Evaluation of $\int^{\infty}_{0} \frac1{1+x^n}dx$ with the use of Residue theorem

Question

Could anyone advise me on how to show$\begin{align} \int^{\infty}_{0}\end{align} \dfrac{1}{1+x^n}dx=\dfrac{\pi}{n\text{sin}\dfrac{\pi}{2}} ,\ $ for all integers $n \geq 2 \ ?$ Thank you.

Here is my attempt: Let $f:\mathbb{C} \to \mathbb{C}$ be defined by $f(z)=\dfrac{1}{1+z^n}.$

$z_k=e^{\dfrac{(2k+1)\pi i}{n}}, k=0,...,n-1,$ are all the roots of $z^n+1$ and they are simple poles of $f.$

For $R>1,$ let $\gamma_R(t)=Re^{it}, t\in[0,\frac{\pi}{2}].$ Then,$\ \gamma_{R} +[0,R]$ is a positively oriented closed contour whose interior contains $z_k,$ where $k \leq \dfrac{\frac{n}{2}-1}{2}=w.$

$\text{Res}_{z=z_k}f(z)=\text{lim}_{\ z \to z_k}(z-z_k)f(z)=\dfrac{1}{nz_k^{n-1}}.$

By Cauchy Residue theorem, $\begin{align}2\pi i\sum_{k \leq w}\text{Res}_{z=z_k}f(z)=\int^{R}_{0} \dfrac{1}{x^n+1}dx+ \int_{\gamma_{R}}f(z)dz\end{align}.$

But how do I evaluate $\begin{align}\sum_{k \leq w}\text{Res}_{z=z_k}f(z) \ ? \end{align} $

Since $\sum \frac{1}{n z_k^{n-1}}$ is a geometric sequence, you may simply use the summation formula for geometric sequence. — Golbez, Aug 27 '14 at 12:05
Your formula is wrong, surely? $\sin \dfrac{\pi}{2}$ is a well-known constant! Do you mean $\dfrac{\pi}{n\sin\dfrac{\pi}{n}}$? — TonyK, Aug 27 '14 at 12:29
@AlexyVincenzo Unless $\exp(2\pi i/n)=1$, you can use the summation formula, since the proof is easy.$1+z+z^2+\ldots+z^n=f(z)$, then $zf(z)-f(z)=z^n-1$, the formula follows as Yssub pointed out. — Golbez, Aug 27 '14 at 12:43
You can also evaluate it without using the residue theorem, by letting $t=\dfrac1{1+x^n}$, and recognizing the expression of the beta function in the new integral. Then, by applying Euler's reflection formula for the $\Gamma$ function, the desired result is established. — Lucian, Aug 27 '14 at 13:21
@Golbez: Hence I will obtain: $z_0\left(\dfrac{1-\left(\exp^{\dfrac{2 \pi i}{n}}\right)^{m+1}}{1-\exp^{\dfrac{2 \pi i}{n}}}\right),$ where $m= \left|{k\in \mathbb{N}: k \leq \dfrac{\frac{n}{2}-1}{n}}\right|.$ But how do I evaluate $m\ ?$ — Alexy Vincenzo, Aug 27 '14 at 13:31
You may consider two cases, one for $n$ to be even and one for $n$ to be odd. — Golbez, Aug 27 '14 at 13:43
Here is what you need. Maybe this post should be marked as duplicate? — Golbez, Aug 27 '14 at 13:47

score 1 · Answer 1 · answered Aug 27 '14 at 12:08

1

Hint: write $$\text{Res}_{z=z_k}f(z)=\dfrac{1}{nz_k^{n-1}}=\dfrac{z_k}{nz_k^{n}}=\dfrac{z_k}{n}$$ and use the sum $$ \sum_{k=0}^n\mu^k =\frac{1-\mu^{n+1}}{1-\mu}, \quad \mu\neq 1 $$ with $\mu^k=z_k=...$.

Then $$\sum_{k=0}^n\text{Res}_{z=z_k}f(z)= \dfrac{1}{n}\sum_{k=0}^n\mu^k=...$$

answered Aug 27 '14 at 12:08

Yssub

61

Thank you. Could you elaborate your last step? – Alexy Vincenzo Aug 27 '14 at 12:24
Shouldn't you sum from $k=0 \ $ to $\left \lfloor \dfrac{\frac{n}{2}-1}{2} \right \rfloor \ ?$ – Alexy Vincenzo Aug 27 '14 at 12:26
@AlexyVincenzo Better distinguish between the two cases. – Yssub Aug 27 '14 at 12:32
Noted. But we still need to derive an expression for the number of $z_k$ with $k \leq \dfrac{\frac{n}{2}-1}{2}...$ – Alexy Vincenzo Aug 27 '14 at 13:11

Evaluation of $\int^{\infty}_{0} \frac1{1+x^n}dx$ with the use of Residue theorem

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