$\int_0^\infty {dx\over (x^2+a^2)(x^2+b^2)}={\pi\over2ab(a+b)}$ Show that $$\int_0^\infty {dx\over (x^2+a^2)(x^2+b^2)}={\pi\over2ab(a+b)}$$ where $a,b>0$. I'm not sure how to simplify this. Any solutions or hints are greatly appreciated.
 A: Fill in details: take the contour
$$C_R:=[-R,R]\cup\gamma_R\;,\;\;R>>0\;,\;\;\gamma_R:=\{z=Re^{it}\in\Bbb C\;:\;0\le t\le \pi\}\;$$
For $\;a\neq b\;$ :
and $\;R\;$ big enough as to be $\;a,b<R\;$ . Observe there are two simple poles of $\;f(z)=\cfrac1{(z^2+a^2)(z^2+b^2)}\;$ within the domain enclosed by the above contour:
$$\begin{align}&\text{Res}_{z=ai}(f)=\lim_{z\to ai}(z-ai)f(z)=\frac1{2ai(b^2-a^2)}\\{}\\
&\text{Res}_{z=bi}(f)=\lim_{z\to bi}(z-bi)f(z)=\frac1{2bi(a^2-b^2)}\end{align}$$
and from the Residue Theorem:
$$\oint_{C_R}f(z)dz=\frac\pi{a^2-b^2}\left(\frac1b-\frac1a\right)=\frac\pi{ab(a+b)}$$
Now show (for example, Jordan's Lemma) that
$$\lim_{R\to\infty}\int_{\gamma_R}f(z)dz=0$$
and use the fact that the real function $\;f(x)\;$ is even to obtain the result.
For $\;a=b\;$ : we now have the function $\;f(z)=\cfrac1{(z^2+a^2)^2}=\cfrac1{(z+ai)^2(z-ai)^2}\;$ . Thus, using the same contour as above and taking a little circle $\;|z-ai|=r\;,\;\;0<r<<R\;$ , we get  from Cauchy Integral Formulae
$$\oint_{C_R}f(z)dz=\oint_{|z|=r}\frac{\frac1{(z+ai)^2}}{(z-ai)^2}dz=2\pi i\left(\frac1{(z+ai)^2}\right)'_{z=ai}=$$
$$=-2\pi i\frac{2}{(2ai)^3}=\frac\pi{a^3}$$
and continue asin the first part.
A: You should take the partial fraction of $\dfrac{1}{(x^2+a^2)(x^2+b^2)}$.
This way, it is easier for you to integrate the expression and you should expect the $\arctan$ function in the numerator.
Recall:
$\dfrac{1}{(x^2+a^2)(x^2+b^2)}=\dfrac{Ax+B}{x^2+a^2}+\dfrac{Cx+D}{x^2+b^2}$.
