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

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}

## marked as duplicate by Alexy Vincenzo, Jean-Claude Arbaut, drhab, user147263, Antonio VargasAug 27 '14 at 14:54

• 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
• How is that a geometric sequence? – Alexy Vincenzo Aug 27 '14 at 12:08
• Since $z_k$ is $\exp(2\pi i/n)$ times of $z_{k-1}$. – Golbez Aug 27 '14 at 12:10
• @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
• Here is what you need. Maybe this post should be marked as duplicate? – Golbez Aug 27 '14 at 13:47

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