Relation between integral, gamma function, elliptic integral, and AGM The integral $\displaystyle\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}}$ is equal to $\displaystyle \frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{\pi }}$.
It is calculated or verified with a computer algebra system that $\displaystyle \frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{\pi }} = K\left(\frac{1}{2}\right)$ , where $K(m)$ is the complete elliptic integral of the first kind. This is in relation to what is called the elliptic integral singular value.
It is also known or verified that
 $\displaystyle K\left(\frac{1}{2}\right) =\displaystyle  \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{1-\frac{\sin ^2(t)}{2}}} \, dt= \frac{1}{2} \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{\sin (t) \cos (t)}} \, dt$.
Can one prove directly or analytically that
$\displaystyle\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}} =\frac{1}{2} \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{\sin (t) \cos (t)}} \, dt =\displaystyle  \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{1-\frac{\sin ^2(t)}{2}}} \, dt = K\left(\frac{1}{2}\right) $ ?

Let $t=\frac{1}{{1+x^4}}$ and then
$$ dx=-\frac14(1-t)^{-3/4}t^{5/4} $$
So
\begin{eqnarray}
&&\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}}\\
&=&\frac14\int\limits_0^{1}(1-t)^{-3/4}t^{1/4}\mathrm dx\\
&=&\frac14B(\frac14,\frac54)\\
&=&\sqrt{\frac{\pi}{2}}\frac{\Gamma(\frac54)}{\Gamma(\frac34)}\\
&=&\frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{\pi}}.
\end{eqnarray}
