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.

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    $\begingroup$ I guess I found it: the trick is just to enforce the substitution $$ x \mapsto \frac{1-t}{1+t}.$$ $\endgroup$ – Jack D'Aurizio Sep 27 '17 at 23:08
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    $\begingroup$ The LHS turns out to be a multiple of a Beta function and we are done. $\endgroup$ – Jack D'Aurizio Sep 27 '17 at 23:08
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    $\begingroup$ ...you know, you should let some of us have a chance to answer your questions before you do... =P $\endgroup$ – Simply Beautiful Art Sep 27 '17 at 23:12
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    $\begingroup$ @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 :) $\endgroup$ – Jack D'Aurizio Sep 27 '17 at 23:13
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    $\begingroup$ :'( welp... guess we shall await for your self-answer and hopefully some nice alternative proofs (which may be a suitable tag) $\endgroup$ – Simply Beautiful Art Sep 27 '17 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 $$ 2\mathfrak{I} = \frac{\pi}{4}\int_{0}^{1} x^{-1/2}(1-x^2)^{-1/2}\,dx =\tfrac{\pi}{8}\,B\left(\tfrac{1}{4},\tfrac{1}{2}\right).$$

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    $\begingroup$ 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. $\endgroup$ – Frank W May 16 '18 at 23:49
  • $\begingroup$ @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 $\mathbb{R}^+$. $\endgroup$ – Jack D'Aurizio May 17 '18 at 0:02
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    $\begingroup$ 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. $\endgroup$ – Frank W Jun 3 '18 at 23:40

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