How Do You Solve This Integral for Positive Integer n?

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Here is this week's POTW:

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If $n$ is a positive integer, evaluate

$$\int_{0}^\infty \frac{dx}{1 + x^n}$$

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This week’s problem was solved correctly by Opalg. Both castor28 and Ackbach receive honorable mention for handling all cases except for the $n = 2$ case correctly. Here is Opalg’s solution.

Spoiler

Let \(\displaystyle I_n = \int_0^\infty \frac{dx}{1+x^n}.\)

If $n=1$ then \(\displaystyle I_1 = \lim_{r\to\infty}\ln(1+x)\bigr|_0^r\), which diverges.

If $n=2$ then \(\displaystyle I_2 = \lim_{r\to\infty}\arctan x\bigr|_0^r = \frac\pi2.\)

For $n\geqslant3$, integrate the function \(\displaystyle \frac1{1+z^n}\) round a 'pizza slice' contour going from $0$ to $r$ on the real axis, then along an arc of the circle $|z|=r$ to the point $re^{2\pi i/n}$, and back along the line $\arg z = 2\pi i/n$ to $0$. Then let $r\to\infty.$

The integral along the axis is then $I_n$. The integral round the arc goes to $0$.

For the remaining segment of the contour, the substitution $z= xe^{2\pi i/n}$ converts it to \(\displaystyle -\int_0^\infty\frac{e^{2\pi i/n}}{1+x^n}dx = -e^{2\pi i/n}I_n.\)

The only singularity of \(\displaystyle \frac1{1+z^n}\) inside the contour is a simple pole at $e^{\pi i/n}$, where the residue is \(\displaystyle \frac1{n(e^{\pi i/n})^{n-1}} = \frac{-1}{ne^{-\pi i/n}}.\)

It follows from Cauchy's theorem that \(\displaystyle (1 - e^{2\pi i/n})I_n = \frac{-2\pi i}{ne^{-\pi i/n}}\), and therefore $$I_n = \frac{2\pi i}{n(e^{\pi i/n} - e^{-\pi i/n})} = \frac{2\pi i}{2ni\sin(\pi/n)} = \boxed{\frac{\pi }{n\sin(\pi/n)}}.$$

As a check, this gives the correct value when $n=2$. (When $n=1$ the denominator is zero, which sort of agrees with the fact that $I_1$ diverges.)

As a further check, it also gives the expected answer as $n\to\infty$. To see that, notice that when $n$ is very large the function \(\displaystyle \frac1{1+x^n}\) is close to $1$ when $0<x<1$, and close to $0$ when $x>1$. So the area under its graph should be close to $1$. But \(\displaystyle \lim_{n\to\infty}\frac{\pi }{n\sin(\pi/n)} = \lim_{n\to\infty}\frac{\pi }{n(\pi/n)} = 1\), as expected.