Closed form for $\int _0^1\int _0^1 \frac{\sqrt{x^2+y^2}}{1-xy} dxdy$

Question

Consider a unit square integral $$I=\int _0^1\int _0^1 \frac{\sqrt{x^2+y^2}}{1-xy} dxdy \approx 1.4843079$$ Many unit square integrals admit nice closed forms. But with this one, standard things like expressing $\frac{1}{1-xy}$ as a geometric series or a change to polar coordinates don't seem to work, so I wonder if this integral actually does have a closed form.

Both Mathematica and Maple's identify functions couldn't guess that hypothetical closed form.

Answer 1

This is not a complete answer; instead, it improves upon L. F.'s previous answer by giving a more approachable form and providing a full derivation for it (see below for a progress update). Previously, L. F. derived

$$I=\sum_{n = 1}^{\infty} \frac{2^{1-\frac{n}{2}}}{n(2n + 1)} \, {_2F_1 \Bigl(\frac{n}{2}, \frac{n + 3}{2}; \frac{n}{2} + 1; \frac{1}{2}\Bigr)}.\label{1}\tag{1}$$

Using this hypergeometric identity, we can rewrite the hypergeometric function within $(1)$ as

$${\left(1-\frac{1}{2}\right)^{-n/2}{_2F_1 \Bigg(\frac{n}{2}, \frac{n}{2}+1-\frac{n+3}{2}; \frac{n}{2} + 1; \frac{\frac{1}{2}}{\frac{1}{2}-1}\Bigg)}=2^{n/2}{_2F_1 \Bigl(\frac{n}{2}, -\frac{1}{2}; \frac{n}{2} + 1; -1\Bigr)}.}$$

Substituting this back into $(1)$, we obtain

$$I={\sum_{n=1}^{\infty}\frac{2}{n(2n+1)}{_2F_1 \Bigl(\frac{n}{2}, -\frac{1}{2}; \frac{n}{2} + 1; -1\Bigr)}.}\tag{2}$$

We can rewrite $(2)$ with the series definition for the hypergeometric function.

$${\sum_{n=1}^{\infty}\frac{2}{n(2n+1)}{_2F_1 \Bigl(\frac{n}{2}, -\frac{1}{2}; \frac{n}{2} + 1; -1\Bigr)}=\sum_{n=1}^{\infty}\frac{2}{n(2n+1)}\sum_{k=0}^{\infty}\frac{(-1)^k\left(\frac{n}{2}\right)_k\left(-\frac{1}{2}\right)_k}{\left(\frac{n}{2}+1\right)_k k!}}$$

Here, $(x)_n$ denotes the rising Pochhammer symbol. (There are various other notations for the rising factorial with $n$ as a superscript, but I chose this convention to help distinguish $(-1)^k$ as a unique factor.) It immediately follows from its definition that $\frac{(x)_n}{(x+1)_n}=\frac{x}{x+n}$; hence, $\frac{(n/2)_k}{(n/2+1)_k}=\frac{n}{2k+n}$. We can use this to rewrite $(2)$ even further.

$${\sum_{n=1}^{\infty}\frac{2}{n(2n+1)}\sum_{k=0}^{\infty}\frac{(-1)^k\left(\frac{n}{2}\right)_k\left(-\frac{1}{2}\right)_k}{\left(\frac{n}{2}+1\right)_k k!}=\sum_{n=1}^{\infty}\frac{2}{2n+1}\sum_{k=0}^{\infty}\frac{(-1)^k\left(-\frac{1}{2}\right)_k}{(2k+n)k!}}\tag{3}$$

Since there are no issues with absolute convergence, the order of summation can be swapped.

