How to evaluate $\int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{1+x^4}}dx$? Mathematica gave me the result
$$\int\limits_{-\infty}^{\infty}\frac{1}{\sqrt{1+x^4}}dx = \frac{8 \Gamma\left(\frac{5}{4}\right)^2}{\sqrt{\pi}}$$,
but I have no idea how the Gamma Function appeared in the result.
Generally, how do we integrate such real integrals with functions needing branch cuts using complex analysis?
 A: First, we exploit the evenness of the integrand and write
$$\int_{-\infty}^\infty \frac{1}{\sqrt{1+x^4}}\,dx=2\int_0^\infty \frac{1}{\sqrt{1+x^4}}\,dx$$
Next, we enforce the substitution $x\to x^{1/4}$ to obtain
$$\begin{align}
\int_{-\infty}^\infty \frac{1}{\sqrt{1+x^4}}\,dx&=\frac12 \int_0^\infty \frac{x^{-3/4}}{(1+x)^{1/2}}\,dx\\\\
&=\frac12 B\left(1/4,1/4\right)\tag 1\\\\
&=\frac12 \frac{\Gamma^2(1/4)}{\Gamma(1/2)}\tag2\\\\
&=\frac12\frac{\Gamma^2(1/4)}{\sqrt \pi}\tag3\\\\
&=8\frac{\left(\frac14 \Gamma(1/4)\right)^2}{\sqrt \pi}\tag4\\\\
&=8\frac{\Gamma^2(5/4)}{\sqrt \pi}\tag5
\end{align}$$
as was to be shown!

NOTES:
In arriving at $(1)$, we used the integral representation $B(x,y)=\int_0^\infty \frac{t^{x-1}}{(1+t)^{x+y}}\,dt$ for the Beta function.
In going from $(1)$ to $(2)$, we used the relationship $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ between the Beta and Gamma functions (See Here).
In going from $(2)$ to $(3)$, we used the fact that $\Gamma(1/2)=\sqrt \pi$ ([See Here]https://en.wikipedia.org/wiki/Gamma_function#Particular_values()).
In going from $(3)$ to $(4)$, we simply rewrote $(3)$ using $\frac12=8\left(\frac14\right)^2$.
And finally, in going from $(4)$ to $(5)$, we used the functional equation $\Gamma(1+x)=x\Gamma(x)$.
