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When I first encounter the integral $$\int_{0}^{\infty} \frac{d x}{x^{n}+1},$$ I am trying to resolve the integrand into partial fractions. Then I found it is very tedious and complicated and look for infinite series. I first split the integral into 2 parts. $$ \begin{aligned} \int_{0}^{\infty} \frac{d x}{x^{n}+1} =& \underbrace{\int_{0}^{1} \frac{d x}{x^{n}+1}}_{J} +\underbrace{\int_{1}^{\infty}\frac{d x}{x^{n}+1}}_{K} \end{aligned} $$

$$ \begin{aligned} J &=\int_{0}^{1} \sum_{k=0}^{\infty}(-1)^{k} x^{n k} d x =\sum_{k=0}^{\infty}\left[\frac{(-1)^{k} x^{n k+1}}{n k+1}\right]_{0}^{1} =\sum_{k=0}^{\infty} \frac{(-1)^{k}}{n k+1}=\frac{1}{n} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k+\frac{1}{n}} \end{aligned} $$

Similarly, $$ \begin{aligned} K&=\int_{1}^{\infty} \frac{1}{x^{n}+1} d x =\int_{1}^{\infty} \frac{x^{-n}}{1+x^{-n}} d x =\int_{1}^{\infty} x^{-n} \sum_{k=0}^{\infty}(-1)^{k}x^{-k n} d x=\sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{-(k+1) n+1} \end{aligned}$$

Rearranging and re-indexing yields $$K=\frac{1}{n} \sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{-(k+1)+\frac{1}{n}}=\frac{1}{n} \sum_{k=-1}^{-\infty} \frac{(-1)^{k}}{k+\frac{1}{n}} $$

Grouping $J$ and $K$ yields $$\int_{0}^{\infty} \frac{d x}{x^{n}+1} =\frac{1}{n}\left(\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k+\frac{1}{n}}+\sum_{k=-1}^{-\infty} \frac{(-1)^{k}}{k+\frac{1}{n}}\right) $$

Using the theorem

$$ \pi \csc (\pi \alpha)=\lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{(-1)^{k}}{k+\alpha} $$ and putting $\alpha=\frac{1}{n}$ yields

$$\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).$$

My question: Is there any elementary proof?

Lai
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  • There are pretty and swift proofs using contour integration or the beta function. Neither is exactly elementary, depending on your sense of elementary, but certainly I’d argue more straightforward than these. This integral has been discussed on the site before – FShrike Jan 31 '22 at 07:29
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    See here for a list of solutions. – FShrike Jan 31 '22 at 07:31
  • Or here https://math.stackexchange.com/questions/48740/int-0-infty-fracdx1xn – LL 3.14 Jan 31 '22 at 08:26
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    Maybe combine the answer by xpaul (similar to yours) here https://math.stackexchange.com/a/876940/399263 and this one by clathratus https://math.stackexchange.com/a/3302113/399263. It is not simple stuff, but it doesn't use complex analysis nor beta/gamma functions if this is what you meant by more elementary. – zwim Jan 31 '22 at 10:34
  • Thank you for your help. I had learnt something from them. – Lai Jan 31 '22 at 11:07

1 Answers1

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Below is an elementary proof via partial fractions, avoiding complex analysis, infinite series, or advanced functions routinely employed for this integral

\begin{align} & \int_{0}^{\infty} \frac{1}{1+x^{n}} dx\\ =&\int_{0}^{1} \frac{1}{1+x^{n}}dx+ \int_{1}^{\infty} \frac{1}{1+x^{n}}\overset{x\to 1/x}{dx } = \int_{0}^{1} \frac{1+x^{n-2}}{1+x^{n}}dx \\ =& \int_0^1 \sum_{k=0}^{n-1} \frac{\frac2n \sin^2 \frac{(1+2k)\pi}n}{x^2-2x \cos \frac{(1+2k)\pi}n +1 }\ dx =-\sum_{k=0}^{n-1} \frac{2\pi k}{n^2} \sin \frac{(1+2k)\pi}n\\ =& \ \frac d{da}\bigg(\sum_{k=0}^{n-1} \frac1n \cos \frac{(1+2k)\pi a}n \bigg)_{a=1} =\frac d{da}\bigg(\frac{\sin 2\pi a}{2n\sin\frac{\pi a}n} \bigg)_{a=1} =\frac{\pi}{n} \csc \frac{\pi}{n} \end{align}

Quanto
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