Prove: $\int_0^1 \int_0^1 \frac{\ln{\left(2+yx\right)}}{1+yx} \; \mathrm{d}y\; \mathrm{d}x = \frac{13}{24} \zeta(3)$

Question

Prove: $$\int_0^1 \int_0^1 \frac{\ln{\left(2+yx\right)}}{1+yx} \; \mathrm{d}y\; \mathrm{d}x = \frac{13}{24} \zeta(3)$$ My attempt: \begin{align} \int_0^1 \int_0^1 \frac{\ln{\left(1+(1+yx)\right)}}{1+yx} \; \mathrm{d}y\; \mathrm{d}x \\ \int_0^1 \int_0^1 \sum_{n=1}^{\infty} \frac{{(-1)}^n{(1+xy)}^{n-1}}{n} \; \mathrm{d}y\; \mathrm{d}x \\ \sum_{n=1}^{\infty} \frac{{(-1)}^n}{n} \int_0^1 \int_0^1 {(1+xy)}^{n-1}\; \mathrm{d}y\; \mathrm{d}x \\ \sum_{n=1}^{\infty} \frac{{(-1)}^n}{n} \int_0^1 \frac{(y+1)^n-1}{yn}\; \mathrm{d}y \\ \sum_{n=1}^{\infty} \frac{{(-1)}^n}{n} \int_0^1 \sum_{i=1}^{n} \frac{{n \choose i} y^{n-1}}{n}\; \mathrm{d}y \\ \sum_{n=1}^{\infty} \frac{{(-1)}^n}{n} \sum_{i=1}^{n} \frac{{n \choose i}}{n}\int_0^1 y^{n-1}\; \mathrm{d}y \\ \sum_{n=1}^{\infty} \frac{{(-1)}^n}{n^3} \sum_{i=1}^{n} {n \choose i} \\ \sum_{n=1}^{\infty} \frac{{(-1)}^n\left(2^n-1\right)}{n^3} \\ \end{align}

This diverges. I realize now that this is because $1+xy$ here is outside of the radius of convergence.

Credit to @Varun Vejalla: The double integral equals $$-\sum_{m=1}^{\infty} \frac{1}{m \cdot 2^m} \sum_{n=1}^m \frac{{(-1)}^n}{n^2}$$ How can we prove this equals the result?

Answer 1

$$\int_0^1\int_0^1\frac{\ln(2+xy)}{1+xy}dydx=\int_0^1\int_0^y\frac{\ln(2+t)}{y(1+t)}dydt$$

$$=\int_0^1\frac{\ln(2+t)}{1+t}\left(\int_t^1\frac{dy}{y}\right)dt=-\int_0^1\frac{\ln t\ln(2+t)}{1+t}dt\overset{t=\frac{1-x}{x}}{=}-\int_{1/2}^1\frac{\ln(\frac{1-x}{x})\ln(\frac{1+x}{x})}{x}dx$$

$$=\underbrace{\int_{1/2}^1\frac{\ln x\ln(1+x)}{x}dx}_{I_1}+\underbrace{\int_{1/2}^1\frac{\ln x\ln(1-x)}{x}dx}_{I_2}-\underbrace{\int_{1/2}^1\frac{\ln^2x}{x}dx}_{I_3}-\underbrace{\int_{1/2}^1\frac{\ln(1-x)\ln(1+x)}{x}dx}_{I_4}$$

By integration by parts we have

$$I_1=-\text{Li}_2(-x)\ln x|_{1/2}^1+ \int_{1/2}^1\frac{\text{Li}_2(-x)}{x}dx=\boxed{-\ln2\text{Li}_2\left(-\frac12\right)-\text{Li}_3\left(-\frac12\right)-\frac34\zeta(3)}$$

$$I_2=-\text{Li}_2(x)\ln x|_{1/2}^1+ \int_{1/2}^1\frac{\text{Li}_2(x)}{x}dx=\boxed{\frac13\ln^32+\frac18\zeta(3)}$$

$$I_3=\boxed{\frac13\ln^32}$$

For $I_4$, by using the algebraic identity

$$ab=\frac14(a+b)^2-\frac14(a-b)^2$$

we have

$$I_4=\frac14\underbrace{\int_{1/2}^1\frac{\ln^2(1-x^2)}{x}dx}_{x^2\to x}-\frac14\underbrace{\int_{1/2}^1\frac{\ln^2(\frac{1-x}{1+x})}{x}dx}_{(1-x)/(1+x)\to x}$$

$$=\frac18\int_{1/4}^1\frac{\ln^2(1-x)}{x}dx-\frac12\int_0^{1/3}\frac{\ln^2x}{1-x^2}dx$$

