TITLE: Is the integral $\int_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$ equal for all $a \neq 0$?
QUESTION [42 upvotes]: Let $a$ be a non-zero real number. Is it true that the value of $$\int\limits_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$$  is independent on $a$?

REPLY [14 votes]: $\displaystyle I=\int_0^\infty \frac{dx}{(1+x^2)(1+x^a)}$
Substitution:
$\displaystyle x=\tan\theta$
$\displaystyle dx=\sec^2\theta d\theta$
$\displaystyle I=\int_0^{\pi/2}\frac{d\theta}{1+\tan^a \theta}$
$\displaystyle I=\int_0^{\pi/2}\frac{\cos^a \theta d\theta}{\sin^a \theta + \cos^a \theta}$ 
$\displaystyle  I=\int_0^{\pi/2}\frac{\cos^a(\pi/2-\theta) d\theta}{\sin^{a}(\pi/2-\theta) + \cos^a (\pi/2-\theta)}$
$\displaystyle I=\int_0^{\pi/2}\frac{\sin^a\theta d\theta}{\sin^a\theta + \cos^a \theta}$ 
Therefore,
$\displaystyle 2I=\int_0^{\pi/2}d\theta$
$\displaystyle I=\pi/4$