Calculate $\displaystyle\int_0^\infty \frac{x^2}{1+x^7} \, dx$

Question

Calculate $\displaystyle\int_0^\infty \frac{x^2}{1+x^7} \, dx$

This showed up on a complex analysis qualification exam. First, I will write

$$\displaystyle\int_0^\infty \frac{x^2}{1+x^7} \, dx = \lim_{R \to \infty} \int_0^R \frac{x^2}{1+x^7} \, dx $$

Now, if $\frac{x^2}{1+x^7}$ were even, I would write the above as half of the integral from $-R$ to $R$ and then create a semicircular arc $C_R$, use Jordan's lemma to show that it vanishes, and then get the result. But $\frac{x^2}{1+x^7}$ is not even (nor odd). What can be done here to sidestep this issue? (I'm open to the use of monotone/dominated/bounded convergence theorems as well if that's easier)

Answer 1

\begin{align} \color{#44f}{\large\int_{0}^{\infty}{x^{2} \over 1 + x^{7}}{\rm d}x} & = {1 \over 7}\int_{0}^{\infty}{x^{-4/7} \over 1 + x}{\rm d}x = {1 \over 7}\int_{0}^{\infty}{x^{3/7 - 1} \over \left(1 + x\right)^{3/7 + 4/7}}{\rm d}x \\[5mm] & = {1 \over 7}\,{\Gamma\left(3/7\right)\Gamma\left(4/7\right) \over \Gamma\left(3/7 + 4/7\right)} \\[5mm] & = \color{#44f}{\large{1 \over 7}\,\pi\csc\left(3\pi \over 7\right)} \approx 0.4603 \end{align} See this link.