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I am practicing for a preliminary exam in complex analysis. I have struggled with the following problem for a while. I was hoping someone would have suggestions for contours to integrate over. (Or maybe even a whole new approach to the problem!)

The problem

Compute $$\int_{0}^{\infty} \frac{1}{1 + x^{n}} d x,$$ where $n$ is a positive even integer.

My (limited) observations and ideas so far

1st-, since our integrand is even, the first thing to do is \begin{equation*} \int_{0}^{\infty} \frac{1}{1+x^n} d x = \frac{1}{2} \int_{-\infty}^{\infty} \frac{1}{1 + x^n} d x. \end{equation*}

Since $x^n \neq -1$ fora ll real $x$ and even $n$, we don't have to worry about intersecting poles on this part of our contour.

2nd-, I assumed I should use a reside of our function. Since $|z|^n = |-1| \implies z = e^{i \theta}$, for some $\theta \in [0, 2 \pi)$, we compute $$ z^n = e^{i \theta n} = - 1 = e^{i \pi}, $$ then by taking $n$th roots attain $\theta = \frac{\pi}{n}$. However, since $e^{2 \pi i} = 1$ we see that $e^{i \frac{\pi(2k+1)}{n}}$ is an $n$th root of $-1$ for each $k = 0, 1, 2, ..., n-1$.

My conundrum- Since $n$ is an arbitrary positive even integer, I can't imagine we want to take a "typical" loop, (like the half circle of radius $R$, and then compute $n/2$ different residues), but instead we would want some kind of loop that singles out just one pole.

Any thoughts or suggests are greatly appreciated! Thanks!

mlg4080
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1 Answers1

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Take the following contour: $0 \to R \to Re^{i2\pi/n} \to 0$. In this case you have only one root inside. enter image description here

  • Should that be $R \to R e^{ i \pi/n}$? Would that not intersect one of the poles? (Namely $e^{i \pi/n}$?) Although if I did $R \to R{e^{2\pi i/n}}$ that might work quite well. Something to try, thanks! – mlg4080 May 05 '15 at 06:03
  • You are right, I've changed. – Nikita Evseev May 05 '15 at 06:08