Evaluating $\int_0^{\infty}\frac1{x^9+1}~\mathrm{d}x$ How can we evaluate this integral?
$$\int_0^{\infty}\frac1{x^9+1}~\mathrm{d}x$$
 A: Let me extend my comment and obtain the answer for a more general integral 
$$I(\gamma)=\int_0^{\infty}\frac{dx}{1+x^{\gamma}},\qquad \gamma>1.$$
The change of variables
$$y=\frac{1}{1+x^{\gamma}},\qquad x=\left(\frac{1-y}{y}\right)^{1/\gamma},\qquad dx=-\frac{1}{\gamma y^2}\left(\frac{1-y}{y}\right)^{1/\gamma-1}dy$$
transforms it into
\begin{align}
I(\gamma)&=\frac{1}{\gamma}\int_0^1 y^{-1/\gamma}\left(1-y\right)^{1/\gamma-1}dy=\\
&=\frac1\gamma B\left(1-\frac1\gamma,\frac1\gamma\right)=\\
&=\frac1\gamma\Gamma\left(1-\frac1\gamma\right)\Gamma\left(\frac1\gamma\right)=\\
&=\frac{\pi}{\gamma \sin\frac{\pi}{\gamma}},
\end{align}
where at the first step we use the beta function, at the second its expression in terms of gamma functions, and at the third Euler's reflection formula.
A: 
How can we evaluate this integral ?

By letting $t=\dfrac1{x^9+1}$ , and recognizing the expression of the beta function in the new integral, 
then applying Euler's reflection formula for the $\Gamma$ function to that expression in order to finally 
arrive at $\displaystyle\int_0^\infty\frac{x^{n-1}}{x^m+1}dx=\frac\pi m\cdot\csc\bigg(n\cdot\frac\pi m\bigg)$, which for $n=1$ and $m=9$ becomes $\dfrac\pi{9\cdot\sin\dfrac\pi9}$
A: Using contour integration, in this answer, it is shown that
$$
\frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right)=\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x
$$
Plugging in $n=0$ and $m=9$ yields
$$
\int_0^\infty\frac{\mathrm{d}x}{1+x^9}=\frac\pi9\csc\left(\frac\pi9\right)
$$
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
&\color{#00f}{\large\int_{0}^{\infty}{\dd x \over x^{9} + 1}}
=\int_{0}^{\infty}\pars{\int_{0}^{\infty}\expo{-\pars{x^{9} + 1}\xi}\,\dd\xi}\,\dd x
\\[3mm]&=\int_{0}^{\infty}\expo{-\xi}
\pars{\overbrace{\int_{0}^{\infty}\expo{-\xi x^{9}}\,\dd x}
^{\ds{\mbox{Set}\ t \equiv \xi x^{9}\ \imp\ x = \xi^{-1/9}t^{1/9}}}}\,\dd\xi
\\[3mm]&=\int_{0}^{\infty}\expo{-\xi}\pars{\int_{0}^{\infty}\expo{-t}\xi^{-1/9}\,
{1 \over 9}\,t^{-8/9}\,\dd t}\,\dd\xi
\\[3mm]&={1 \over 9}\pars{\int_{0}^{\infty}\xi^{-1/9}\expo{-\xi}\,\dd\xi}
\pars{\int_{0}^{\infty}t^{-8/9}\expo{-t}\,\dd t}
={1 \over 9}\,\Gamma\pars{8 \over 9}\Gamma\pars{1 \over 9}
\end{align}
where $\ds{\Gamma\pars{z}}$ is the Gamma Function.

By using the
  Euler Reflection Formula
  ${\bf\mbox{6.1.17}}$:
  \begin{align}
&\color{#00f}{\large\int_{0}^{\infty}{\dd x \over x^{9} + 1}}
=\color{#00f}{\large{1 \over 9}\,\pi\,\csc\pars{\pi \over 9}}
\end{align}

