Integrate $1/(x^5+1) $from $0$ to $\infty$? How can I calculate the integral $\displaystyle{\int_{0}^{\infty} \frac{1}{x^5+1} dx}$?
 A: Consider the contour integral of the integrand from 0 to R on the real axis, then a circle segment of radius R from R to R exp(2 pi i/5) and then along a straight line back from there to 0. The integral along this line is proportional to the one you want to evaluate. There is one pole at z =  exp(pi i/5) located inside the contour, the residue is trivial to compute. 
A: The change of variable $\frac{1}{1+x^{5}}=u$ gives $dx=-\frac{u^{-1-\frac{1}{5}}(1-u)^{-\frac{4}{5}}}{5}du$
Hence our integral reduces to $$\frac{1}{5}\int_{0}^{1}{u^{\frac{4}{5}-1}(1-u)^{\frac{1}{5}-1}du}=\frac{1}{5}\beta(1-\frac{1}{5},\frac{1}{5})=\frac{1}{5}\Gamma(1-\frac{1}{5})\Gamma(\frac{1}{5})=\frac{1}{5}\frac{\pi}{\sin(\frac{\pi}{5})}$$ we can actually simplify things a bit more by considering the following identity, deduced by De'Moivres formula $$\sin(5x)=16\sin^{5}(x)-20\sin^{3}(x)+5\sin(x)$$ pluggin in $x=\frac{\pi}{5}$ combined with the fact that $\sin(\frac{\pi}{5})>0$ We get $$16\sin^{4}(\frac{\pi}{5})-20\sin^{2}(\frac{\pi}{5})+5=0$$ which we can begin by solving for $\sin^{2}(\frac{\pi}{5})$ and then taking the positive root. We will end up with $\sin(\frac{\pi}{5})=\sqrt{5-\sqrt{5}}$ hence your integral will finally evaluate at $$\frac{1}{5}\frac{\pi}{\sqrt{5-\sqrt{5}}}$$
A: In this answer, it is shown, using contour integration, that
$$
\frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right)=\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x
$$
Plug in $n=0$ and $m=5$ to get
$$
\int_0^\infty\frac1{1+x^5}\,\mathrm{d}x=\frac\pi5\csc\left(\frac\pi5\right)
$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
\overbrace{\int_{0}^{\infty}{\dd x \over x^{5} + 1}}
^{\ds{\mbox{Set}\ x^{5} \equiv t}\ \imp\ x = t^{1/5}}&=
\color{#00f}{{1 \over 5}\int_{0}^{\infty}{t^{-4/5} \over t + 1}\,\dd t}
={1 \over 5}\,2\pi\ic\expo{-4\pi\ic/5}
-{1 \over 5}\int_{\infty}^{0}{t^{-4/5}\expo{-8\pi\ic/5} \over t^{5} + 1}\,\dd t
\\[3mm]&={1 \over 5}\,2\pi\ic\expo{-4\pi\ic/5}
+ \expo{-8\pi\ic/5}\,
\color{#00f}{{1 \over 5}\int_{0}^{\infty}{t^{-4/5} \over t^{5} + 1}\,\dd t}
\end{align}

\begin{align}
\int_{0}^{\infty}{\dd x \over x^{5} + 1}&
=\color{#00f}{{1 \over 5}\int_{0}^{\infty}{t^{-4/5} \over t^{5} + 1}\,\dd t}
={1 \over 5}\,2\pi\ic\expo{-4\pi\ic/5}\,{1 \over 1 - \expo{-8\pi\ic/5}}
={\pi \over 5}\,{2\ic \over \expo{4\pi\ic/5} - \expo{-4\pi\ic/5}}
\\[3mm]&={\pi \over 5}\,{2\ic \over 2\ic\sin\pars{4\pi/5}}
\end{align}

$$
\color{#66f}{\large\int_{0}^{\infty}{\dd x \over x^{5} + 1}}
={\pi \over 5}\,\csc\pars{4\pi \over 5}
=\color{#66f}{\large{\root{50 + 10\root{5}} \over 25}\,\pi}\approx {\tt 1.0690}
$$
