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?
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 $\Gamma(1+x)=x\Gamma(x)$.