6

Question: Prove that $$ \int_{0}^{1} \frac{K^{\prime}\left ( \frac{1-x^2}{2} \right ) }{\sqrt{1+x^2} } \text{d} x =\frac{\Gamma\left ( \frac{1}{24} \right ) \Gamma\left ( \frac{5}{24} \right)\Gamma\left ( \frac{7}{24} \right ) \Gamma\left ( \frac{11}{24}\right ) }{48\pi\sqrt{3}}, $$ where $K^\prime(x)=K\left(\sqrt{1-x^2}\right)$ and $K(x)$ is known as a complete elliptic integral of the first kind with the elliptic modulus $x$. $\Gamma(z)=\int_{0}^{\infty}t^{z-1}e^{-t}\text{d}t$ is the Euler Gamma function.

The equality's right hand side is probably given by an elliptic integral singular value $K(k_6)$. Here are my thoughts.

  1. We can determine a similar one $$ \int_{0}^{1}\frac{K\left ( \frac{1+x^2}{2} \right ) }{ \sqrt{1-x^2} }\text{d}x=\frac{1}{\pi\sqrt{2}}\int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\pi} \frac{1}{3-\cos x-\cos y-\cos z}\text{d}x\text{d}y\text{d}z\\ =\frac{\Gamma\left ( \frac{1}{24} \right ) \Gamma\left ( \frac{5}{24} \right)\Gamma\left ( \frac{7}{24} \right ) \Gamma\left ( \frac{11}{24}\right ) }{32\pi\sqrt{3}}. $$ The direct calculation shows the first equality. The second one is refered to the third of Watson's triple integrals. So we only need to show $$ \int_{0}^{1}\frac{K\left ( \frac{1+x^2}{2} \right ) }{ \sqrt{1-x^2} }\text{d}x =\frac32\int_{0}^{1} \frac{K^\prime\left ( \frac{1-x^2}{2} \right ) }{\sqrt{1+x^2} } \text{d}x. $$
  2. So far, however, I can't see any connections between two integrals. Some subtle transformations and substitutions may be helpful. But I am not able to come up with them by myself.
  3. By integrating the function $f(x)=\frac{K\left ( \frac{1+x^2}{2} \right ) }{ \sqrt{1-x^2} }$ along the real axis, we deduce $$ \int_{0}^{\sqrt{2} } \frac{K^\prime\left (1-x^2\right ) }{ \sqrt{1+x^2} }\text{d}x =\frac{\sqrt{2}}{2}\int_{0}^{1}\frac{K\left ( \frac{1+x^2}{2} \right ) }{ \sqrt{1-x^2} }\text{d}x\\ =\frac{\sqrt{6}\,\Gamma\left ( \frac{1}{24} \right ) \Gamma\left ( \frac{5}{24} \right)\Gamma\left ( \frac{7}{24} \right ) \Gamma\left ( \frac{11}{24}\right ) }{192\pi}. $$

The desired integral is actually represented as follows $$ \int_{-1}^{1} \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{\sqrt{x^2-1}\sqrt{y^2-1} \sqrt{1-z^2} } \frac{1}{3+x+y+z} \text{d}x\text{d}y\text{d}z $$ which is equivalent to $\int_{0}^{\infty}e^{-3z}K_0(z)^2I_0(z)\text{d}z=\frac{\pi^2}3\int_{0}^{\infty}e^{-3z}I_0(z)^3\text{d}z$ after some complex integral technique. Detailed explanations are writing now.

