24
$\begingroup$

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.

$\endgroup$
7
  • 1
    $\begingroup$ I guess I found it: the trick is just to enforce the substitution $$ x \mapsto \frac{1-t}{1+t}.$$ $\endgroup$ Sep 27 '17 at 23:08
  • 1
    $\begingroup$ The LHS turns out to be a multiple of a Beta function and we are done. $\endgroup$ Sep 27 '17 at 23:08
  • 5
    $\begingroup$ ...you know, you should let some of us have a chance to answer your questions before you do... =P $\endgroup$ Sep 27 '17 at 23:12
  • 2
    $\begingroup$ @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 :) $\endgroup$ Sep 27 '17 at 23:13
  • 3
    $\begingroup$ :'( welp... guess we shall await for your self-answer and hopefully some nice alternative proofs (which may be a suitable tag) $\endgroup$ Sep 27 '17 at 23:16
19
$\begingroup$

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

$\endgroup$
3
  • 1
    $\begingroup$ 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 $\mathbb{R}^+$. $\endgroup$ May 17 '18 at 0:02
  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.