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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.

Jack D'Aurizio
  • 353,855

3 Answers3

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

Jack D'Aurizio
  • 353,855
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    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. – Frank W 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 $\mathbb{R}^+$. – Jack D'Aurizio May 17 '18 at 00:02
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    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. – Frank W Jun 03 '18 at 23:40
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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 $$ \,_4F_3\left ( \frac34,1,1,1; \frac32,\frac32,2;1 \right )=\frac{\Gamma\left ( \frac14 \right )^4 }{8\pi}-\frac{3\pi^2}{8}-2G. $$

Setness Ramesory
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$$I=\int_{0}^{1}\frac{\arctan x}{\sqrt{x(1-x^2)}}\,dx\stackrel{x=\tan\theta}{=}\frac18\int_0^{\pi/4}\frac{2\theta\,d\theta}{\sqrt{\sin4\theta}}\stackrel{4\theta=\phi}{=}\frac18\int_0^\pi\frac{\phi\,d\phi}{\sqrt{\sin\phi}}.$$ By King's property, $\phi=\pi-\alpha$, and a definition of beta function $$I=\frac\pi{16}\int_0^\pi\frac{d\alpha}{\sqrt{\sin\alpha}}=\frac\pi{8}\int_0^{\pi/2}\sin^{-1/2}\alpha\,d\alpha=\frac\pi{16}B(\tfrac14,\tfrac12)=\frac{\sqrt{2\pi}}{32}\Gamma\left(\tfrac14\right)^2.$$

Bob Dobbs
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