# 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 '17 at 23:08
• The LHS turns out to be a multiple of a Beta function and we are done. Sep 27 '17 at 23:08
• ...you know, you should let some of us have a chance to answer your questions before you do... =P Sep 27 '17 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 '17 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 '17 at 23:16

## 1 Answer

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 '18 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 '18 at 0:02
• Okay... but how did you know that the denominator would stay the same? I can see how you would arrive at the substitution for the arctan function, but it seems kind of coincidental that the denominator was unchanged. Jun 3 '18 at 23:40