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Can we possibly compute the following integral in terms of known constants?

$$\int_0^1 \int_0^1 \frac{dx\,dy}{1-xy(1-x)(1-y)}$$ Some progress was already done here http://integralsandseries.prophpbb.com/topic279.html but still we have a hypergeometric function. What's your thoughts on it?

UPDATE: The question was also posted on mathoverflow here https://mathoverflow.net/questions/206253/calculate-in-closed-form-int-01-int-01-fracdx-dy1-xy1-x1-y#

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  • $\begingroup$ Plouffe's Inverse Symbolic Calculator doesn't seem to find any evidence that there's a better representation for the answer than as a hypergeometric value. Though that in itself is nothing conclusive. $\endgroup$ Commented May 11, 2015 at 16:17

4 Answers 4

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Let $x=\sin^2(t)$ and $y=\sin^2(w)$, we obtain \begin{align} I & = \int_0^{\pi/2}\int_0^{\pi/2} \dfrac{4\sin(t)\cos(t)\sin(w)\cos(w)dtdw}{1-\sin^2(t)\cos^2(t)\sin^2(w)\cos^2(w)}\\ & = 4\sum_{k=0}^{\infty}\int_0^{\pi/2}\int_0^{\pi/2} \sin^{(2k+1)}(t) \cos^{(2k+1)}(t)\sin^{(2k+1)}(w) \cos^{(2k+1)}(w)dtdw\\ & = 4 \sum_{k=0}^{\infty} \left(\int_0^{\pi/2}\sin^{(2k+1)}(t) \cos^{(2k+1)}(t)dt\right)^2\\ & = 4 \sum_{k=0}^{\infty} \dfrac1{4^{2k+1}}\left(\int_0^{\pi/2}\sin^{(2k+1)}(2t)dt\right)^2 = \sum_{k=0}^{\infty} \dfrac1{4^{2k}}\left(\int_0^{\pi}\sin^{(2k+1)}(t)\dfrac{dt}2\right)^2\\ & = \sum_{k=0}^{\infty} \dfrac1{4^{2k}}\left(\int_0^{\pi/2}\sin^{(2k+1)}(t)dt\right)^2 \end{align} Recall that $$\int_0^{\pi/2}\sin^{(2k+1)}(t)dt=\dfrac{4^k}{2k+1}\dfrac1{\dbinom{2k}k}$$ This gives us $$I = \sum_{k=0}^{\infty} \dfrac1{(2k+1)^2} \dfrac1{\dbinom{2k}k^2} = _3F_2\left(1,1,1;\frac32,\frac32;\frac1{16}\right)$$

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  • $\begingroup$ Wolfram Alpha evaluates the hypergeometric in the last line to: 1.0289425662942443315755244192977092068490... , which cannot be identified by the Inverse Symbolic Calculator even in "advanced" mode. So a more "closed" solution than the given seems to be unlikely. $\endgroup$ Commented May 11, 2015 at 13:06
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    $\begingroup$ @JohannesTrost the fact that the inverse symbolic calculator in advanced mode doesn't find anything means almost nothing (I tell you that from my experience). $\endgroup$ Commented May 11, 2015 at 13:56
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An incomplete answer.

Tackling the inner integral with elementary methods, \begin{align} \int^1_0\frac{{\rm d}x}{1-ax+ax^2} &=\int^1_0\frac{{\rm d}x}{a\left(x-\frac{1}{2}\right)^2+1-\frac{a}{4}}\\&=\int^\frac{1}{2}_{-\frac{1}{2}}\frac{{\rm d}x}{ax^2+1-\frac{a}{4}}\\ &=\frac{1}{\sqrt{a\left(1-\frac{a}{4}\right)}}\int^\frac{\sqrt{a}}{2\sqrt{(1-\frac{a}{4})}}_{-\frac{\sqrt{a}}{2\sqrt{(1-\frac{a}{4})}}}\frac{{\rm d}x}{1+x^2}\\&=\frac{4}{\sqrt{a(4-a)}}\arctan\left(\sqrt{\frac{a}{4-a}}\right) \end{align} Thus \begin{align} \iint\limits_{[0,1]^2}\frac{{\rm d}A}{1-xy(1-x)(1-y)} &=4\int^1_0\arctan\left(\sqrt{\frac{x(1-x)}{x^2-x+4}}\right)\frac{{\rm d}x}{\sqrt{x(1-x)(x^2-x+4)}}\\ &=16\int^1_0\arctan\left(\sqrt{\frac{1-x^2}{15+x^2}}\right)\frac{{\rm d}x}{\sqrt{(1-x^2)(15+x^2)}}\\ &=\frac{16}{\sqrt{15}}\int^1_0\operatorname{F}\left(\arcsin{x}\right)\frac{x\ {\rm d}x}{\sqrt{(1-x^2)(15+x^2)}}\\ &=\frac{8}{15}\int^{\operatorname{K}}_{-\operatorname{K}}u\operatorname{sn}(u)\ {\rm d}u \end{align} where $\operatorname{sn}(u)$ is a Jacobi elliptic function with modulus $k=15^{-1/2}i$.

