Integrating $\int_0^\infty\frac{1}{1+x^6}dx$ $$I=\int_0^\infty\frac{1}{1+x^6}dx$$
How do I evaluate this?
 A: Define
$$f(z)=\frac1{z^6+1}\;,\;\;C_R=[-R,R]\cup\gamma_R\;,\;\;\gamma_R:=\{z\in\Bbb C\;;\;z=Re^{it}\,,\,0<t<\pi\}$$
Since
$$-1=e^{\pi i+2k\pi}=e^{\pi i(1+2k)}\implies z_k=e^{\frac{\pi i}6(1+2k)}\;,\;\;k=0,1,...,5$$
So the poles of $\;f\;$ within the domain enclosed by $\;C_R\;$ are $\;z_0,z_1,z_2\;$ , and
$$\text{Res}_{z=z_n}(f)=\lim_{z\to z_n}(z-z_n)f(z)\stackrel{\text{l'H}}=\frac1{6z_n^5}$$
And thus CRT gives
$$\oint_{C_R}f(z)dz=2\pi i\frac16\left(e^{-5\pi i/6}+e^{-5\pi i/ 2}+e^{-\pi i/6}\right)$$
And since
$$\left|\;\int_{\gamma_R}f(z)dz\;\right|\le \pi R\frac1{R^6-1}\xrightarrow[R\to\infty]{}0$$
We finally get (the function is an even one):
$$\int_0^\infty \frac{dx}{x^6+1}=\frac12\text{Re}\left(\lim_{R\to\infty}\oint_{C_R}f(z)dz\right)=$$
$$=\frac12\text{Re}\left(\frac{\pi i}3\left(\frac{\sqrt3}2-\frac i2+0-i+\frac{\sqrt3}2-\frac i2\right)\right)=\frac\pi3$$
A: Let $t=\dfrac1{1+x^6}$ , then recognize the expression of the Beta function in the new integral, and apply the reflection formula for the $\Gamma$ function. In general, all integrals of the form $\displaystyle\int_0^\infty\frac{x^a}{(1+x^b)^c}dx$ can be evaluated in this same way.
A: Here is a somewhat unusual way to evaluate this integral. By splitting up the integral over $[0,1]$ and $[1,\infty]$ and making the change of variables $x \mapsto 1/x$ in the second integral, we find that your integral equals
$$\int_0^1 \left(\frac{1}{1+x^6} + \frac{x^4}{1+x^6}\right) dx = \sum_{n=1}^\infty \left(\frac{(-1)^n}{6n+1} + \frac{(-1)^n}{6n+5}\right) = L(\chi, 1)$$
where $\chi$ is the odd Dirichlet character of conductor $12$ defined by $\chi(1)=\chi(5)=1$ and $\chi(-1)=\chi(-5)=-1$. This character is not primitive, being induced by the primitive quadratic Dirichlet character $\eta$ of conductor $4$. The formula for the special value at $s=1$ of the $L$-function of an odd primitive Dirichlet character $\eta$ of conductor $f$ is
$$L(1, \eta) = -\frac{\tau(\eta)}{f}\frac{\pi}{if} \sum_{a=1}^f \overline{\eta}(a) a$$
where $\tau$ is the Gauss sum. In this case, the Gauss sum evaluates to $e^{2\pi i/4} - e^{6 \pi i /4} = 2i$ and the sum over $a$ evaluates to $1-3=-2$, so we get
$$L(1, \eta) = 1 - \frac{1}{3} + \frac{1}{5} - \dots = \frac{\pi}{4},$$
a formula which is anyways well-known. The $L$-function of the imprimitive character $\chi$ is related to the $L$-function of $\eta$ by removing the Euler factor at $3$,
$$L(\chi, s) = (1-\eta(3)3^{-s})L(\eta, s) = (1+3^{-s})L(\eta, s).$$
Evaluating both sides at $s=1$ we have
$$L(\chi, 1) = \frac{4}{3} \frac{\pi}{4} = \frac{\pi}{3}.$$
A: The easiest way, if you know the residue theorem, is to consider the following integral:
$$\oint_C \frac{dz}{1+z^6} $$
where $C$ is a wedge of radius $R$ of angle $\pi/3$ in the upper half plane in the complex plane.  The integral over the circular arc vanishes as $R\to\infty$, and we have that
$$\left ( 1-e^{i \frac{\pi}{3}} \right ) \int_0^{\infty} \frac{dx}{1+x^6} = i 2 \pi \frac1{6 e^{i 5 \pi/6}} = \frac{\pi}{3} e^{-i \pi/3}$$
by the residue theorem.  (Pole at $z=e^{i \pi/6}$.)  Therefore,
$$\int_0^{\infty} \frac{dx}{1+x^6} = \frac{\pi}{3}$$
A: Here's a rabbit-from-a-hat approach that avoids doing anything messy or complex:
Note that $1+x^6=(1+x^2)(1-x^2+x^4)$ implies
$${1\over1+x^6}={1\over2}\left({1-x^2+x^4\over1+x^6}+{x^2\over1+x^6}+{1-x^4\over1+x^6}\right)={1\over2}\left({1\over1+x^2}+{x^2\over1+x^6}+{1-x^4\over1+x^6}\right)$$
Now we know that 
$$\int_0^\infty{dx\over1+x^2}={\pi\over2}$$  
The change of variable $u=x^3$, so that $du=3x^2dx$, gives
$$\int_0^\infty{x^2dx\over1+x^6}={1\over3}\int_0^\infty{du\over1+u^2}={\pi\over6}$$
Finally, letting $x=1/u$, so that $dx=-du/u^2$, shows
$$I=\int_0^\infty{1-x^4\over1+x^6}dx=-\int_\infty^0{1-(1/u^4)\over1+(1/u^6)}{du\over u^2}=\int_0^\infty{u^4-1\over u^6+1}du=-I$$
which implies $I=0$.  Putting everything together, we have
$$\int_0^\infty{dx\over1+x^6}={1\over2}\left({\pi\over2}+{\pi\over6}+0\right)={\pi\over3}$$
I'd rather not say how long it took me to find this "simple" solution!
A: Without complex analysis nor special functions: 
$$x^6+1=(x^2+1)(x^4-x^2+1)=(x^2+1)(x^2+\sqrt3x +1)(x^2-\sqrt3x+1).$$
($x=\pm i$ are obviously roots and $(x^4-x^2+1)$ is biquadratic)
And now you have the following partial fraction decomposition:
$${1\over x^6+1} =
{Ax+B\over x^2+1} + {Cx+D\over x^2+\sqrt3x +1} + {Ex+F\over x^2 -\sqrt3x + 1}.$$
$\cdots$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\dd x \over 1 + x^{6}}} =
{1 \over 6}\int_{0}^{\infty}{x^{1/6 - 1} \over 1 + x}\,\dd x =
{1 \over 6}\,\
\overbrace{\Gamma\pars{1 \over 6}\Gamma\pars{1 - {1 \over 6}}}
^{\ds{\substack{\ds{Ramanujan's} \\[0.5mm] \ds{Master}\\[0.75mm] \ds{Theorem}}}}
\\[5mm] = &\
{1 \over 6}\,{\pi \over \sin\pars{\pi/6}} =
{1 \over 6}\,\pi\ \underbrace{\csc\pars{\pi \over 6}}_{\ds{2}}\
=\ \bbx{\pi \over 3} \approx 1.0472 \\ &
\end{align}
