An elementary proof of $\int_{0}^{1}\frac{\arctan x}{\sqrt{x(1-x^2)}}\,dx = \frac{1}{32}\sqrt{2\pi}\,\Gamma\left(\tfrac{1}{4}\right)^2$

Question

When playing with the complete elliptic integral of the first kind and its Fourier-Legendre expansion, I discovered that a consequence of $\sum_{n\geq 0}\binom{2n}{n}^2\frac{1}{16^n(4n+1)}=\frac{1}{16\pi^2}\,\Gamma\left(\frac{1}{4}\right)^4 $ is:

$$\int_{0}^{1}\frac{\arctan x}{\sqrt{x(1-x^2)}}\,dx = \tfrac{1}{32}\sqrt{2\pi}\,\Gamma\left(\tfrac{1}{4}\right)^2\tag{A}$$

which might be regarded as a sort of Ahmed's integral under steroids.

I already have a proof of this statement (through Fourier-Legendre expansions), but I would be happy to see a more direct and elementary proof of it, also because it might have some consequences about the moments of $K(x)$ of the form $\int_{0}^{1}K(x)\,x^{m\pm 1/4}\,dx$, which are associated with peculiar hypergeometric functions.

Answer 1

A possible way is to enforce the substitution $x\mapsto\frac{1-t}{1+t}$, giving:

$$ \mathfrak{I}=\int_{0}^{1}\frac{\arctan(x)}{\sqrt{x(1-x^2)}}\,dx = \int_{0}^{1}\frac{\tfrac{\pi}{4}-\arctan t}{\sqrt{t(1-t^2)}}\,dt $$ and $$ 2\mathfrak{I} = \frac{\pi}{4}\int_{0}^{1} x^{-1/2}(1-x^2)^{-1/2}\,dx =\tfrac{\pi}{8}\,B\left(\tfrac{1}{4},\tfrac{1}{2}\right).$$

Answer 2

Other approaches are possible. Making the substitution $t=2\arctan(x)$, we have $$ \int_{0}^{1} \frac{\arctan(x)}{\sqrt{x(1-x^2)} } \text{d}x =\frac{1}{2} \int_{0}^{\frac\pi2} \frac{x}{\sqrt{\sin(2x)} } \text{d} x. $$ I prefer using contour integration. It directly evaluates from the integral $$ f(z)=\frac{\ln(z)}{\sqrt{1-z^4} } $$ whose path goes along the quarter circle with radius $1$. Through some slight modifications like $$ f(z)=\frac{(1+z^2)^m}{(1-z^2)^n} \frac{\ln(z)^p}{z^a\left ( 1-z^4\right)^b }, $$ we have $$ \,_4F_3\left ( \frac34,1,1,1; \frac32,\frac32,2;1 \right )=\frac{\Gamma\left ( \frac14 \right )^4 }{8\pi}-\frac{3\pi^2}{8}-2G. $$