx}{(x^2+a^2)^2}=\frac{\pi}{2a^3}$.