\begin{align*} \sum_{n=1}^{\infty}\frac{2}{2n+1}\sum_{k=0}^{\infty}\frac{(-1)^k\left(-\frac{1}{2}\right)_k}{(2k+n)k!}&{=\sum_{k=0}^{\infty}\frac{(-1)^k\left(-\frac{1}{2}\right)_k}{k!}\sum_{n=1}^{\infty}\frac{1}{\left(n+\frac{1}{2}\right)\!(n+2k)}}\\ &{=2\sum_{k=0}^{\infty}\frac{(-1)^k\left(-\frac{1}{2}\right)_k}{(4k-1)k!}\sum_{n=1}^{\infty}\left(\frac{1}{n+\frac{1}{2}}-\frac{1}{n+2k}\right)}\tag{4} \end{align*}

With this identity for the digamma function, the inner series in $(4)$ can be expressed as $\psi(2k+1)-\psi(3/2)=\psi(2k+1)+\gamma-2+2\ln{2}$, where $\gamma$ denotes the Euler-Mascheroni constant. Additionally, since $\psi(n+1)+\gamma=H_n$, where $H_n$ is the $n$-th harmonic number, we have $\psi(2k+1)+\gamma-2+2\ln{2}=H_{2k}-2+2\ln{2}$. We can now write $(4)$ as a single sum.

$${2\sum_{k=0}^{\infty}\frac{(-1)^k\left(-\frac{1}{2}\right)_k}{(4k-1)k!}\sum_{n=1}^{\infty}\left(\frac{1}{n+\frac{1}{2}}-\frac{1}{n+2k}\right)=2\sum_{k=0}^{\infty}\frac{(-1)^k\left(-\frac{1}{2}\right)_k}{(4k-1)k!}(H_{2k}-2+2\ln{2})}\label{5}\tag{5}$$

Since $(x)_n:=\frac{\Gamma(x+n)}{\Gamma(x)}$, we have $(-1/2)_k=\frac{\Gamma(k-1/2)}{\Gamma(-1/2)}=-\frac{2\sqrt{\pi}(2k-3)!!}{2^{k+1}\sqrt{\pi}}=-2^{-k}(2k-3)!!$. Hence, we can write

$${2\sum_{k=0}^{\infty}\frac{(-1)^k\left(-\frac{1}{2}\right)_k}{(4k-1)k!}\left(H_{2k}-2+2\ln{2}\right)=-2\sum_{k=0}^{\infty}\frac{\left(-\frac{1}{2}\right)^k(2k-3)!!}{(4k-1)k!}(H_{2k}-2+2\ln{2}).}$$

Jumping back to $\eqref{1}$, we can conclude

\begin{align*} I &= \sum_{n = 1}^{\infty} \frac{2^{1-\frac{n}{2}}}{n(2n + 1)} \, {_2F_1 \Bigl(\frac{n}{2}, \frac{n + 3}{2}; \frac{n}{2} + 1; \frac{1}{2}\Bigr)} \\ &= -2\sum_{k=0}^{\infty}\frac{\left(-\frac{1}{2}\right)^k(2k-3)!!}{(4k-1)k!}(H_{2k}-2+2\ln{2}).\tag{6} \end{align*}

This is as far as I can go, as I'm not well-versed with series involving the harmonic numbers and don't immediately see a way to approach the non-harmonic terms (see below). However, this form looks much more pliable than a hypergeometric. It converges very slowly (it's only accurate to 14 decimal digits after a million terms), but it is equivalent to $I$, as shown.

---

UPDATE: I have found a closed form for the non-harmonic terms and will write my process below.

We want to evaluate

$$4(1-\ln{2})\sum_{k=0}^{\infty}\frac{\left(-\frac{1}{2}\right)^k(2k-3)!!}{(4k-1)k!}.\tag{7}$$

We will write this in terms of the rising factorial to obtain a hypergeometric function again. Note that $\frac{1}{4k-1}=-\frac{-1/4}{k-1/4}=-\frac{(-1/4)_k}{(3/4)_k}$; given that $-2^{-k}(2k-3)!!=(-1/2)_k$ (derived below \eqref{5}), we have

$$\sum_{k=0}^{\infty}\frac{\left(-\frac{1}{2}\right)^k(2k-3)!!}{(4k-1)k!}=\sum_{k=0}^{\infty}\frac{(-1)^k\left(-\frac{1}{2}\right)_k\left(-\frac{1}{4}\right)_k}{\left(\frac{3}{4}\right)_k k!}={}_2F_1\biggl(-\frac{1}{2},-\frac{1}{4};\frac{3}{4};-1\biggr).$$

