Closed form for $\int_0^1\int_0^1 \frac{x y \sqrt{1-x} \sqrt{1-y}}{(x+y)^2} dxdy$

Question

Consider a unit square integral $$I =\int_0^1\int_0^1 \frac{x y \sqrt{1-x} \sqrt{1-y}}{(x+y)^2} dxdy \approx 0.0826$$ Many integrals over the unit square have nice closed form expressions. I wonder if this one does too. Using an idea suggested by richrow in the comments we can write $$\frac{1}{(x+y)^2} = \int _0 ^{\infty}t e^{-(x+y)t} dt$$ and so the integral "decouples" into $$\int _0 ^{\infty}tdt \int_0^1 x e^{-tx}\sqrt{1-x}\cdot dx\int_0^1 y e^{-ty}\sqrt{1-y}\cdot dy$$ For the little integrals Wolfram Mathematica gives $$\int_0^1 x e^{-tx}\sqrt{1-x}\cdot dx = \frac{4}{15} \mbox{}_{1}\!\operatorname{F}_{1}\left(2;\ \frac{7}{2}; -t\right)$$ where $\mbox{}_{1}\!\operatorname{F}_{1}$ is Kummer's hypergeometric function and so the original integral is $$I=\frac{16}{225}\int_0^{\infty}t\cdot \mbox{}_{1}\!\operatorname{F}_{1}\left(2;\ \frac{7}{2}; -t\right)^2dt$$ We can use the following property: $$\mbox{}_{1}\!\operatorname{F}_{1}\left(b-a;\ b; t\right)=e^t\cdot\mbox{}_{1}\!\operatorname{F}_{1}\left(a;\ b; -t\right)$$ We can now represent the hypergeometric function in the integrand as an absolutely convergent series and after applying Cauchy's product formula we can integrate term by term to get: $$I=16\sum_{k=0}^{\infty} \frac{k+1}{2^{k+2}}\sum_{m=0}^{k}\binom{k}{m}\frac{1}{(2m+3)(2m+5)(2k-2m+3)(2k-2m+5)}$$ which is a good place to get stuck!

