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For $f(x)=\int_{-\infty}^{\infty}{\frac{1}{x^n+1}\, dx}$ I tried using a semi-circle $\gamma$ in the upper half plane. So the integral over the whole contour $C$ is :

$$\oint_{C}=\int_{-R}^{R}{\frac{1}{x^n+1}\, dx} + \int_{\gamma}{\frac{1}{z^n+1}\, dz}=2\pi i\sum{\mathrm{Res}(f(z))}$$

The issue is the residues :

$$\begin{align}z^n+1&=0\\re^{in\theta}&=e^{i\pi}\\n\theta&=\pi +2k\pi\\z&=e^{\frac{\pi}{n}\left(2k+1\right)}\, \, \, k=0\dots (n-1)\end{align}$$

Using the limit formula :

$$\begin{align}\lim_{z\to e^{\frac{\pi}{n}\left(2k+1\right)}}{(z-e^{\frac{i\pi}{n}\left(2k+1\right)})}{\frac{1}{z^n+1}}&=\lim_{z\to e^{\frac{i\pi}{n}\left(2k+1\right)}}\frac{1}{nz^{n-1}}\\&=\frac{1}{n(e^{\frac{i\pi}{n}\left(2k+1\right)})^{n-1}}\\&=\frac{e^{\frac{i\pi}{n}\left(2k+1\right)}}{ne^{{i\pi}\left(2k+1\right)}}\cdot\frac{e^{-i\pi\left(2k+1\right)}}{e^{-i\pi\left(2k+1\right)}}\\&=\frac{1}{n}\left(e^{i\pi(2k+1)(\frac{1}{n}-1)}\right)\end{align}$$

Putting this into the sum :

$$2\pi i\sum_{k=0}^{n-1}{\frac{1}{n}e^{i\pi(2k+1)(\frac{1}{n}-1)}}=\frac{2\pi i}{n}\left[e^{i\pi(\frac{1}{n}-1)}+e^{3i\pi(\frac{1}{n}-1)}+e^{5i\pi(\frac{1}{n}-1)}+...\right]$$

Can this be continued to arrive at an answer? Could a different contour yield better results?

AnthonyML
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    You can simplify the given series with a bit of algebra using the result $q^r + \cdots + q + 1 = (q^{r+1} - 1)/(q - 1)$. (But make sure that you're just looking at the poles in the upper half-plane.) It might be easier to use a contour made up of the lines $L_\theta = {re^{i\theta}: 0\leq r \leq R}$ for $\theta = 0, 2\pi/n$ and a circular arc ${Re^{i\theta}:, 0\leq \theta\leq 2\pi/n}$, which bounds a sector containing only one pole. – anomaly Mar 11 '23 at 13:31
  • Does this answer your question? Closed form for $ \int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$ – Captain Chicky Mar 11 '23 at 13:39
  • @CaptainChicky thanks for the link. It seems I chose a contour which isn't the easiest way, but I'm still interested if this method of using the whole upper half plane can arrive at the same answer. – AnthonyML Mar 11 '23 at 13:47
  • Hm I think you can probably condense that into a closed form partial sum of some sort and solve but I'm not sure if its feasible. Good luck xd – Captain Chicky Mar 11 '23 at 14:30
  • https://math.stackexchange.com/a/110465/155629 – Travis Willse Mar 11 '23 at 17:40

1 Answers1

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It does work out, granted with more work than with the wedge contour that only encloses a single pole. Continuing that sum :

$$\begin{align}2\pi i\sum_{k=0}^{n-1}{\frac{1}{n}e^{i\pi(2k+1)(\frac{1}{n}-1)}}&=\frac{2\pi i}{n}\left[e^{i\pi(\frac{1}{n}-1)}+e^{3i\pi(\frac{1}{n}-1)}+e^{5i\pi(\frac{1}{n}-1)}+...\right]\\&=\frac{2\pi i}{n}\left[\frac{e^{i\pi(\frac{1}{n}-1)}(1-e^{2i\pi n(\frac{1}{n}-1)})}{1-e^{2i\pi(\frac{1}{n}-1)}}\right]\\&=\frac{2\pi i}{n}{\left[\frac{e^{\frac{i\pi}{n}}}{e^{\frac{2i\pi}{n}}-1}\right]}\\&=\frac{2\pi i}{n}{\left[\frac{-2i\sin(\frac{\pi}{n})}{4\sin^2(\frac{\pi}{n})}\right]}\\&=\frac{\pi}{n}\csc\left(\frac{\pi}{n}\right)\end{align}$$

AnthonyML
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