This is of the form ${}_2F_1(a,b;1+a-b;-1)$, with $a=-1/2$ and $b=-1/4$, so we can use this formula from Kummer to rewrite this as

$${}_2F_1\!\biggl(-\frac{1}{2},-\frac{1}{4};\frac{3}{4};-1\biggr)=\frac{\Gamma\big(1-\frac{1}{2}+\frac{1}{4}\big)\Gamma\big(1-\frac{1}{4}\big)}{\Gamma\big(1-\frac{1}{2}\big)\Gamma\big(1-\frac{1}{4}+\frac{1}{4}\big)}=\frac{\Gamma\big(\frac{3}{4}\big)^2}{\sqrt{\pi}}=\frac{2\pi^{3/2}}{\Gamma\big(\frac{1}{4}\big)^2}.\tag{8}$$

Which form to use comes down to personal preference. We can now return to $(7)$ and write

$$4(1-\ln{2})\sum_{k=0}^{\infty}\frac{\left(-\frac{1}{2}\right)^k(2k-3)!!}{(4k-1)k!}=\frac{8\pi^{3/2}}{\Gamma\big(\frac{1}{4}\big)^2}(1-\ln{2}).$$

Thus, we can conclude

$$\boxed{I=\frac{8\pi^{3/2}}{\Gamma\big(\frac{1}{4}\big)^2}(1-\ln{2})+\sum_{k=1}^{\infty}\frac{\left(-\frac{1}{2}\right)^{k-1}H_{2k}(2k-3)!!}{(4k-1)k!}}\tag{9}$$

(the zeroeth term can be removed since $H_0=0$). The problem has now been reduced to a series involving harmonic numbers. It may be useful to convert it to the rising factorials, as I did above, and try to obtain a closed form that way, but the index of $2k$ has thus far made it resistant to my efforts. There is some literature close to what is needed, e.g. Junesang Choi's and H.M. Srivastava's paper here, but I haven't been able to find anything closer. The identity $H_k+H_{k+1/2}=2H_{2k+1}-2\ln{2}$ strikes me as potentially useful, but the $k+1/2$ index again leads me nowhere. Perhaps someone may have further insight.

Answer 2

Polar integration shows that $$ I = 2 \int_0^{\pi/4} \frac{\operatorname{artanh} (\sqrt{\tan \theta}) - \sqrt{\tan \theta}}{(\cos \theta \sin \theta)^{3/2}} \, d\theta. $$ Using the series expansion $$ \operatorname{artanh} x - x = \sum_{n = 1}^{\infty} \frac{x^{2n + 1}}{2n + 1}, $$ we have \begin{align*} I &= \sum_{n = 1}^{\infty} \frac{2}{2n + 1} \int_0^{\pi/4} \tan^{n - 1} \theta \sec^3 \theta \, d \theta \\ &= \boxed{\sum_{n = 1}^{\infty} \frac{2^{1-\frac{n}{2}}}{n(2n + 1)} \, {_2F_1 \Bigl(\frac{n}{2}, \frac{n + 3}{2}; \frac{n}{2} + 1; \frac{1}{2}\Bigr)}}, \end{align*} where $_2F_1(a, b; c; z)$ is the hypergeometric function.

