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.
 A: Note that we can write 
$$\begin{align}
\int_{-\infty}^\infty \frac{x^2}{(x^2+a^2)(x^2+b^2)}\,dx&=\frac12\int_{-\infty}^\infty \frac{(x^2+b^2)+(x^2+a^2)}{(x^2+a^2)(x^2+b^2)}\,dx-\frac{1}2\int_{-\infty}^\infty\frac{a^2+b^2}{(x^2+a^2)(x^2+b^2)}\,dx\\\\
&=\frac12\int_{-\infty}^\infty \frac1{x^2+a^2}\,dx+\frac12\int_{-\infty}^\infty \frac1{x^2+b^2}\,dx\\\\
&-\frac{a^2+b^2}2\int_{-\infty}^\infty\frac{1}{(x^2+a^2)(x^2+b^2)}\,dx\\\\
&=\frac\pi {2a}+\frac\pi {2b} -\frac{a^2+b^2}2\int_{-\infty}^\infty\frac{1}{(x^2+a^2)(x^2+b^2)}\,dx\tag1
\end{align}$$
Now apply Parseval to the integral on the right-hand side of $(1)$ with $f(x)=\frac{1}{x^2+a^2}$ and $g(x)=\frac1{x^2+b^2}$ and $F(k)=\frac{\pi}{|a|}e^{-|a|k}$ and $G(k)=\frac\pi{|b|}e^{-|b|k}$.
A: HINT: Unless you’re dead-set on using the Fourier transform, I would try using that
$$\frac{x^2}{(x^2+a^2)(x^2+b^2)}=\frac{1}{a^2-b^2}\bigg(\frac{a^2}{x^2+a^2}-\frac{b^2}{x^2+b^2}\bigg)$$
A: If you are able to use theorems from complex analysis (although it is likely you can not) then this integral is easily solved using the residue theorem. Since such an answer might be useful to others, I'll put it here for posterity.
First, note that
$$\int_{-\infty}^{\infty}\frac{z^2}{(z^2+a^2)(z^2+b^2)}dz=\lim_{R\to\infty}\left(\int_{0}^{R}\frac{z^2}{(z^2+a^2)(z^2+b^2)}dz+\int_{-R}^{0}\frac{z^2}{(z^2+a^2)(z^2+b^2)}dz\right)$$
Define $\gamma$ to be the counter-clockwise path traveled on the upper-plain semicircle of radius $R$. That is, $\gamma=\{Re^{i\theta}:0\leq \theta\leq \pi\}$. Turned into a line integral, this is
$$\int_\gamma \frac{z^2}{(z^2+a^2)(z^2+b^2)}dz=\int_{0}^{\pi} \frac{(Re^{i\theta})^2}{((Re^{i\theta})^2+a^2)((Re^{i\theta})^2+b^2)} Rie^{i\theta}d\theta$$
However, since the numerator has $R^3$ and the denominator has $R^4$, we know that
$$\lim_{R\to\infty}\left(\int_{0}^{\pi} \frac{(Re^{i\theta})^2}{((Re^{i\theta})^2+a^2)((Re^{i\theta})^2+b^2)} Rie^{i\theta}d\theta\right)=0$$
This implies
$$\lim_{R\to\infty}\left(\int_{0}^{R}\frac{z^2}{(z^2+a^2)(z^2+b^2)}dz+\int_{-R}^{0}\frac{z^2}{(z^2+a^2)(z^2+b^2)}dz\right)$$
$$=\lim_{R\to\infty}\left(\int_{0}^{R}\frac{z^2}{(z^2+a^2)(z^2+b^2)}dz+\int_{-R}^{0}\frac{z^2}{(z^2+a^2)(z^2+b^2)}dz+\int_\gamma \frac{z^2}{(z^2+a^2)(z^2+b^2)}dz\right)$$
Putting these three path integrals together, we get a simple closed curve (call it $\beta$) which starts at $(-R,0)$, goes to $(R,0)$, and then follows $\gamma$ back to $(-R,0)$. This implies the integral is equal to
$$=\lim_{R\to\infty}\left(\int_\beta \frac{z^2}{(z^2+a^2)(z^2+b^2)}dz\right)$$
Since this is a simple, closed, positively-oriented curve, the residue theorem applies. Now, for $R>\text{max}\{a,b\}$ the function 
$$\frac{z^2}{(z^2+a^2)(z^2+b^2)}$$
