Integral $\int_{-\infty}^{\infty} \frac{1}{a^2 x^2 + 1} \frac{1}{x^2 + 1} \mathrm{d}x,\quad a \in \mathbb{R}$

Question

I struggling to find the trick one should use to compute this integral:

$$\int_{-\infty}^{\infty} \frac{1}{a^2 x^2 + 1} \frac{1}{x^2 + 1} \mathrm{d}x,\quad a \in \mathbb{R}$$

any idea?

I know that the result is $$\frac{\pi}{1 + |a|}$$

Thank you

Answer 1

Note that: $$\frac{1}{a^2x^2+1}\frac{1}{x^2+1}=\frac{1}{a^2-1}\left(\frac{a^2}{a^2x^2+1}-\frac{1}{x^2+1}\right)$$ So, you can split the original integral into two simples.

Answer 2

Maybe we can use residue.
Let:
$$f(z)=\frac1{a^2z^2+1}\frac1{z^2+1}\\ =\frac1{a^2}\frac1{z+\frac ia}\frac1{z-\frac ia}\frac1{z+i}\frac1{z-i}$$ We turn the integration into: $$\int_{-\infty}^\infty{f(z)dz}$$ Consider the integration limit above: $$\lim_{n\to\infty}(\int_{-n}^n{f(z)dz}+\int_0^\pi{f(ne^{i\theta})d\theta})\\ =2\pi i(\operatorname{Res}[f(\frac i{|a|})]+\operatorname{Res}[f(i)])\\ =2\pi i(\frac{i|a|}{2-2a^2}+\frac i{-2+2a^2})=\frac \pi{1+|a|}$$ As$$\lim_{n\to\infty}\int_0^\pi{f(ne^{i\theta})d\theta}=0$$ The integral equals to$\frac \pi{1+|a|}$.