---

This is the closest I can get to a closed form. Experimentation with Wolfram Mathematica (attached below) seems to show that \begin{align*} f_n &= {_2F_1 \Bigl(\frac{n}{2}, \frac{n + 3}{2}; \frac{n}{2} + 1; \frac{1}{2}\Bigr)} \\ &= \begin{cases} a_m \sqrt{2} \operatorname{arsinh}(1) + b_m & \text{for } n = 2m - 1 \\ c_m \sqrt{2} + d_m & \text{for } n = 2m \end{cases} \end{align*} for $a_m, b_m, c_m, d_m \in \mathbb{Q}$. FindSequenceFunction returns \begin{align*} a_m &= \frac{(-1)^{m + 1} 2^{m - 2} \bigl( \frac{3}{2} \bigr)^{\overline{m - 1}}}{m!}, \\ d_m &= \frac{(-2)^m m!}{\bigl( \frac{3}{2} \bigr)^{\overline{m}}}, \end{align*} where $$ x^{\overline{n}} = x(x + 1)(x + 2) \dotsm (x + n - 1) $$ is the rising factorial (a.k.a. Pochhammer symbol). Similarly, FindGeneratingFunction returns \begin{align*} \sum_{m = 1}^{\infty} a_m x^m &= \frac{1}{2} - \frac{1}{2 \sqrt{1 + 2x}}, \\ \sum_{m = 1}^{\infty} d_m x^m &= \frac{\operatorname{arsinh} \sqrt{2x}}{\sqrt{2x(2x + 1)}} - 1. \end{align*} For $b_m$ and $c_m$, FindSequenceFunction and FindGeneratingFunction either freeze or refer back to $_2F_1$.

