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[Math] Calculate in closed form $\int_0^1 \int_0^1 \frac{dx\,dy}{1-xy(1-x)(1-y)}$
integrationreal-analysis
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Suppose $r$ is even, and write $r = 2k$ and $rm/2 = k+j$. Let $$ J(s,j,k) = \int_0^\infty \frac{x^{2k+s-1}}{(1+x)^s(1+x^2)^{k+j}}\ dx$$ The integral converges if $j \ge 1$ and $s \ge 1$. The (ordinary) generating function of this is $$G(u,v,w) = \sum_{j=1}^\infty \sum_{k=0}^\infty \sum_{s=1}^\infty J(s,j,k) u^j v^k w^s$$ Now for $|u|,|v|,|w|<1$ $$ \sum_{j=1}^\infty \sum_{k=0}^\infty \sum_{s=1}^\infty \frac{x^{2k+s-1} u^jv^k w^s}{(1+x)^s(1+x^2)^{k+j}} = \frac{u w (1+x^2)}{ (1-u+x^2)(1 + (1-v)x^2)(1+(1-w)x)}$$ Integrating on $(0,\infty)$, Maple finds (for $|u|,|v|,|w|<1$) a rather formidable closed-form expression for $G(u,v,w)$ which is a rational function in $u$, $v$, $w$, $\sqrt{1-u}$, $\sqrt{1-v}$, $\ln(1-u)$, $\ln(1-v)$ and $\ln(1-w)$.
I sketch the arguments for $C(x)$, the arguments for $L(x)$ are essentially the same.
The specific form of the sum suggests probabilistic arguments. Let $X_x$ be a $\mathrm{Poiss}(x^2)$-distributed random variable and note that $$C(x)\,e^{-x^2}=\mathbb{E}\frac{x}{\sqrt{X_x}}\,1_{\{X_x\geq 1\}}$$ It is known that $\frac{X_x -x^2}{x}\longrightarrow N(0,1)$ (standard normal) in distribution as $x\longrightarrow \infty$ and that that all moments $\mathbb{E}\left(\frac{X_x-x^2}{x}\right)^k$ of $X_x$ converge to the corresponding moments of $N(0,1)$, so that (for large $x$) $X_x$ is concentrated around $x^2$, with deviations of order $x$.
To use that information split $\{X_x\geq 1\}$ on the rhs into (say) the parts $1\leq X_x <\tfrac{1}{2}x^2$, $X_x-x^2 >\tfrac{1}{2} x^{2}$ and $|X_x-x^2| \le \tfrac{1}{2}x^{2}$.
By routine arguments the integrals over the first two parts are asymptotically exponentially small. For the remaining part write \begin{align*} \mathbb{E}\frac{x}{\sqrt{X_x}}\,1_{\{|X_x-x^2|\leq \tfrac{1}{2}x^2\}} &=\mathbb{E}\frac{x}{\sqrt{x^2+(X_x-x^2)}}\,1_{\{|X_x-x^2|\leq \tfrac{1}{2} x^2\}}\\ &=\mathbb{E}\frac{1}{\sqrt{1 +\frac{1}{x^2}(X_x-x^2)}}\,1_{\{|X_x-x^2|\leq \tfrac{1}{2} x^2\}}\\ &=\mathbb{E}\sum_{k=0}^\infty {-\frac{1}{2} \choose k} \frac{1}{x^{2k}}(X_x-x^2)^k \,1_{\{|X_x-x^2|\leq \tfrac{1}{2}x^2\}}\\ %C(x)\,e^{-x^2}&= 1+\frac{3}{8}x^{-2} + \frac{65}{128}x^{-4} + \frac{1225}{1024}x^{-6} + \frac{1619583}{425984}x^{-8}+\mathcal{O}(x^{-10})\\ %L(x)\,x^2\,e^{-x^2}&= 1+\frac{15}{8}x^{-2} + \frac{665}{128}x^{-4}+\frac{19845}{1024}x^{-6}+\frac{37475823}{425984}x^{-8}+\mathcal{O}(x^{-10}) \end{align*} Clearly the series may be integrated termwise, and completing the tails changes it only by asymptotically exponentially small terms. For $k\geq 2$ the central moment $c_k(x):=\mathbb{E}\left(X_x-x^2\right)^k$ is a polynomial in $x^2$ of degree $\lfloor k/2\rfloor$. Thus (after regrouping of terms) the formal series $$\sum_{k=0}^\infty x^{-2k}{-\tfrac{1}{2} \choose k} c_k(x)$$ gives a full asymptotic expansion of $C(x)e^{-x^2}$. Evaluating the first eight terms gives $$C(x)\,e^{-x^2}= 1+\frac{3}{8}x^{-2} + \frac{65}{128}x^{-4} + \frac{1225}{1024}x^{-6} + \frac{131691}{32768}x^{-8}+\mathcal{O}(x^{-10})$$
EDIT: I corrected the coefficient of $x^{-8}$. Thanks to Johannes Trost for pointing out that I miscalculated. Similarly the coefficient of $x^{-8}$ in the asymptotic series for $L(x)$ given in the comment must be corrected.
Best Answer
Not what you're really looking for, but here's a "word answer" in terms of a probabilistic interpretation.
Suppose Alice and Bob are playing a game where they are presented with $2n+1$ indistinguishable bees and have to separate them into
They both have to try to guess what groups the other has chosen. They play this game for increasing $n=0,1,2,3,\dots$
The probability that Alice is right is $$ \frac1{\binom{2n+1}{n}\binom{n}{1}} = \frac{(n!)^2}{(2n+1)!} = B(n+1,n+1) $$
Thus the number you're interested in is the expected number of times Alice and Bob both guess correctly.
For the record here is the usual derivation relating the integral to the infinite series in terms of the beta function $B$.
Let $a(x)=x(1-x)$ and $b(x,y)=a(x)a(y)$, and $$ c(x,y)=\frac1{1-b(x,y)} = \sum_{n=0}^\infty b(x,y)^n. $$ In terms of the beta function $B$ $$ \int_0^1 a(x)^n\,dx = \frac{(n!)^2}{(2n+1)!} = B(n+1,n+1) $$ so $$ \int_{[0,1]^2} c(x,y)\,dx\,dy = \sum_{n=0}^\infty B(n+1,n+1)^2 = \sum_{n=1}^\infty B(n,n)^2 = \sum_{n=1}^\infty \left(\frac{\Gamma(n)^2}{\Gamma(2n)}\right)^2 $$ $$ =\sum_{n=1}^\infty \frac{\Gamma(n)^4}{\Gamma(2n)^2} = {_3}F_2\left(1,1,1; \frac32,\frac32; \frac1{16}\right). $$ (Wolfram Alpha link for the last equation)