<s>[INST] Showing that $\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}$ via Fourier Transform If $a,b>0$, how can I prove this using Fourier Series
$$\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}.$$
I tried to split the product and calculate the integral using Parceval's Theorem, but $\frac{x}{(x^2+a^2)}$ and $\frac{x^2}{(x^2+a^2)}$ aren't in $L^1(\mathbb{R})$.
Any hints hold be appreciated.
 [/INST] I know what is following is not what is asked but it's a simpler way.
$a\neq b$
\begin{align}J&=\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx\\
&=2\int_{0}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx\\
&=\frac{2b^2}{b^2-a^2}\int_0^\infty \frac{1}{x^2+b^2}\,dx-\frac{2a^2}{b^2-a^2}\int_0^\infty \frac{1}{x^2+a^2}\,dx\\
&=\frac{2b^2}{b(b^2-a^2)}\left[\arctan\left(\frac{x}{b}\right)\right]_0^\infty -\frac{2a^2}{a(b^2-a^2)}\left[\arctan\left(\frac{x}{a}\right)\right]_0^\infty\\
&=\frac{\pi b}{b^2-a^2}-\frac{\pi a}{b^2-a^2}\\
&=\boxed{\frac{\pi }{a+b}}
\end{align}
By continuity, the formula is also true for $a=b$.
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