---

$$ \begin{array}{cc} n & f_n \\ \hline 1 & \frac{1}{2} \left(2+\sqrt{2} \operatorname{arsinh}(1)\right) \\ 2 & \frac{4}{3} \left(2 \sqrt{2}-1\right) \\ 3 & -\frac{3}{4} \left(\sqrt{2} \operatorname{arsinh}(1)-6\right) \\ 4 & \frac{32}{15} \left(1+\sqrt{2}\right) \\ 5 & \frac{5}{12} \left(14+3 \sqrt{2} \operatorname{arsinh}(1)\right) \\ 6 & \frac{32}{35} \left(11 \sqrt{2}-4\right) \\ 7 & -\frac{7}{48} \left(15 \sqrt{2} \operatorname{arsinh}(1)-122\right) \\ 8 & \frac{256}{315} \left(8+13 \sqrt{2}\right) \\ 9 & \frac{3}{80} \left(682+105 \sqrt{2} \operatorname{arsinh}(1)\right) \\ 10 & \frac{128}{693} \left(211 \sqrt{2}-64\right) \\ 11 & -\frac{11}{480} \left(315 \sqrt{2} \operatorname{arsinh}(1)-3074\right) \\ 12 & \frac{512 \left(128+271 \sqrt{2}\right)}{3003} \\ 13 & \frac{13 \left(27626+3465 \sqrt{2} \operatorname{arsinh}(1)\right)}{3360} \\ 14 & \frac{512 \left(1919 \sqrt{2}-512\right)}{6435} \\ 15 & \frac{501022-45045 \sqrt{2} \operatorname{arsinh}(1)}{1792} \\ 16 & \frac{8192 \left(1024+2597 \sqrt{2}\right)}{109395} \\ 17 & \frac{187 \left(37862+4095 \sqrt{2} \operatorname{arsinh}(1)\right)}{16128} \\ 18 & \frac{2048 \left(67843 \sqrt{2}-16384\right)}{230945} \\ 19 & -\frac{19 \left(765765 \sqrt{2} \operatorname{arsinh}(1)-9434878\right)}{161280} \\ 20 & \frac{8192 \left(32768+95259 \sqrt{2}\right)}{969969} \\ 21 & \frac{13 \left(11618366+1119195 \sqrt{2} \operatorname{arsinh}(1)\right)}{84480} \\ 22 & \frac{8192 \left(588933 \sqrt{2}-131072\right)}{2028117} \\ 23 & -\frac{23 \left(14549535 \sqrt{2} \operatorname{arsinh}(1)-194991322\right)}{1013760} \\ 24 & \frac{65536 \left(262144+850251 \sqrt{2}\right)}{16900975} \\ 25 & \frac{5 \left(3819921514+334639305 \sqrt{2} \operatorname{arsinh}(1)\right)}{2635776} \\ 26 & \frac{32768 \left(10098967 \sqrt{2}-2097152\right)}{35102025} \\ 27 & -\frac{3 \left(1673196525 \sqrt{2} \operatorname{arsinh}(1)-24084946414\right)}{4100096} \\ 28 & \frac{131072 \left(4194304+14904091 \sqrt{2}\right)}{145422675} \\ 29 & \frac{29 \left(62097106486+5019589575 \sqrt{2} \operatorname{arsinh}(1)\right)}{61501440} \\ 30 & \frac{131072 \left(85806311 \sqrt{2}-16777216\right)}{300540195} \\ 31 & -\frac{31 \left(145568097675 \sqrt{2} \operatorname{arsinh}(1)-2229742283746\right)}{984023040} \\ 32 & \frac{4194304 \left(33554432+128927573 \sqrt{2}\right)}{9917826435} \\ 33 & \frac{19 \left(3150308268566+237505843575 \sqrt{2} \operatorname{arsinh}(1)\right)}{506920960} \\ 34 & \frac{524288 \left(5792144099 \sqrt{2}-1073741824\right)}{20419054425} \\ 35 & \frac{73030431035486-4512611027925 \sqrt{2} \operatorname{arsinh}(1)}{260702208} \\ 36 & \frac{2097152 \left(2147483648+8834766227 \sqrt{2}\right)}{83945001525} \\ 37 & \frac{259 \left(9093229741774+644658718275 \sqrt{2} \operatorname{arsinh}(1)\right)}{4953341952} \\ 38 & \frac{2097152 \left(48605936617 \sqrt{2}-8589934592\right)}{172308161025} \\ 39 & \frac{2838808987540886-166966608033225 \sqrt{2} \operatorname{arsinh}(1)}{2540175360} \\ 40 & \frac{16777216 \left(17179869184+75096287791 \sqrt{2}\right)}{1412926920405} \\ 41 & \frac{41 \left(2488324849033834+166966608033225 \sqrt{2} \operatorname{arsinh}(1)\right)}{53343682560} \\ 42 & \frac{8388608 \left(812156618077 \sqrt{2}-137438953472\right)}{2893136075115} \\ 43 & -\frac{43 \left(6845630929362225 \sqrt{2} \operatorname{arsinh}(1)-121718302325751046\right)}{117356101 6320} \\ 44 & \frac{33554432 \left(274877906944+1268822838961 \sqrt{2}\right)}{11835556670925} \\ 45 & \frac{4610656329982787582+294362129962575675 \sqrt{2} \operatorname{arsinh}(1)}{599820075008} \\ 46 & \frac{33554432 \left(6760265315081 \sqrt{2}-1099511627776\right)}{24185702762325 } \\ 47 & -\frac{423 \left(10902301109725025 \sqrt{2} \operatorname{arsinh}(1)-201950171465282598\right)}{479856060 0064} \\ 48 & \frac{536870912 \left(2199023255552+10665172132163 \sqrt{2}\right)}{395033145117975} \\ 49 & \frac{7 \left(75588052822912144694+461167336941368557 5 \sqrt{2} \operatorname{arsinh}(1)\right)}{17137716428800} \\ 50 & 33554432 \, _2F_1\left(-\frac{1}{2},25;26;-1\right) \\ \end{array} $$