Setness Ramesory
  • 6,789
  • Surely your identity is wrong unless I'm mistaken? Here is your integral which is negative https://www.wolframalpha.com/input?i=Integrate+%5B%282+x+EllipticE%281%2F2+-+x%5E2%2F2%29+-+x+%281+%2B+x%5E2%29+EllipticK%281%2F2+-+x%5E2%2F2%29%29%2F%28-1+%2B+x%5E4%29%5D%2F%28sqrt%281%2Bx%5E2%29%29+from+0+to+1 and here is your gamma identity which is positive https://www.wolframalpha.com/input?i=FullSimplify%5BGamma%5B1%2F24%5DGamma%5B5%2F24%5DGamma%5B7%2F24%5DGamma%5B11%2F24%5D%2F%2848+pi+sqrt3%29%5D – Max0815 Jan 12 '23 at 03:54
  • 1
    @Max0815 Did you mistake $K'(x)$ for the derivative of $K(x)$ whereas the OP defined it as $K'(x)=K\left(\sqrt{1-x^2}\right)$? But even with the correct interpretation of $K'(x)$ numeric evaluation of the integral suggests the identity in the title is incorrect (see https://www.wolframalpha.com/input?i=NIntegrate%5BEllipticK%5BSqrt%5B1-%28%281-x%5E2%29%2F2%29%5E2%5D%5D%2FSqrt%5B1%2Bx%5E2%5D%2C%7Bx%2C0%2C1%7D%5D). – Steven Clark Jan 12 '23 at 03:57
  • 3
    @StevenClark I think there's a fine line between parameter $K(m)=\int_{0}^{1} \frac{1}{\sqrt{1-t^2}\sqrt{1-mt^2} }\text{d}t$ and modulus $K(k)=\int_{0}^{1} \frac{1}{\sqrt{1-t^2}\sqrt{1-k^2t^2} }\text{d}t$. WA uses the former. So in WA, you can input $K\left(1-\frac{(1-x^2)^2}{4}\right)$ instead. – Sakup2485 Jan 12 '23 at 08:30
  • When I first saw the question title, I thought the answer involved some sort of new, multiple Gamma function – Max Lonysa Muller Jan 12 '23 at 12:56
  • @Sakup2485 Thanks, that perhaps explains the problems I was having verifying the formulas. – Steven Clark Jan 12 '23 at 15:56
  • When Maple computes these numerically, they agree to (at least) 50 places. – GEdgar Jan 12 '23 at 16:43
  • Perhaps this is relevant: https://mathoverflow.net/a/365844/454 For example, $$\eta(i\sqrt{6})=\frac{1}{2^{3/2}3^{1/4}}\big(\sqrt{2}-1\big)^{1/12}\frac{\Big(\Gamma\big(\tfrac{1}{24}\big) \Gamma\big(\tfrac{5}{24}\big) \Gamma\big(\tfrac{7}{24}\big) \Gamma\big(\tfrac{11}{24}\big)\Big)^{1/4}}{\pi^{3/4}}$$ – GEdgar Jan 12 '23 at 16:52
  • There is: $$\int_0^1\frac{K'\left(\frac{1-x^2}{2}\right)}{\sqrt{1+x^2}}dx=\int_0^1\frac{1}{\sqrt{1-t^2}}\int_0^1 \frac{\sqrt 2}{\sqrt{\frac{3-t}{2}-x^2}\sqrt{\frac{3+t}{2}-x^2}}dxdt$$ So it would be enough to show that: $$\int_0^1 \frac{\sqrt 2}{\sqrt{\frac{3-t}{2}-x^2}\sqrt{\frac{3+t}{2}-x^2}}dx=\frac{2}{3}K\left(\frac{1+t^2}{2}\right)$$ – Zacky Jan 13 '23 at 15:11
  • @Zacky: putting $t=0$ makes integral in your comment as elementary, but I don't think $K(1/2)$ has an elementary closed form. – Paramanand Singh Jan 14 '23 at 02:28
  • @ParamanandSingh I only checked the correctness of the first equality (the one with the double integral), then I assumed the inner integral must be equal with $\frac{2}{3}K\left(\frac{1+t^2}{2}\right)$ given that OP mentioned about $\int_{0}^{1}\frac{K\left ( \frac{1+x^2}{2} \right ) }{ \sqrt{1-x^2} }\text{d}x =\frac32\int_{0}^{1} \frac{K^\prime\left ( \frac{1-x^2}{2} \right ) }{\sqrt{1+x^2} } \text{d}x$. Since the outer integral is over $(0,1)$ I think the second equality should hold only for $0 < t < 1$, otherwise I'm not sure what goes wrong there. – Zacky Jan 14 '23 at 02:44

0 Answers0