has two singular points inside $\beta$ at $ia$ and $ib$. To calculate the residues at these points, we simply need
$$\text{Res}(ia)=\lim_{z\to ia} (z-ia)\frac{z^2}{(z^2+a^2)(z^2+b^2)}=\lim_{z\to ia} (z-ia)\frac{z^2}{(z-ia)(z+ia)(z^2+b^2)}$$
$$=\lim_{z\to ia}\frac{z^2}{(z+ia)(z^2+b^2)}=\frac{-a^2}{2ia(b^2-a^2)}=\frac{-a}{2i(b^2-a^2)}$$
For $ib$, we get a residue of
$$\text{Res}(ib)=\frac{-b}{2i(a^2-b^2)}$$
Then the residue theorem states
$$\int_\beta \frac{z^2}{(z^2+a^2)(z^2+b^2)}dz=2\pi i\left( \text{Res}(ia)+\text{Res}(ib)\right)$$
$$=2\pi i\left(\frac{-a}{2i(b^2-a^2)}+\frac{-b}{2i(a^2-b^2)}\right)=\frac{\pi}{a+b}$$
We conclude
$$\int_{-\infty}^{\infty}\frac{z^2}{(z^2+a^2)(z^2+b^2)}dz=\lim_{R\to\infty}\left(\int_\beta \frac{z^2}{(z^2+a^2)(z^2+b^2)}dz\right)$$
$$=\lim_{R\to\infty}2\pi i\left( \text{Res}(ia)+\text{Res}(ib)\right)=\lim_{R\to\infty}\frac{\pi}{a+b}=\frac{\pi}{a+b}$$
A: $$\begin{align}
&\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx\\
=&\int^{\infty}_{0}\frac{2dx}{x^2+\frac{a^2b^2}{x^2}+(a^2+b^2)}\\
=& \int^{\infty}_{0}\frac{d(x+\frac{ab }x)}{(x+\frac{ab}{x})^2+(a-b)^2}
+ \int^{\infty}_{0}\frac{d(x-\frac{ab }x)}{(x-\frac{ab}{x})^2+(a+b)^2}\\
=& \int^{\infty}_{\infty}\frac{dt}{t^2+(a-b)^2}
+ \int^{\infty}_{-\infty}\frac{dt}{t^2+(a+b)^2}\\
=& \>0+\frac\pi{a+b}
\end{align}$$
A: 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 $a=b$.
A: Using Plancherel's theorem for the Fourier transform $L^2 \to L^2 $ allows us to do this by recognising $x/(a^2+x^2)$ as a Fourier transform, namely via
$$ \tilde{f}(k) = \int_{-\infty}^{\infty} e^{-ikx} e^{-a\lvert x \rvert } \operatorname{sgn} x \, dx = \frac{2ik}{a^2+k^2} $$
(How do you think of this? You remember the Fourier transform of $e^{-a\lvert x \rvert }$ is $2a/(a^2+k^2)$, and suspect that an odd variant will exist, which will likely include an extra $k$; since the result isn't going to be in $L^1$, we expect our original function to have a discontinuity, which indeed it does.)
Now, for our convention, Plancherel's theorem says for $f,g \in L^2$,
$$ \int_{-\infty}^{\infty} \overline{f(x)} g(x) \, dx = \frac{1}{2\pi} \int_{-\infty}^{\infty} \overline{\tilde{f}(k)} \tilde{g}(k) \, dk . $$
Applying this to
$$ f(x) = e^{-a\lvert x \rvert } \operatorname{sgn} x , \qquad g(x) = e^{-b\lvert x \rvert } \operatorname{sgn} x $$
gives (notice the $i(-i)=1$ cancels)
$$ \begin{align}
\int_{-\infty}^{\infty} \frac{x^2}{(a^2+x^2)(b^2+x^2)} \, dx 
&= \frac{2\pi}{4} \int_{-\infty}^{\infty}  e^{-(a+b)\lvert x \rvert } (\operatorname{sgn} x)^2 \, dx \\
&= \frac{\pi}{2} \int_{-\infty}^{\infty}  e^{-(a+b)\lvert x \rvert } \, dx \\
&= \pi \int_0^{\infty} e^{-(a+b)x} \, dx = \frac{\pi}{a+b} ,
\end{align} $$
as required. (We need $a,b>0$, but this does extend to positive real parts if we are more careful about taking the conjugates.)
