Using the residue theorem to evaluate an integral $$I = \int_0^{\infty} \frac {dx}{x^6+1}$$
My thinking is that I can use the property of even functions to integrate across the whole domain from negative infinity to positive infinity. 
Can the residue theorem be used to evaluate this integral? 
Can you use a pie-shaped contour subtending an angle of: $$\frac {\pi}3$$? 
 A: First approach:
$$\begin{eqnarray*}\int_{\mathbb{R}^+}\frac{dx}{x^6+1}&=&\frac{1}{2}\int_{\mathbb{R}}\frac{dx}{x^6+1}=\pi i\sum_{\xi\in\left\{i,e^{\pi i/6},e^{5\pi i/6}\right\}}\operatorname*{Res}_{z=\xi}\frac{1}{z^6+1}\\&=&\pi i\sum_{\xi\in\left\{i,e^{\pi i/6},e^{5\pi i/6}\right\}}\frac{1}{6\xi^5}=-\frac{\pi i}{6}\sum_{\xi\in\left\{i,e^{\pi i/6},e^{5\pi i/6}\right\}}\xi\\&=&\frac{\pi}{6}\left(1+2\sin\frac{\pi}{6}\right)=\color{red}{\frac{\pi}{3}}.\end{eqnarray*}$$
The contour integration is performed along a semicircle contour in the upper half plane: the ML lemma clearly applies since $\left|\frac{1}{x^6+1}\right|=o\big(|x|^{-2}\big)$ as $|x|\to +\infty$.
Second approach: by splitting $\mathbb{R}^+$ as $(0,1]\cup[1,+\infty)$ and performing the substitution $x\mapsto\frac{1}{x}$ on the second "half" we get
$$\begin{eqnarray*}\int_{\mathbb{R}^+}\frac{dx}{x^6+1}&=&\int_{0}^{1}\frac{1+x^4}{1+x^6}\,dx=\int_{0}^{1}\frac{1+x^4-x^6-x^{10}}{1-x^{12}}\,dx\\&=&\sum_{n\geq 0}\left(\frac{1}{12n+1}+\frac{1}{12n+5}-\frac{1}{12n+7}-\frac{1}{12n+11}\right)\end{eqnarray*}$$
where the reflection formula for the $\psi$ (digamma) function gives
$$ \sum_{n\geq 0}\left(\frac{1}{an+b}-\frac{1}{an+(a-b)}\right)=\frac{\pi}{a}\cot\frac{\pi b}{a} $$
then
$$ \int_{\mathbb{R}^+}\frac{dx}{x^6+1} = \frac{\pi}{12}\left(\cot\frac{\pi}{12}+\cot\frac{5\pi}{12}\right)=\frac{\pi}{12}\left[(2+\sqrt{3})+(2-\sqrt{3})\right]=\color{red}{\frac{\pi}{3}}. $$
Third approach: by setting $\frac{1}{1+x^6}=u$ we have
$$ \int_{0}^{+\infty}\frac{dx}{x^6+1}=\frac{1}{6}\int_{0}^{1}u^{-1/6}(1-u)^{-5/6}\,du =\frac{\Gamma(5/6)\Gamma(1/6)}{6}=\frac{\pi}{6\sin\frac{\pi}{6}}=\color{red}{\frac{\pi}{3}}$$
by the Beta function and the reflection formula for the $\Gamma$ function.
A: The singularities of the integrating function  
$$f(x) = \frac{1}{x^6+1}$$
come by solving the equation : $x^6 + 1 = 0$.
But we're integrating over the bounds $0$ and $\infty$, which means we come upon only the roots that are in the upper plane (there are $3$ there).
We will integrate the function $f(x)$ with respect to $x$, over the partially smooth curve that is consisted of the half-moon of the upper level of $γ_R$, with $z(θ) = Re^{iθ}$, $0 \geq θ \geq π$ and the line segment $[-R,R]$, where we take $R$ to be big enough, so that the singularity points in the upper plane are inside $γ_R$.
From the Residue-Theorem we have :
$$\int_{-R}^R \frac{1}{x^6+1}dx + \int_{γ_R}\frac{1}{z^6+1}dz = 2\pi i \sum_{x=x_1, x_2, x_3} \text{ Res } f(x)$$ 
The function also satisfies Jordan's Lemma, so it is : 
$$\lim_{R\to \infty}\int_{γ_R}\frac{1}{z^6+1}=0$$
Thus, for the initial integral, we have :
$$ \int_0^\infty\frac{1}{x^6+1}dx=\frac{1}{2} \lim_{R\to \infty} \int_{-R}^R \frac{1}{x^6+1}dx= \pi i \sum_{x=x_1, x_2, x_3} \text{ Res } f(x)$$
where $x_1,x_2,x_3$ are the roots of $x^6+1=0$ in the upper-plane.
I'll leave the residue calculation to you!
A: Yes, if we call the disk $D(R)$ of radius $R>0$ centered in the origin "the pie", then we can use the Residue Theorem for either half of the pie or sixth of the pie to compute the integral. We usually use half of the pie. But ok, the OP wanted $1/6$ of it. So the contour is:

and we have to compute $J$ below:
$$
\begin{aligned}
J 
&= \int_0^\infty\frac 1{x^6+1}\; dx
=\lim_R\int_0^R\frac 1{x^6+1}\; dx\ .
\\
&\qquad\text{Then the integrals on the three subpaths are...}\\
\int_{\text{Segment }0\to R}\frac 1{z^6+1}\; dz
&\to
J\ ,\\
\left|\int_{\text{Arc }R\to Re^{2\pi i/6}}\frac 1{z^6+1}\; dz\right|
&\le
\int_0^{2\pi/6}\frac 1{R^6-1}R\; dt
\\
&\to 0\ ,
\\
\int_{\text{Segment }Re^{2pi i/6}\to 0}\frac 1{z^6+1}\; dz
&=
\int_R^0
\frac 1{(t\; e^{2\pi i/6})^6+1}\; d(t\; e^{2\pi i/6})
=
e^{2\pi i/6}\frac 1{t^6+1}\; dt
\\
&\to
-e^{2\pi i/6}\;J\ ,\\
&\qquad\text{so we finally get in the limit with $R\to\infty$}
\\
J+0-Je^{2\pi i/6}
&=
2\pi i\operatorname{Res}_{z=\exp 2\pi i/12}
\frac 1{z^6+1}\ .
\end{aligned}
$$
Using a computer now, sage:
sage: var('z');
sage: RES = ( 1/(z^6+1) ).residue(z==exp(2*pi*i/12))
sage: RES = RES.simplify_full()
sage: RES
1/36*sqrt(3)*(-I*sqrt(3) - 3)
sage: J = 2*pi*i * RES / (1-exp(2*pi*i/6))
sage: J = J.simplify_full()
sage: J
1/3*pi

OK, we have the result $\pi/3$ on this way.
Well, we can also ask for:
sage: integral( 1/(z^6+1), z, 0, Infinity )
1/3*pi

