Calculate integral with help the Euler's integrals It is my first question. In advance please sorry for my bad English!
I need to calculate this integral with help the Euler's integrals:
$$
\int_0^{+\infty} \frac{1}{1+x^5} 
$$
I have tried decompose integrand in Taylor Series but I did not get anything. Also I tried use partial fractions and I got crazy expression.
I'm here to get an elegant solution of this question.
Thank you for help in advance.
 A: Let's directly calculate this:
$$\int_0^{+\infty} \frac{1}{1+x^n}\ \text{d}x$$
At the end you will plug $n = 5$ or whatever. Notice that for $n = 1$ the integral diverges.

Method 1:
Let $x=\sqrt[n]{\tan^{2}\theta }$, then
\begin{align*}
\int_{0}^{\infty }\frac{1}{1+x^{n}}\, \mathrm{d}x&=\frac{2}{n}\int_{0}^{\pi /2}\cos^{1-2/n}\theta \sin^{2/n-1}\theta \, \mathrm{d}\theta \\
&=\frac{1}{n}\mathrm{B}\left ( 1-\frac{1}{n},\frac{1}{n} \right )\\
&=\frac{1}{n}\Gamma \left ( 1-\frac{1}{n} \right )\Gamma \left ( \frac{1}{n} \right )\\
&=\frac{\pi }{n}\mathrm{csc}\frac{\pi }{n}
\end{align*}
where $\mathrm{B}\left ( \cdot  \right )$ is the Beta function and $\Gamma\left ( \cdot  \right )$ is the Gamma function.
Method 2:
\begin{align*}
&{\int_0^\infty \frac{1}{x^n+1} \:{\rm{d}}x}=\int_0^\infty\int_0^\infty e^{-(x^n+1)t} \:{\rm{d}}t\:{\rm{d}}x
\\&=\int_0^\infty e^{-t}\int_0^\infty e^{-x^n t} \:{\rm{d}}t\:{\rm{d}}x
=\int_0^\infty e^{-t}\left(\int_0^\infty e^{-x^n t}\:{\rm{d}}x\right)\:{\rm{d}}t
\\&=\frac1n\int_0^\infty t^{-\frac1n}e^{-t}\left(\int_0^\infty u^{\frac1n-1} e^{-u}{\rm{d}}u\right)\:{\rm{d}}t
=\frac1n \Gamma\left(1-\frac1n\right)\Gamma\left(\frac1n\right)
\\&=\frac{\pi }{n}\mathrm{csc}\frac{\pi }{n}
\end{align*}

More general, using the same way as Method 1 mentioned,we get
$$\int_{0}^{\infty} \frac{x^{\mu-1}}{1+x^{\nu}} \; \mathrm{d}x=\frac{\pi}{\nu} \csc \left( \frac{\pi \mu}{\nu} \right)$$
where $0< \mu < \nu $
A: I would love to add another method.
$$ \int_{0}^{+\infty}\frac{dx}{1+x^5} = \int_{0}^{1}\frac{1+x^3}{1+x^5}\,dx=\int_{0}^{1}\frac{1+x^3-x^5-x^8}{1-x^{10}}\,dx \tag{1}$$
where the first equality follows from splitting $\mathbb{R}^+$ as $(0,1)\cup[1,+\infty)$ and by enforcing the substitution $x\mapsto\frac{1}{x}$ on the second "half". Since $\frac{1}{1-x^{10}}=\sum_{n\geq 0}x^{10n}$ for any $x\in(0,1)$, the RHS of $(1)$ can be written as
$$ \sum_{n\geq 0}\left(\frac{1}{10n+1}+\frac{1}{10n+4}-\frac{1}{10n+6}-\frac{1}{10n+9}\right).\tag{2} $$
On the other hand for any $a,b>0$ with $a\neq b$ we have $\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}$. Due to the reflection formula for the $\psi$ function we also have the following useful lemma:
$$ \forall \alpha\in(0,1),\qquad \sum_{n\geq 0}\left(\frac{1}{n+\alpha}-\frac{1}{n+(1-\alpha)}\right)=\pi\cot(\pi \alpha)\tag{3}$$
which is also a consequence of Herglotz' trick. It is enough to consider $\alpha=\frac{1}{10}$ and $\alpha=\frac{2}{5}$ to crack $(2)$ and so the original integral:
$$ \int_{0}^{+\infty}\frac{dx}{1+x^5}= \frac{\pi}{10}\left(\cot\frac{\pi}{10}+\cot\frac{2\pi}{5}\right)=\frac{\pi}{5\sin\frac{\pi}{5}}.\tag{4} $$
In general, for any $\beta>1$ the same approach leads to
$$ \int_{0}^{+\infty}\frac{dx}{1+x^\beta}= \frac{\pi}{\beta\sin\frac{\pi}{\beta}}.\tag{5} $$
