From the Plancherel theorem I know that: $2\pi\int_{-\infty}^{\infty}f(x)g(x) dx = \int_{-\infty}^{\infty}\frak{F}(f)\frak{F}(g) d\xi $
I need to prove the equality below by using the Plancherel theorem :
$\int_{-\infty}^{\infty}\frac{ab}{(x^2+a^2)(x^2+b^2)} dx = \frac{\pi}{a+b} $ with a, b > 0
What I did :
Let $f(x) = \frac{a}{x^2 +a^2}$ and $g(x) = \frac{b}{x^2 +b^2} $
$\frak{}F(f) = \sqrt{\frac{\pi}{2}}$$e^{-a|\xi|} $
$\frak{}F^{-1}(g) = \sqrt{\frac{\pi}{2}}$$e^{-b|\xi|} $
And now, using the Pancherel theorem, can I say that the following is correct?
$2\pi\int_{-\infty}^{\infty}\frac{ab}{(x^2+a^2)(x^2+b^2)} dx = \int_{-\infty}^{\infty}\sqrt{\frac{\pi}{2}}$$e^{-a|\xi|} \sqrt{\frac{\pi}{2}}$$e^{-b|\xi|} d\xi = \frac{\pi}{2}\int_{-\infty}^{\infty}e^{(-a-b)|\xi|} d\xi $
if it is correct, then i get the right answer.
What troubles my mind is that it implies that:
$2\pi\int_{-\infty}^{\infty}\frac{ab}{(x^2+a^2)(x^2+b^2)} dx = \frac{\pi}{a+b} = \int_{-\infty}^{\infty}\frac{ab}{(x^2+a^2)(x^2+b^2)} dx$
then it gets insane and:
$2\pi \frac{\pi}{a+b} = \frac{\pi}{a+b} $
therefore: $\pi = \frac{1}{2} $
Why do i get $\frac{\pi}{a+b}$ calculating the right side integral from the Plancherel Theorem but it still looks like it isn't the right answer?