<s>[INST] Solving $\int\limits_{-\infty}^\infty \frac{1}{x^8+1}dx$ through Glasser's Master Theorem Trying to find a way to solve 

$$\int_{-\infty}^\infty \frac{1}{x^8+1}dx$$ 

through Glasser's Master Theorem, more specifically the Cauchy–Schlömilch substitution. Preferably, I'm looking for the closed form solution, and I am already aware of how to attain this through contour integration. 

Solution: $$\frac{\pi}{4\sin(\frac{\pi}{8})}$$ 

Link to general closed form solution: solutions to $\int_{-\infty}^\infty \frac{1}{x^n+1}dx$ for even $n$
 [/INST] I believe the most direct way for tackling such integrals is to exploit Euler's Beta function and the reflection formula for the $\Gamma$ function: assuming $m>1$,
$$ \int_{0}^{+\infty}\frac{dx}{1+x^m}\stackrel{\frac{1}{1+x^m}\to u}{=}\frac{1}{m}\int_{0}^{1}u^{-\frac{m-1}{m}}(1-u)^{-\frac{1}{m}}\,du =\frac{\Gamma\left(\tfrac{1}{m}\right)\Gamma\left(\tfrac{m-1}{m}\right)}{m\,\Gamma(1)}=\frac{\pi/m}{\sin(\pi/m)}.$$
Another way is to exploit the reflection formula for the digamma function, in the form
$$ \sum_{n\geq 0}\left[\frac{1}{an+b}+\frac{1}{an+(a-b)}\right]=\frac{\pi}{a}\,\cot\left(\frac{\pi b}{a}\right)$$
via
$$\begin{eqnarray*} \int_{0}^{+\infty}\frac{dx}{1+x^m}&=&\int_{0}^{1}\frac{1+x^{m-2}}{1+x^m}\,dx=\int_{0}^{1}\frac{1+x^{m-2}-x^m-x^{2m-2}}{1-x^{2m}}\,dx\\
&=& \sum_{n\geq 0}\left[\color{blue}{\frac{1}{2mn+1}}+\color{red}{\frac{1}{2mn+m-1}}-\color{red}{\frac{1}{2mn+m+1}}-\color{blue}{\frac{1}{2mn+2m-1}}\right]\\&=&\frac{\pi}{2m}\left[\cot\left(\frac{\pi}{2m}\right)+\tan\left(\frac{\pi}{2m}\right)\right].\end{eqnarray*}$$
Through Glasser's master theorem, in the $m=8$ case we may state
$$\begin{eqnarray*} \int_{-\infty}^{+\infty}\frac{dx}{1+x^8}&\stackrel{x\mapsto z\cdot 2^{1/4}}{=}&\int_{-\infty}^{+\infty}\frac{dz}{(1-\sqrt{2}z^2+z^4)(1+\sqrt{2}z^2+z^4)} \\&=&2\int_{0}^{+\infty}\frac{1}{2\sqrt{2}z^2}\left[\frac{1}{1-\sqrt{2}z^2+z^4}-\frac{1}{1+\sqrt{2}z^2+z^4}\right]\,dz
\\&\stackrel{z\mapsto 1/z}{=}&2\int_{0}^{+\infty}\frac{z^4}{2\sqrt{2}}\left[\frac{1}{1-\sqrt{2}z^2+z^4}-\frac{1}{1+\sqrt{2}z^2+z^4}\right]\,dz
\\&=&\frac{1}{\sqrt{2}}\int_{0}^{+\infty}\frac{1}{z^4}\left[\frac{1}{\frac{1}{z^2}-\sqrt{2}+z^2}-\frac{1}{\frac{1}{z^2}+\sqrt{2}+z^2}\right]\,dz
\\&=&\frac{1}{\sqrt{2}}\int_{0}^{+\infty}z^2\left[\frac{1}{\frac{1}{z^2}-\sqrt{2}+z^2}-\frac{1}{\frac{1}{z^2}+\sqrt{2}+z^2}\right]\,dz\end{eqnarray*}$$
then invoke integration by parts and averaging in order to convert the original integral into 
$$\begin{eqnarray*}\int_{0}^{+\infty}r\left(x^2+\frac{1}{x^2}\right)\,dx &=& \frac{1}{2}\int_{-\infty}^{+\infty}r\left(\left(x-\frac{1}{x}\right)^2+2\right)\,dx\\&\stackrel{\text{GMT}}{=}&\frac{1}{2}\int_{-\infty}^{+\infty}r(x^2+2)\,dx=\int_{0}^{+\infty}r(x^2+2)\,dx\end{eqnarray*}$$
with $r$ being a rational function. On the other hand, this approach looks pretty forced/artificial, especially if compared to the previous ones.
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