Answer 1

A pedestrian solution. \begin{align}J&=\int_0^1\int_0^1 \frac{x y \sqrt{1-x} \sqrt{1-y}}{(x+y)^2} dxdy\\ &=\frac{1}{2}\int_0^1\int_0^1\frac{\Big((x+y)^2-x^2-y^2\Big)\sqrt{1-x}\sqrt{1-y}}{(x+y)^2}dxdy\\ &\overset{\text{Fubini+symmetry}}=\left(\int_0^1\sqrt{1-x}dx\right)^2-\int_0^1\int_0^1\frac{x^2\sqrt{1-x}\sqrt{1-y}}{(x+y)^2}dxdy\\ &=\frac29-\int_0^1\int_0^1\frac{x^2\sqrt{1-x}\sqrt{1-y}}{(x+y)^2}dxdy\\ &\overset{\text{IBP}}=\frac29+\int_0^1\sqrt{1-y}\left(\left[\frac{x^2\sqrt{1-x}}{x+y}\right]_0^1-\frac12\int_0^1\frac{x(4-5x)}{(x+y)\sqrt{1-x}}dx\right)dy\\ &=\frac29-\frac12\int_0^1\frac{x(4-5x)(1-y)}{(x+y)\sqrt{1-x}\sqrt{1-y}}dxdy\\ &=\frac29-\frac12\int_0^1\frac{4x+5x^2y}{(x+y)\sqrt{1-x}\sqrt{1-y}}dxdy+\frac52\int_0^1\frac{x^2}{(x+y)\sqrt{1-x}\sqrt{1-y}}dxdy+\\ &2\int_0^1\frac{xy}{(x+y)\sqrt{1-x}\sqrt{1-y}}dxdy\\ &\overset{\text{Symmetry}}=\frac29-\frac14\int_0^1\frac{4+5xy}{\sqrt{1-x}\sqrt{1-y}}dxdy+\frac54\int_0^1\frac{x^2+y^2}{(x+y)\sqrt{1-x}\sqrt{1-y}}dxdy+\\ &2\int_0^1\frac{xy}{(x+y)\sqrt{1-x}\sqrt{1-y}}dxdy\\ &\overset{\text{Fubini}}=\frac29-\frac{56}9+\frac54\int_0^1\frac{(x+y)^2-2xy}{(x+y)\sqrt{1-x}\sqrt{1-y}}dxdy+2\int_0^1\frac{xy}{(x+y)\sqrt{1-x}\sqrt{1-y}}dxdy\\ &=-6+\frac54\int_0^1\frac{x+y}{\sqrt{1-x}\sqrt{1-y}}dxdy-\frac12\int_0^1\frac{xy}{(x+y)\sqrt{1-x}\sqrt{1-y}}dxdy\\ &\boxed{J=\frac23-\frac12\underbrace{\int_0^1\frac{xy}{(x+y)\sqrt{1-x}\sqrt{1-y}}dxdy}_{=K}}\\ K&=\int_0^1 \frac{2y}{\sqrt{1-y}}\left[\frac{\text{arctanh}\left(\frac{\sqrt{1-x}}{\sqrt{1+y}}\right)}{\sqrt{1+y}}-\sqrt{1-x}\right]_0^1dy\\&=-\int_0^1 \frac{2y}{\sqrt{1-y}}\left(\frac{y\text{arctanh}\left(\frac1{\sqrt{1+y}}\right)}{\sqrt{1+y}}-1\right)dy\\ &=\frac83-2\int_0^1 \frac{y^2\text{arctanh}\left(\frac1{\sqrt{1+y}}\right)}{\sqrt{1-y^2}}\\ &\overset{\text{IBP}}=\frac83+\left[\left(y\sqrt{1-y^2}-\arcsin y\right)\text{arctanh}\left(\frac1{\sqrt{1+y}}\right)\right]_0^1+\\&\!\!\frac12\int_0^1\frac{y\sqrt{1-y^2}-\arcsin y}{y\sqrt{1+y}}dy\\ &=\frac83-\frac{\pi\text{arctanh}\left(\frac1{\sqrt{2}}\right)}2+\frac13-\frac12\underbrace{\int_0^1\frac{\arcsin y}{y\sqrt{1+y}}dy}_{=L}\\ &\boxed{K=3+\frac{\pi}2\ln\left(\sqrt{2}-1\right)-\frac L2}\\ L&\overset{y=\sin t}=\int_0^{\frac{\pi}2}\frac{t\cos t}{\sqrt{2}\sin\left(\frac{t}2+\frac{\pi}4\right)\sin t}dt=\frac12\int_0^{\frac{\pi}2}\frac{t\left(\cos^2\left(\frac t2\right)-\sin^2\left(\frac t2\right)\right)}{\sin\left(\frac t2\right)\cos\left(\frac t2\right)\left(\cos\left(\frac t2\right)+\sin\left(\frac t2\right)\right)}dt\\ &=\frac12\int_0^{\frac{\pi}2}\frac{t}{\sin\left(\frac t2\right)}dt-\frac12\int_0^{\frac{\pi}2}\frac{t}{\cos\left(\frac t2\right)}dt\\ &\overset{u=\frac t2}=2\int_0^{\frac{\pi}4}\frac{u}{\sin u}du-2\underbrace{\int_0^{\frac{\pi}4}\frac{u}{\cos u}du}_{z=\frac{\pi}2-u}=2\underbrace{\int_0^{\frac{\pi}2}\frac{u}{\sin u}du}_{{\text{IBP}}}-\pi\int_{\frac{\pi}4}^{\frac{\pi}2}\frac1{\sin u}du\\ &=2\left[u\ln\left(\tan\left(\frac u2\right)\right)\right]_0^{\frac{\pi}2}-2\underbrace{\int_0^{\frac{\pi}2}\ln\left(\tan\left(\frac u2\right)\right)du}_{z=\frac u2}-\pi\left[\ln\left(\tan\left(\frac u2\right)\right)\right]_{\frac{\pi}4}^{\frac{\pi}2}\\&=4\text{G}+\pi\ln\left(\tan\left(\frac{\pi}8\right)\right)\\ &\boxed{L=\int_0^1\frac{\arcsin y}{y\sqrt{1+y}}dy=4\text{G}+\pi\ln\left(\sqrt{2}-1\right)}\\ &\boxed{K=\int_0^1\frac{xy}{(x+y)\sqrt{1-x}\sqrt{1-y}}dxdy=3-2\text{G}}\\ &\boxed{J=\int_0^1\int_0^1 \frac{x y \sqrt{1-x} \sqrt{1-y}}{(x+y)^2} dxdy=\text{G}-\frac56} \end{align}

Answer 2

Using only Mathematica and Mellin Transform with Inverse Mellin Transform to solve:

