How to evaluate $\int_0^\infty \frac{1}{x^n+1} dx$ Noticed that the integral $$\int_0^\infty \frac{1}{x^n+1} dx$$ is often approached with partial fraction decomposition, but this gets increasingly ugly as $n$ gets bigger.  Is there a neat trick to do these all in one fell swoop? Or a famous name for these integrals that I can look up for more info?
 A: This can  also be done  using complex variables  when $n\ge 2.$  Use a
pizza slice contour consisting of the line $\Gamma_1$ from zero to $R$
on the real axis, an arc  $\Gamma_2$ along the circle $|z|=R$ from the
real   axis   to  $\theta =  2\pi/n$   and  a  line $\Gamma_3$ back to
the  origin parameterized by $z=\exp(2\pi i/n)t.$

Set $$f(z) = \frac{1}{z^n+1}$$ We integrate $f(z)$ around this contour
and use  the fact  that the  integral is equal  to the  residues times
$2\pi i$ of the poles contained therein. Along $\Gamma_2$ we have
$$|f(z)| \in \Theta(R^{-n})$$ so that the contribution is bounded by
$$\frac{2\pi R}{n} \times R^{-n} \to 0$$
as $R\to\infty$ when $n\ge 2$ and the contribution from $\Gamma_2$ 
vanishes.

Along $\Gamma_1$ the integral goes to the value $I(n)$ being sought as
$R\to\infty.$ Along $\Gamma_3$ we get in the limit
$$\int_\infty^0 \frac{1}{(e^{2\pi i/n} t)^n + 1} e^{2\pi i/n} dt
= - e^{2\pi i/n}\int_0^\infty \frac{1}{t^n+1} dt
= - e^{2\pi i/n} I(n).$$

There is just one pole inside the pizza slice contour at $z=e^{i\pi/n}$
and we get
$$I(n) (1 - e^{2\pi i/n}) =
2\pi i \times \mathrm{Res}\left(f(z); z=e^{i\pi/n}\right).$$
The pole is simple and hence the residue is
$$\lim_{z\to e^{i\pi/n}}\frac{z-e^{i\pi/n}}{z^n+1}
= \lim_{z\to e^{i\pi/n}}\frac{1}{n z^{n-1}}
= \lim_{z\to e^{i\pi/n}}\frac{z}{n z^n}
= \frac{e^{i\pi/n}}{n e^{i\pi}}
= - \frac{e^{i\pi/n}}{n}.$$
This finally yields
$$I(n) = - 2\pi i\frac{e^{i\pi/n}}{n(1 - e^{2\pi i/n})}
= - 2\pi i\frac{1}{n(e^{-i\pi/n} - e^{\pi i/n})}
\\ = \pi\frac{2i}{n(e^{\pi i/n} -  e^{-i\pi/n})}
= \frac{\pi}{n\sin(\pi/n)}.$$
A: The given integral indeed has a closed-form:
$$\int_0^\infty\frac{1}{1+x^n}\ dx=\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}.$$
For the complete proof, you may refer to this OP: Prove $\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^n}=\frac{\pi}{n \sin\frac{\pi}{n}}$ using real analysis techniques only
A: Hint. You may recall the celebrated $\Gamma$ function defined by 
$$
\Gamma(\alpha)=\int_0^\infty u^{\alpha-1} e^{-u}\:{\rm{d}}u, \quad \alpha>0
$$
and you may write
$$
\begin{align}
\int_0^\infty \frac{1}{x^n+1} \:{\rm{d}}x&=\int_0^\infty\int_0^\infty e^{-(x^n+1)t} \:{\rm{d}}t\:{\rm{d}}x
\\&=\int_0^\infty e^{-t}\int_0^\infty e^{-x^n t} \:{\rm{d}}t\:{\rm{d}}x
\\&=\int_0^\infty e^{-t}\left(\int_0^\infty e^{-x^n t}\:{\rm{d}}x\right)\:{\rm{d}}t
\\&=\frac1n\int_0^\infty t^{-\frac1n}e^{-t}\left(\int_0^\infty u^{\frac1n-1} e^{-u}{\rm{d}}u\right)\:{\rm{d}}t
\\&=\frac1n \Gamma\left(1-\frac1n\right)\Gamma\left(\frac1n\right)
\\&=\frac{\pi}{n\sin \frac{\pi}{n}}
\end{align}
$$ 
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
&\overbrace{\color{#66f}{\large\int_{0}^{\infty}{\dd x \over x^{n} + 1}}}
^{\ds{\color{#c00000}{t \equiv {1 \over x^{n} + 1}\
\imp\ x = \pars{{1 \over t} - 1}^{1/n}}}}\ =\
\int_{1}^{0}t\,{1 \over n}\,\pars{{1 \over t} - 1}^{1/n - 1}
\pars{-\,{1 \over t^{2}}}\,\dd t
\\[5mm]&={1 \over n}\int_{0}^{1}t^{-1/n}\pars{1 - t}^{1/n - 1}\,\dd t
={1 \over n}\,{\rm B}\pars{1 - {1 \over n},{1 \over n}}
={1 \over n}\,{\Gamma\pars{1 - 1/n}\Gamma\pars{1/n} \over \Gamma\pars{1}}
\\[5mm]&=\color{#66f}{\large{1 \over n}\,{\pi \over \sin\pars{\pi/n}}}
\end{align}

The result is valid when $\ds{\Re\pars{1 - {1 \over n}} > 0}$ and
  $\ds{\Re\pars{1 \over n} > 0}$
  $\ds{\imp\ \color{#c00000}{0 < \Re\pars{1 \over n} < 1}}$.