Answer 3

I will contribute a simpler way to derive the harmonic sums and the following closed form: $$\int_0^1 \int_0^1 \frac{\sqrt{x^2+y^2}}{1-xy} dxdy = 2\left(-5-\sqrt{2} + \operatorname{arcsinh}(1)\right.$$ $$\left.+\frac{2\pi^{3/2}(-8+\operatorname{arcsinh}(1))}{\Gamma \left(\frac14\right)^2}+8\times 2^{1/4} \mathcal{H}_1 + 4\sqrt{\sqrt{2}-1} \mathcal{H}_2\right)$$ $$=1.4843079462309457237929418228902024\!\ldots$$ where $$\mathcal{H}_1 = {}_3F_2\left(-\frac14,-\frac14,\frac14;\,\frac34,\frac34;\,\frac12\right) = \frac{1}{\Gamma \left(\frac14\right)}\sum_{k=0}^\infty \frac{\Gamma \left(k+\frac14\right)}{2^k k! (4k-1)^2} $$ $$\mathcal{H}_2 = {}_3F_2\left(\frac14,\frac12,\frac12;\,\frac32,\frac32;\,2(\sqrt{2}-1)\right) = \frac{1}{\Gamma \left(\frac14\right)}\sum_{k=0}^\infty \frac{\Gamma \left(k+\frac14\right)2^k(\sqrt{2}-1)^k}{k! (2k+1)^2}$$ Since the integrand is invariant with respect to exchange $x \leftrightarrow y$ we can only consider the case $x \leq y$. Also, by the change of variables $v^2=xy$, $u^2=y/x$ and integrating with respect to $v$ we can reduce the dimension of the integral immediately: $$\int_0^1 \int_0^1 \frac{\sqrt{x^2+y^2}dxdy}{1-xy} = 2\int_{x=0}^1 \int_{y=0}^x \frac{\sqrt{x^2+y^2}dxdy}{1-xy}$$ $$=4\int_{u=0}^1 \int_{v=0}^u \frac{\sqrt{1+u^4}}{1-v^2} \frac{v^2 du dv}{u^2}=4\int_0^1 \frac{\sqrt{1+u^4}}{u^2} (\operatorname{arctanh} u - u) du$$ Next, we expand the integrand $$=4\int_0^1 u \left[ \begin{pmatrix}1/2\\0\end{pmatrix} + \begin{pmatrix}1/2\\1\end{pmatrix} u^4 + \begin{pmatrix}1/2\\2\end{pmatrix} u^8 + \cdots\right] \left[ \frac13 + \frac{u^2}{5} + \frac{u^4}{7} + \cdots \right] du$$ $$=2\sum_{k=0}^\infty \begin{pmatrix}1/2\\k\end{pmatrix} \sum_{j=0}^{\infty} \frac{1}{(2j+3)(j+2k+1)} = 2\sum_{k=0}^\infty \begin{pmatrix}1/2\\k\end{pmatrix} \frac{1}{4k-1} \sum_{j=0}^\infty \left( \frac{2}{2j+3} - \frac{1}{j+2k+1}\right)$$ $$=2\sum_{k=0}^\infty \begin{pmatrix}1/2\\k\end{pmatrix} \frac{\psi(2k+1) - \psi(3/2)}{4k-1} = 2\sum_{k=0}^\infty \begin{pmatrix}1/2\\k\end{pmatrix} \frac{H_{2k}}{4k-1} - 2\psi(3/2) \sum_{k=0}^\infty \begin{pmatrix}1/2\\k\end{pmatrix} \frac{1}{4k-1}$$ The first sum can be decomposed into $$\sum_{k=0}^\infty \begin{pmatrix}1/2\\k\end{pmatrix} \frac{H_{2k}}{4k-1} = \sum_{k=0}^\infty \frac{(-1)^{k+1}}{4^k(2k-1)} \begin{pmatrix}2k\\k\end{pmatrix} \frac{H_{2k}}{4k-1} = -\underbrace{\sum_{k=0}^\infty \frac{(-1)^{k}}{4^k} \begin{pmatrix}2k\\k\end{pmatrix} \frac{H_{2k}}{2k-1}}_{S_1} + 2\underbrace{\sum_{k=0}^\infty \frac{(-1)^{k}}{4^k} \begin{pmatrix}2k\\k\end{pmatrix} \frac{H_{2k}}{4k-1}}_{{S}_2}$$ where $S_1$ and $S_2$ are evaluated in another question.

The second sum is relatively easy to find: $$\sum_{k=0}^\infty \begin{pmatrix}1/2\\k\end{pmatrix} \frac{1}{4k-1} = {}_2F_1 \left(-\frac12,-\frac14;\,\frac34;\,-1\right) = \frac{2\pi^{3/2}}{\Gamma \left(\frac14\right)^2}$$