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?
Best Answer
$$\int_0^1\int_0^1\frac{\ln(2+xy)}{1+xy}dydx=\int_0^1\int_0^y\frac{\ln(2+t)}{y(1+t)}dydt$$
$$=\int_0^1\frac{\ln(2+t)}{1+t}\left(\int_t^1\frac{dy}{y}\right)dt=-\int_0^1\frac{\ln t\ln(2+t)}{1+t}dt\overset{t=\frac{1-x}{x}}{=}-\int_{1/2}^1\frac{\ln(\frac{1-x}{x})\ln(\frac{1+x}{x})}{x}dx$$
$$=\underbrace{\int_{1/2}^1\frac{\ln x\ln(1+x)}{x}dx}_{I_1}+\underbrace{\int_{1/2}^1\frac{\ln x\ln(1-x)}{x}dx}_{I_2}-\underbrace{\int_{1/2}^1\frac{\ln^2x}{x}dx}_{I_3}-\underbrace{\int_{1/2}^1\frac{\ln(1-x)\ln(1+x)}{x}dx}_{I_4}$$
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.