<s>[INST] 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: $$\frac {\pi}3$$? 
 [/INST] 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.
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