Let $R$ be the parallelogram with vertices $-\operatorname{K}-2i\operatorname{K}'$, $\operatorname{K}-2i\operatorname{K}'$, $\operatorname{K}+2i\operatorname{K}'$, $-\operatorname{K}+2i\operatorname{K}'$. Parameterising along the contour, \begin{align} \int_{\partial R}z^{\color{red}{2}}\operatorname{sn}(z)\ {\rm d}z &=-8i\operatorname{K}\left(\frac{4}{\sqrt{15}}\right)\int^{\operatorname{K}}_{-\operatorname{K}}u\operatorname{sn}(u)\ {\rm d}u+2i\int^{2\operatorname{K}'}_{-2\operatorname{K}'}(\operatorname{K}^2-t^2)\operatorname{cd}(it)\ {\rm d}t\\ &=2\pi i\sum_{\pm}\operatorname*{Res}_{z=\pm i\operatorname{K}'}z^2\operatorname{sn}(z)=-4\pi\sqrt{15}\operatorname{K}\left(\frac{4}{\sqrt{15}}\right)^2 \end{align} $\operatorname{cd}(t)$ has an anti-derivative, namely $$2i\operatorname{K}^2\int^{2\operatorname{K}'}_{-2\operatorname{K}'}\operatorname{cd}(it)\ {\rm d}t=-4i\sqrt{15}\operatorname{K}^2\left.\ln\left(\operatorname{nd}(t)+\frac{i}{\sqrt{15}}\operatorname{sd}(t)\right)\right|^{2i\operatorname{K}'}_0$$ Otherwise, this seems to be as far as I can go for now.

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    $\begingroup$ (+1) Well done. Now we just need to find the bravery to compute $I'(\alpha)$ :) $\endgroup$ Commented May 4, 2015 at 12:00
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    $\begingroup$ (+1) A very good starting point! This is the way to go (at least for the first part of the way). However it might be an illusion that the last integral is easier than the initial one. :-) $\endgroup$ Commented May 4, 2015 at 12:31
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    $\begingroup$ Thank you very much for your work to my question. $\endgroup$ Commented May 13, 2015 at 19:29
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From hypergeometric function to elliptic integrals

The hypergeometric function $_3F_2$ from your link (or user17762's answer) may be rewritten (using following integral representation and $_2F_1\left(1,1;\frac32; t\right)=\frac d{dt}\left[\left(\arcsin\sqrt{t}\,\right)^2\right]$) as : \begin{align} \tag{1}I&=\;_3F_2\left(1,1,1;\frac32,\frac32;\frac1{16}\right)=\frac 12\int_0^1\frac{_2F_1\left(1,1;\frac32;\frac x{16}\right)}{\sqrt{1-x}}dx\\ &=2\int_0^1\frac{\arcsin\frac{\sqrt{x}}4}{\sqrt{x\,(1-x)\left(1-\frac x{16}\right)}}dx\\ \tag{2}&=16\int_0^{1/4}\frac{\arcsin u}{\sqrt{(1-16 u^2)\left(1-u^2\right)}}\,du\\ &\tag{3}=16\int_0^{\arcsin(1/4)} \frac {y}{\sqrt{1-4^2\sin(y)^2)}}\;dy\\ &=\left.16\;y\;F\left(y, 4\right)\right|_0^{\arcsin(1/4)}-16\int_0^{\arcsin(1/4)}\;F\left(y, 4\right)\;dy\\ \tag{4}&=4\;\arcsin(1/4)\;K\left(\frac 1{4}\right)-16\int_0^{\arcsin(1/4)}\;F\left(y, 4\right)\;dy\\ \end{align}

With $\;\displaystyle F(\phi,k):=\int_0^{\phi} \frac {dy}{\sqrt{1-k^2\sin(y)^2)}},\;K(k):=F\left(\frac{\pi}2,k\right)\;$ the incomplete and complete elliptic integral of the first kind $F$ and $K$ (one should take care of replacing $F(\phi,k)$ by EllipticF[$\phi,k^2$] and $K(k)$ by EllipticK[$k^2$] while using Alpha/Mathematica).

