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

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.

• I guess I found it: the trick is just to enforce the substitution $$x \mapsto \frac{1-t}{1+t}.$$ Sep 27, 2017 at 23:08
• The LHS turns out to be a multiple of a Beta function and we are done. Sep 27, 2017 at 23:08
• ...you know, you should let some of us have a chance to answer your questions before you do... =P Sep 27, 2017 at 23:12
• @SimplyBeautifulArt: sorry, I didn't do it on purpose, I just realized it a few minutes after writing the question. I guess that happens, quite often :) Sep 27, 2017 at 23:13
• :'( welp... guess we shall await for your self-answer and hopefully some nice alternative proofs (which may be a suitable tag) Sep 27, 2017 at 23:16

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 $$2\mathfrak{I} = \frac{\pi}{4}\int_{0}^{1} x^{-1/2}(1-x^2)^{-1/2}\,dx =\tfrac{\pi}{8}\,B\left(\tfrac{1}{4},\tfrac{1}{2}\right).$$
• If I may, can I ask what was your line of thinking that made you realize, "You know what, substituting $x=(1-t)/(1+t)$ is the perfect way to evaluate this problem!" I fail to see how someone even gets there in the first place. May 16, 2018 at 23:49
• @FrankW.: the geometry of the arctangent function made me realize it. $\arctan\left(\frac{1-t}{1+t}\right)$ is a nice object; indeed the substitution $x=\frac{1-t}{1+t}$ removes the arctangent from the integrand function. Given the relation between the arctangent and the logarithm, this is more or less the same thing as $$\int_{0}^{+\infty}\frac{\log(x)}{p(x)}\,dx=0$$ for any quadratic and palindromic polynomial $p(x)$, non-vanishing over $\mathbb{R}^+$. May 17, 2018 at 0:02
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 $$\,_4F_3\left ( \frac34,1,1,1; \frac32,\frac32,2;1 \right )=\frac{\Gamma\left ( \frac14 \right )^4 }{8\pi}-\frac{3\pi^2}{8}-2G.$$