Integral $\int_0^\infty \frac{1}{(1+x^m)(1+x^2)}\,dx$

Question

I saw somewhere that the above integral is equal to $\pi/4$ for all real number $m$.

This seems to be surprising. Does anyone have a nice proof?

Answer 1

Let

$$\begin{align} I(m) &= \int_0^\infty \frac{dx}{(1+x^m)(1+x^2)}\tag{$x = t^{-1}$}\\ &= \int_0^\infty \frac{t^{-2}\,dt}{(1+t^{-m})(1+t^{-2})}\\ &= \int_0^\infty \frac{dt}{(1+t^{-m})(1+t^2)}\\ &= I(-m). \end{align}$$

But

$$\begin{align} I(m) + I(-m) &= \int_0^\infty \left(\frac{1}{1+x^m} + \frac{1}{1+x^{-m}}\right)\frac{dx}{1+x^2}\\ &= \int_0^\infty \left(\frac{1}{1+x^m} + \frac{x^m}{1+x^{m}}\right)\frac{dx}{1+x^2}\\ &= \int_0^\infty \frac{dx}{1+x^2}\\ &= \frac{\pi}{2}. \end{align}$$