'$I$' may be expressed with elliptic integrals but this doesn't seem to help for a closed form... Observing that $\;\displaystyle K\left(\frac 1{4}\right)=\frac{\pi}2\sum_{k=0}^\infty\frac{\binom{2k}{k}^2}{2^{8\,k}}\;$ let's start it all again using the

Squared central binomial series

We want $\;\displaystyle I=\sum_{k=0}^\infty \frac 1{(2k+1)^2\binom{2k}{k}^2}\;$ but Nected Batir proved in 2004 following more general formula (for $n\in\mathbb{N},\; n\ge3$) :

\begin{align} \tag{5}I_n(x)&:=\sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)^n\,\binom{2k}{k}^2}\\ &=\frac{4\,(-1)^{n-3}}{(n-3)!}\int_0^{\pi/2}\int_0^{\arcsin(x\cos(y)/4)}\frac t{\sin(t)}\;\log^{n-3}\left(\frac{4\sin(t)}{x\cos(y)}\right)\;dt\,dy\\ \end{align} For $\,n=3\,$ this is simply $\;\displaystyle I_3(x)=\int_0^{\pi/2}\int_0^{\arcsin(x\cos(y)/4)}\frac t{\sin(t)}\;dt\,dy\;\;$ and after derivation \begin{align} I_2(x)&=x\,I_3(x)'=4\,x\int_0^{\pi/2} \frac {\arcsin(x\cos(y)/4)}{x\cos(y)/4} \frac {\partial}{\partial x}\left[\arcsin(x\cos(y)/4)\right]\;dy\\ \tag{6}I_2(x)&=4\int_0^{\pi/2} \frac {\arcsin(x\cos(y)/4)}{\sqrt{1-(x\cos(y)/4)^2}} \;dy\\ \end{align} Unfortunately this is merely $(2)$ in the case $x=1$ (the integral for $x=4$ is divergent and $I_3(4)=8\pi G-14\zeta(3)\;$ but this won't help here...) so nothing complete here either...

The journey may be of interest anyway...

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    $\begingroup$ Thank you for giving insights on the problem (+1). Very nice the formula by Nected Batir I wasn't aware of. $\endgroup$ Commented May 4, 2015 at 23:27
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    $\begingroup$ Thanks @Chris'ssis and interesting question btw! Is this question linked to another of your quests? Excellent continuation, $\endgroup$ Commented May 5, 2015 at 15:41
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    $\begingroup$ @ Raymond Manzoni Thank you! :-) Indeed, I'm deeply involved in personal research that covers such integrals, series and limits. $\endgroup$ Commented May 5, 2015 at 16:40
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Our integral equals: $$\begin{eqnarray*} I=\sum_{n\geq 0}\iint_{[0,1]^2}x^n y^n(1-x)^n (1-y)^n\,dx\,dy &=& \sum_{n\geq 0}\left(\int_{0}^{1}x^n(1-x)^n\,dx\right)^2\\&=&\sum_{n\geq 0}\left(\frac{\Gamma(n+1)^2}{\Gamma(2n+2)}\right)^2\\&=&\sum_{n\geq 0}\frac{1}{(2n+1)^2\binom{2n}{n}^2}\tag{1}\end{eqnarray*}$$ and: $$ \sum_{n\geq 0}\frac{x^n}{(2n+1)\binom{2n}{n}}=\frac{4\arcsin\frac{\sqrt{x}}{2}}{\sqrt{x(4-x)}}\tag{2}$$ hence $I$ is the squared $L^2$-norm of a quite complicated function: $$ I = \frac{16}{\pi}\int_{0}^{\pi}\frac{\arcsin\frac{e^{it/2}}{2}\cdot\arcsin\frac{e^{-it/2}}{2}}{\sqrt{(4-e^{it})(4-e^{-it})}}\,dt.\tag{3}$$

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  • $\begingroup$ @Chris'ssis Good to see you back on the main site. I remember you were taking a break from the site. $\endgroup$
    – Adhvaitha
    Commented May 2, 2015 at 13:26
  • $\begingroup$ @user17762 hehe, indeed :-). I actually never take a break from doing math. What was your username in the past? I might recollect who you are. :-) $\endgroup$ Commented May 2, 2015 at 13:29
  • $\begingroup$ @Chris'ssis marvis in my earlier avatar. $\endgroup$
    – Adhvaitha
    Commented May 2, 2015 at 14:09
  • $\begingroup$ @user17762 long time I haven't seen you around. Nice to meet you again! :-) $\endgroup$ Commented May 2, 2015 at 14:41
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    $\begingroup$ I've checked your (1) by hand and found it to be correct. Meanwhile M.N.C.E. has changed his answer. $\endgroup$ Commented May 10, 2015 at 19:30

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