$\int _0^1\int _0^1\frac{x y \sqrt{1-x} \sqrt{1-y}}{(x+y)^2}dydx=\\\mathcal{M}_s^{-1}\left[\int _0^1\int _0^1\mathcal{M}_A\left[\frac{x y \sqrt{1-x} \sqrt{1-y}}{(x+A y)^2}\right](s)dxdy\right](1)=\\\mathcal{M}_s^{-1}\left[\int _0^1\int _0^1-\pi (-1+s) \sqrt{1-x} x^{-1+s} \sqrt{1-y} y^{1-s} \csc (\pi s)dxdy\right](1)=\\\mathcal{M}_s^{-1}\left[\frac{\pi ^2 \csc (\pi s) \Gamma (2-s) \Gamma (s)}{4 \Gamma \left(\frac{7}{2}-s\right) \Gamma \left(\frac{3}{2}+s\right)}-\frac{\pi ^2 s \csc (\pi s) \Gamma (2-s) \Gamma (s)}{4 \Gamma \left(\frac{7}{2}-s\right) \Gamma \left(\frac{3}{2}+s\right)}\right](1)=\underset{A\to 1}{\text{lim}}\frac{1}{4} \pi \left(G_{3,3}^{2,2}\left(A\left| \begin{array}{c} -1,0,\frac{3}{2} \\ 0,0,-\frac{5}{2} \\ \end{array} \right.\right)-G_{3,3}^{2,2}\left(A\left| \begin{array}{c} -1,0,\frac{3}{2} \\ 0,1,-\frac{5}{2} \\ \end{array} \right.\right)\right)=\\\underset{A\to 1}{\text{lim}}\frac{-4 \sqrt{A} (9+A (2+9 A))-6 \left(3+A+A^2+3 A^3\right) \Im\left(\text{Li}_2\left(-\frac{i}{\sqrt{A}}\right)\right)+6 \left(3+A+A^2+3 A^3\right) \Im\left(\text{Li}_2\left(\frac{i}{\sqrt{A}}\right)\right)+3 (1+A) \left(-6 (-1+A) \sqrt{A}+(3+A (-2+3 A)) \pi -2 (3+A (-2+3 A)) \tan ^{-1}\left(\sqrt{A}\right)\right) \log (A)}{96 A^{5/2}}=-\frac{5}{6}+C$

where: $C$ is Catalan constant.

Answer 3

Start by substituting $(x,y)=(\cos^2u,\cos^2v)$ and consult @Quanto's evaluation, which includes a handful of closed forms of $I(a,b)$ when $a=b$, where

$$I(a,b) = \int_0^\tfrac\pi2 \int_0^\tfrac\pi2 \frac{(\cos u)^a (\cos v)^b}{\left(\cos^2u+\cos^2v\right)^2} \, du \, dv$$

---

$$\begin{align*} I &= 4 \int_0^\tfrac\pi2 \int_0^\tfrac\pi2 \frac{\sin^2u\cos^3u \cdot \sin^2v \cos^3v}{\left(\cos^2u + \cos^2v\right)^2} \, du \, dv \\ &= 4 \int_0^\tfrac\pi2 \int_0^\tfrac\pi2 \frac{\left(\cos^3u-\cos^5u\right)\left(\cos^3v-\cos^5v\right)}{\left(\cos^2u + \cos^2v\right)^2} \, du \, dv \\ &= 4 \bigg[I(3,3) - I(5,3) - I(3,5) + I(5,5)\bigg] \\ &= \frac{13}6 - G - 4 \bigg[I(3,5)+I(5,3)\bigg] \\ &= \boxed{G - \frac56} \end{align*}$$

---

$$\begin{align*} & I(3,5)+I(5,3) \\ &= \int_0^\tfrac\pi2 \int_0^\tfrac\pi2 \cdots \,du\,dv = 2 \int_0^\tfrac\pi2 \int_0^v \frac{\cos^3u \cos^3v}{\cos^2u+\cos^2v} \, du \, dv & \text{symmetry} \\ &= 2 \int_0^\tfrac\pi2 \sin u \cos^3u \, du \\ &\qquad - 2\int_0^\tfrac\pi2 \frac{\cos^5u}{\sqrt{1+\cos^2u}} \operatorname{artanh}\frac{\sin u}{\sqrt{1+\cos^2u}} \, du & \text{by parts} \\ &= \frac12 - 2 \int_0^1 \frac{\left(1-w^2\right)^2}{\sqrt{2-w^2}} \operatorname{artanh}\frac w{\sqrt{2-w^2}} \, dw & w=\sin u \\ &= \frac12 - 2 \int_0^1 \frac{\left(1-\omega^2\right)^2}{\left(1+\omega^2\right)^3} \operatorname{artanh} \omega \, d\omega & \omega=\frac w{\sqrt{2-w^2}} \\ &= \frac12 + 4 \int_0^1 \frac{\Omega^2 \log\Omega}{\left(1+\Omega^2\right)^3} \, d\Omega & \Omega=\frac{1-\omega}{1+\omega} \\ &= \frac12 + \int_0^1 \frac{1+\log \Omega}{\left(1+\Omega^2\right)^2} \, d\Omega & \text{by parts} \\ &= \frac34 - \frac G2 \end{align*}$$

The last integral can be evaluated in a number of ways; here's one.