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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 Vargas Aug 27 '14 at 14:54

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  • $\begingroup$ Since $\sum \frac{1}{n z_k^{n-1}}$ is a geometric sequence, you may simply use the summation formula for geometric sequence. $\endgroup$ – Golbez Aug 27 '14 at 12:05
  • $\begingroup$ How is that a geometric sequence? $\endgroup$ – Alexy Vincenzo Aug 27 '14 at 12:08
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    $\begingroup$ Since $z_k$ is $\exp(2\pi i/n)$ times of $z_{k-1}$. $\endgroup$ – Golbez Aug 27 '14 at 12:10
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    $\begingroup$ @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. $\endgroup$ – Golbez Aug 27 '14 at 12:43
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    $\begingroup$ Here is what you need. Maybe this post should be marked as duplicate? $\endgroup$ – 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 $$\sum_{k=0}^n\text{Res}_{z=z_k}f(z)= \dfrac{1}{n}\sum_{k=0}^n\mu^k=...$$

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

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