Since

\begin{align} \int\frac{\ln^2(1-x)}{x}\ dx&=\ln(x)\ln^2(1-x)+2\int\frac{\ln(x)\ln(1-x)}{1-x}\ dx\nonumber\\ &=\ln(x)\ln^2(1-x)+2\left(\ln(1-x)\text{Li}_2(1-x)+\int\frac{\text{Li}_2(1-x)}{1-x}\ dx\right)\nonumber\\ &=\ln(x)\ln^2(1-x)+2\ln(1-x)\text{Li}_2(1-x)-2\text{Li}_3(1-x)+C. \end{align}

we have

$$\int_{1/4}^1\frac{\ln^2(1-x)}{x}dx=2\ln(2)\ln^2\left(\frac34\right)-2\ln\left(\frac34\right)\text{Li}_2\left(\frac34\right)+2\text{Li}_3\left(\frac34\right)$$

and since

$$\int\frac{\ln^2x}{1-x^2}dx=\text{Li}_3(x)-\text{Li}_3(-x)-\ln x(\text{Li}_2(x)-\text{Li}_2(-x))-\frac12\ln^2x\ln\left(\frac{1-x}{1+x}\right)$$

we have

$$\int_0^{1/3}\frac{\ln^2x}{1-x^2}dx=\text{Li}_3\left(\frac13\right)-\text{Li}_3\left(-\frac13\right)+\ln3\left(\text{Li}_2\left(\frac13\right)-\text{Li}_2\left(-\frac13\right)\right)+\frac12\ln^23\ln2$$

Collect the two integrals we have

$$I_4=\boxed{\ln^32-\ln^22\ln3-\frac14\ln\left(\frac34\right)\text{Li}_2\left(\frac34\right)+\frac14\text{Li}_3\left(\frac34\right)-\frac12\text{Li}_3\left(\frac13\right)+\frac12\text{Li}_3\left(-\frac13\right)}$$

$$\boxed{-\frac12\ln3\left(\text{Li}_2\left(\frac13\right)-\text{Li}_2\left(-\frac13\right)\right)}$$

and whats left is only simplification.

