Finding closed form of $\int_{-\infty}^{\infty}\frac{dx}{x^{2n}+1}$ So, I would like to find a closed form (if it exists?) for the following integral:
$$\int_{-\infty}^{\infty}\frac{dx}{x^{2n}+1}$$ for $n\in\Bbb{N}$. I tried complex analysis approach. Define function $$f(z)=\frac{1}{z^{2n}+1}$$
$f$ has simple poles at $z_k=e^{i\frac{\pi+2k\pi}{2n}}$. Define contour $\gamma$ to be a half circle in the upper part of the complex plane: $\gamma:(0,\pi)\rightarrow\Bbb{C},z=Re^{i\psi}$, where we will let $R\to\infty$ later on. By residue theorem we have:
$$\int_{-R}^{R}f(x)dx+\int_\gamma f(z)dz=2 \pi i\sum_{k=0}^{n-1}\operatorname{Res}(f,z_k)$$ Now I can factor the polynomial denominator of the function $f(z)$:
$$f(z)=\prod_{l=0}^{2n-1}\frac{1}{z-z_l}$$
So $$\sum_{k=0}^{n-1}\operatorname{Res}(f,z_k)=\sum_{k=0}^{n-1}\prod_{l=0,l\neq k}^{2n-1}\frac{1}{z-z_l}$$
But from this point, I have no idea how to proceed, not even if this is the correct approach. Any useful hints, tips, links are appreciated, thanks in advance.
 A: Note that $z^{2n}+1=0$ when $z=e^{i(2k+1)\pi/2n}$.  For $k=0$, $z=e^{i\pi/2n}$ and for $k=1$, $z=e^{i3\pi/2n}$.
Let $C_R$ be the wedge-shaped contour that is comprised of $(1)$ the line segment from $0$ to $R$, $(2)$ the circular arc from $R$ to $Re^{i\pi/n}$, and $(3)$ the line segment from $Re^{i\pi/n}$ to $0$.
From the residue theorem, we have
$$\begin{align}
\oint_{C_R} \frac{1}{1+z^{2n}}\,dz&=2\pi i \text{Res}\left(\frac{1}{1+z^{2n}}, z=e^{i\pi/2n}\right)\\\\
&=2\pi i \left(\frac{1}{2n(e^{i\pi/2n})^{2n-1}}\right)\\\\
&=-\frac{i\pi e^{i\pi/2n}}{n}\tag1
\end{align}$$
We also have 
$$\begin{align}
\oint_{C_R} \frac{1}{1+z^{2n}}\,dz&=\int_0^R \frac{1}{1+x^{2n}}\,dx\\\\
&+\int_0^{\pi/n}\frac{iRe^{i\phi}}{1+R^{2n}e^{i2n\phi}}\,d\phi\\\\
&+\int_R^0\frac{e^{i\pi/n}}{1+x^{2n}}\,dx\tag2
\end{align}$$
Letting $R\to \infty$, the second integral on the right-hand side of $(2)$ vanishes and we find 
$$\lim_{R\to \infty}\oint_{C_R} \frac{1}{1+z^{2n}}\,dz=(1-e^{i\pi/n})\int_0^\infty \frac{1}{1+x^{2n}}\,dx\tag3$$
Setting $(1)$ equal to $(3)$ yields
$$\int_0^\infty \frac{1}{1+x^{2n}}\,dx=-\frac{i\pi e^{i\pi/2n}}{n(1-e^{i\pi/n})}=\frac{\pi}{2n\sin(\pi/2n)}$$
Therefore, we find that 
$$\int_{-\infty}^\infty \frac{1}{1+x^{2n}}\,dx=\frac{\pi}{n\sin(\pi/2n)}$$
A: Call your integral $I$. If $n\in\mathbb{N}$, then
$$I=2\int_0^{+\infty}\frac{1}{x^{2n}+1}\,dx$$
Substitute $u=x^{2n}$. Also, let $2n=m$.
\begin{align}
I&=\frac{1}{n}\int_0^{+\infty} \frac{u^{1/m -1}}{u+1}\,du \\ \\
&=\frac{1}{n} \mathcal{B}\left(\frac{1}{m},1-\frac{1}{m}\right) \\ \\ 
&=\frac{1}{n} \frac{\Gamma\left(\frac{1}{m}\right)\Gamma\left(1-\frac{1}{m}\right)}{\Gamma(1)} \\ \\ 
&=\frac{\pi}{n}\csc\left(\frac{\pi}{m}\right) \\ \\ 
I&=\frac{\pi}{n}\csc\left(\frac{\pi}{2n}\right)
\end{align}
A: Use $x=\tan^{1/n}t$ to obtain a Beta function. The final result is $\dfrac{2}{\operatorname{sinc}\frac{\pi}{2n}}$.
A: $\frac {1}{z^2n + 1}$ has poles at $e^{\frac{(2k-1)pi}{2n} i}$
We only care about the poles in the upper half plane.
The residue at  $e^{\frac{\pi}{2n} i}$
We can factor the denominator:
$z^{2n} + 1 = (e^{\frac{\pi}{2n} i}-z)\left(e^{\frac{(2n-1)\pi}{2n} i}+e^{\frac{(2n-2)\pi}{2n} i}z + e^{\frac{(2n-3)\pi}{2n} i}z^2 + \cdots + z^{2n-1}\right)$
$\lim_\limits{z\to e^{\frac{\pi}{2n} i}} \frac{z - e^{\frac{\pi}{2n} i}}{z^{2n}+1} = \frac{-1} {\left(e^{\frac{(2n-1)\pi}{2n} i}+e^{\frac{(2n-1)\pi}{2n} i} + e^{\frac{(2n-1)\pi}{2n} i} + \cdots + e^{\frac{(2n-1)\pi}{2n} i}\right)} = \frac {-e^{\frac{\pi}{2n} i}}{2n}$
And the residues at the other poles is similar.
$2\pi i\sum Res f(z)  = 2\pi i\sum_\limits{k=1}^{n} \frac {-i\sin \frac {(2k-1)\pi}{2n}}{2n} = \frac {\pi}{n} \sum_\limits{k=1}^{n} \sin \frac {(2k-1)\pi}{2n}$
