# Evaluation of a tough double integral

This is an integral coming from personal research, and very important to me, but it does not
seem an easy job to do. If a solution is not possible then I'd be glad with a closed form only.

$$\int_{[0,1]^2} \frac{(1-x-y+x y+x \log(x)-x y\log(x)+y \log(y)- x y\log(y)+x y\log(x)\log(y))\log(1+x y)}{x y (1-x) (1-y)\log(x)\log(y)} \ dx \ dy$$

Hopefully this will be seen by Cleo too and maybe we'll find its closed form.

Write the integral as $$\int_{[0,1]^2}h(x)h(y)\log(1+xy)\,dx\,dy = \sum_{n\geq1}(-1)^{n+1}\frac{a_n^2}{n}, \qquad h(x) = \frac{1-x+x\log x}{x(1-x)\log x},$$ where the coefficients $a_n$ are given by $$a_n = \int_0^1 h(x)x^n\,dx = \gamma - H_n + \log n.$$

Using the sums $$\sum_{n\geq1} \frac{(-1)^{n+1}}{n} = \log2$$ $$\sum_{n\geq1} \frac{H_n(-1)^{n+1}}{n} = \tfrac12\big(\zeta(2)-\log^22\big),$$ $$\sum_{n\geq1}\frac{H_n^2(-1)^{n+1}}{n} = -\tfrac12\zeta(2)\log2+\tfrac13\log^22+\tfrac34\zeta(3),$$ $$\sum_{n\geq1}\frac{(-1)^{n+1}\log n}{n} = -\gamma\log2+\tfrac12\log^22,$$ $$\sum\frac{(-1)^{n+1}\log^2n}{n} = -\gamma\log^22+\tfrac13\log^32-2\gamma_1\log2,$$ where $\gamma_1$ is a Stieltjes gamma constant, and noting that the sum $$\sum_{n\geq 1}(-1)^{n+1}\frac{H_n\log n}{n} = \int_0^1 \frac{du}{u}\big(\lambda_1(u-1)-\lambda_1(-1)\big), \qquad \lambda_s(t) = \frac{\partial \mathrm{Li}_s}{\partial s}(t)$$ has no closed form at all, the integral can be written as $$-\gamma\zeta(2)-\gamma^2\log2-\tfrac12\zeta(2)\log2+\gamma\log^22+\tfrac23\log^32-2\gamma_1\log2\\+\tfrac34\zeta(3) - 2\sum_{n\geq1}\frac{(-1)^{n+1}H_n\log n}{n}.$$

• The last series is as least as tough as the integral I posted (unfortunately). – user 1357113 Oct 31 '14 at 16:27
• @Chris'ssis Isn't it the same sum as in one of your bountied questions? It doesn't have a closed form. – Kirill Oct 31 '14 at 16:28
• Yeah, I just saw it gets reduced to that series. Are you sure that series doesn't have a closed form? – user 1357113 Oct 31 '14 at 16:29
• @Chris'ssis Quite sure. robjohn's answer is exactly how far I got, and the only comparable sums I could find in the literature were left unevaluated. – Kirill Oct 31 '14 at 16:31

I can simplify the integrand into;

$$\frac{(y\log y + (1-y))(x \log x +(1-x)}{(1-x)(1-y)}$$

The computation of $$\int_{0}^{1} \int_{0}^{1} \frac{(y\log y + (1-y))(x \log x +(1-x))}{(1-x)(1-y)} dx dy$$ is proving to be tougher than expected!

• Are they numerically the same? – user 1357113 Oct 31 '14 at 16:18
• @Chris's $\int _0^1\int _0^1\frac{(-x+x \log (x)+1) (-y+y \log (y)+1) \log (x y+1)}{(x-1) x (y-1) y \log (x) \log (y)}dydx$ is. – user142198 Nov 1 '14 at 9:23
• @Chris'ssis But the above most certainly is not. Yours: $0.15825342193404878562153727387791277452748654970709833318847300007209\dots$ His: $=\frac{1}{36} \left(\pi ^2-12\right)^2$ – user142198 Nov 1 '14 at 9:31