Answer 2

Here's my solution. It's somewhat similar to what Ali Shather has done. Let's start by reducing the double integral into a single integral. $$ \begin{align*} \int_0^1 \int_0^1 \frac{\log(2+xy)}{1+xy}dy\; dx &= \int_0^1 \frac{1}{x}\int_{-1}^{-1-x}\frac{\log(1-y)}{y}dy\; dx\quad (y\mapsto -(1+y)/x) \\ &= \int_0^1 \frac{\text{Li}_2(-1)-\text{Li}_2(-1-x)}{x}dx \\ &= \log(x)\left[ \text{Li}_2(-1)-\text{Li}_2(-1-x)\right]\Big|_0^1 - \int_0^1 \frac{\log(x)\log(2+x)}{1+x}dx \quad (\text{IBP})\\ &= - \int_0^1 \frac{\log(x)\log(2+x)}{1+x}dx \end{align*} $$ Let $I = \int_0^1 \frac{\log(x)\log(2+x)}{1+x}dx$. With the help of some algebra, we can write: $$I = -\frac{1}{2}\int_0^1 \frac{\log^2\left( \frac{x}{2+x}\right)}{1+x}dx + \frac{1}{2}\int_0^1 \frac{\log^2(x)}{1+x}dx + \frac{1}{2}\int_0^1 \frac{\log^2(2+x)}{1+x}dx \tag{1}$$ The leftmost integral can be dealt with the substitution $x \mapsto \frac{2x}{1-x}$. This gives us: \begin{align*} \int_0^1 \frac{\log^2\left( \frac{x}{2+x}\right)}{1+x}dx &= 2 \int_0^{\frac{1}{3}} \frac{\log^2(x)}{1-x^2}dx \\ &= 2\sum_{n=0}^\infty \int_0^{1\over 3} x^{2n} \log^2(x) \; dx \\ &= 2\sum_{n=0}^\infty \frac{1}{3^{2n+1}}\left\{\frac{2}{(2n+1)^3} + \frac{2\log(3)}{(2n+1)^2}+ \frac{\log^2(3)}{2n+1}\right\} \\ &= 2\left[\text{Li}_3\left(\frac{1}{3} \right) -\text{Li}_3\left(-\frac{1}{3} \right)\right] + 2\log(3)\left[\text{Li}_2\left(\frac{1}{3} \right) - \text{Li}_2\left(-\frac{1}{3} \right)\right] \\ &\quad +\log^2(3)\log(2)\tag{2} \end{align*} For the middle integral, we have: \begin{align*} \int_0^1 \frac{\log^2(x)}{1+x}dx &= \sum_{n=0}^\infty (-1)^{n}\int_0^1 x^{n}\log^2(x)\; dx \\ &= 2\sum_{n=0}^\infty \frac{(-1)^{n}}{(1+n)^3} \\ &= \frac{3\zeta(3)}{2} \tag{3} \end{align*} In the rightmost integral make the substitution $x\mapsto \frac{1-2x}{x}$: \begin{align*} \int_0^1 \frac{\log^2(2+x)}{1+x}dx &= \int_{1\over 3}^{1\over 2}\frac{\log^2(x)}{x(1-x)}dx \\ &= \int_{1\over 3}^{1\over 2} \log^2(x)\left(\frac{1}{x}+\frac{1}{1-x} \right)dx \\ &= \frac{\log^3(3)-\log^3(2)}{3}+\sum_{n=0}^\infty \int_{1\over 3}^{1\over 2} x^n \log^2(x) dx \\ &= \frac{\log^3(3)-\log^3(2)}{3} + \sum_{n=0}^\infty \frac{1}{2^{n+1}}\left(\frac{2}{(n+1)^3} +\frac{2\log(2)}{(n+1)^2}+\frac{\log^2(2)}{n+1}\right)\\ &\quad - \sum_{n=0}^\infty \frac{1}{3^{n+1}}\left(\frac{2}{(n+1)^3} +\frac{2\log(3)}{(n+1)^2}+\frac{\log^2(3)}{n+1}\right) \\ &= \frac{-2\log^3(3)+2\log^3(2)}{3} + 2\text{Li}_3\left(\frac{1}{2} \right)+2\log(2) \text{Li}_2\left(\frac{1}{2} \right) \\\ &\quad - 2\text{Li}_3\left(\frac{1}{3} \right)-2\log(3) \text{Li}_2\left(\frac{1}{3} \right)+\log^2(3)\log(2) \tag{4} \end{align*} Plugging the results equations (2), (3), and (4) into (1) gives: $$ \begin{align*} I &= -\left[2\text{Li}_3\left(\frac{1}{3} \right)-\text{Li}_3\left(-\frac{1}{3} \right)\right] - \log(3)\left[ 2\text{Li}_2\left(\frac{1}{3} \right)-\text{Li}_2\left(-\frac{1}{3} \right)\right] + \text{Li}_3\left(\frac{1}{2} \right) \\ &\quad +\log(2)\text{Li}_2\left(\frac{1}{2} \right) + \frac{\log^3(2)-\log^3(3)}{3} +\frac{3\zeta(3)}{4}\tag{5} \end{align*} $$ To simplify equation (5), we need the following polylogarithm identities: \begin{align*} \text{Li}_2\left(\frac{1}{2}\right) &= \frac{\pi^2}{12}-\frac{\log^2(2)}{2} \tag{6} \\ \text{Li}_3\left(\frac{1}{2}\right) &= -\frac{\pi^2 \log(2)}{12}+\frac{\log^3(2)}{6}+\frac{7}{8}\zeta(3)\tag{7}\\ 2\text{Li}_2\left(\frac{1}{3} \right)-\text{Li}_2\left(-\frac{1}{3} \right) &= \frac{\pi^2}{6}-\frac{1}{2}\log^2(3) \tag{8}\\ 2\text{Li}_3\left(\frac{1}{3} \right)-\text{Li}_3\left(-\frac{1}{3} \right) &= \frac{\log^3(3)}{6}-\frac{\pi^2}{6}\log(3)+\frac{13\zeta(3)}{6} \tag{9} \end{align*} Equation (9) was derived in this thread by the user @Start wearing purple. The others can also be derived with the help of polylogarithm functional equations. After plugging the above equations into (5), a lot of terms cancel out and we get: $$I=-\frac{13}{24}\zeta(3)$$ Hence, the original double integral is equal to $ \frac{13}{24}\zeta(3)$.