# 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

$\displaystyle\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}} =\frac{1}{2} \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{\sin (t) \cos (t)}} \, dt =\displaystyle \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{1-\frac{\sin ^2(t)}{2}}} \, dt = K\left(\frac{1}{2}\right)$ ?

• It is enough to factor $x^4+1$ as $(x^2+i)(x^2-i)$, exploit the relation between $K$ and the $\text{AGM}$ mean, then the identity $$\text{AGM}(\sqrt{i},\sqrt{-i}) = \text{AGM}\left(1,\tfrac{1}{\sqrt{2}}\right)$$ – Jack D'Aurizio Dec 25 '17 at 18:29
• – Jack D'Aurizio Dec 25 '17 at 18:31
• In your last line, the second integral is equal to the first after the substitution $w=\tan(t)$ followed by $u=w^{\frac{1}{2}}$ – Countingstuff Dec 25 '17 at 18:31
• See also for more gamma functions at fractional values. – Simply Beautiful Art Dec 25 '17 at 18:52
• You can find an alternative, very short proof through Fourier-Legendre series expansions at page $29$ here – Jack D'Aurizio Dec 25 '17 at 21:25

This one is a cakewalk. Just use the definition $$K(1/2)=\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-(1/2)\sin^{2}x}}=\frac{\sqrt{2}}{2}\int_{0}^{\pi}\frac{dx}{\sqrt{2-\sin^{2}x}}$$ and then put $t=\tan (x/2)$ so that $dx=2/(1+t^2)\,dt$ and $$2-\sin^{2}x=2-\frac{4t^2}{(1+t^2)^{2}}=2\cdot\frac{1+t^{4}}{(1+t^{2})^{2}}$$ and hence we have $$K(1/2)=\frac{1}{\sqrt{2}}\int_{0}^{\infty}\frac{1+t^{2}}{\sqrt{2}\sqrt{1+t^{4}}}\cdot\frac{2}{1+t^2}\,dt=\int_{0}^{\infty}\frac{dt}{\sqrt{1+t^{4}}}$$ Similarly put $\tan t=x^2$ to get $$\frac{1}{2}\int_{0}^{\pi/2}\frac{dt}{\sqrt{\sin t\cos t}} =\int_{0}^{\infty} \frac{dx} {\sqrt{1+x^{4}}}$$ This avoids the more complicated approaches from the theory of elliptic and theta functions and AGM.
We have $$\int_{0}^{+\infty}\frac{dx}{\sqrt{(x^2+a^2)(x^2+b^2)}}=\frac{\pi}{2\,\text{AGM}(a,b)}=\frac{\pi}{2\,\text{AGM}\left(\frac{a+b}{2},\sqrt{ab}\right)}\tag{1}$$ as a consequence of Lagrange's identity $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$.
On the other hand $$K\left(\tfrac{1}{2}\right)=\int_{0}^{+\infty}\frac{dx}{\sqrt{(1+x^2)(1+x^2/2)}} = \frac{\pi}{2\,\text{AGM}\left(1,\frac{1}{\sqrt{2}}\right)}\tag{2}$$ and $$\text{AGM}\left(\sqrt{i},\sqrt{-i}\right)=\text{AGM}\left(1,\tfrac{1}{\sqrt{2}}\right) \tag{3}$$ so $$K\left(\tfrac{1}{2}\right)=\int_{0}^{+\infty}\frac{dx}{\sqrt{(1+x^2)(1+x^2/2)}} =\int_{0}^{+\infty}\frac{dx}{\sqrt{(x^2+i)(x^2-i)}} = \int_{0}^{+\infty}\frac{dx}{\sqrt{x^4+1}}$$ as claimed. You can find an alternative, very short proof through Fourier-Legendre series expansions at page $29$ here.
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}
• If I understood correctly, the OP is aware of the connection between the integral and the $\Gamma$ / Beta function, the missing link is the connection between the integral and $K(1/2)$. – Jack D'Aurizio Dec 25 '17 at 22:51
• the last step uses the reflection formula $\Gamma(s) \Gamma(1-s) = \frac{\pi}{\sin(\pi s)}$ – reuns Dec 26 '17 at 3:49