Solve $\int_{0}^{1} \log(x)\log(1-x) dx$ without convolution Maybe it's too much to ask for, but is there a way to solve $\int \limits_{0}^{1} \log(x)\log(1-x) dx$ without convolution?
Note that $\log x =\log_e x$.
 A: Step 1: We prove that for each $n\in \mathbb{N}$, $$\int_{0}^{1} x^n \log(x) \, dx = -\frac{1}{(n+1)^2}$$ Starting with the identity $$\int_{0}^{1} x^n \, dx = \frac{1}{n+1}$$ differentiate both sides with respect to $n$ to get the desired result.
Step 2: Expanding $\log(1-x)$ as its Taylor series, we have $$\log(x)\log(1-x) = -x\log(x) - \frac{x^2 \log(x)}{2} - \frac{x^3 \log(x)}{3} - \cdots$$ Sweeping issues of convergence under the rug, integrating both sides over the interval $[0,1]$ gives $$\int_{0}^{1} \log(x) \log(1-x) \, dx = \frac{1}{2^2} + \frac{1}{2\cdot 3^2} + \frac{1}{3\cdot 4^2} + \cdots$$
Step 3: We must evaluate the sum $$\sum_{k=1}^{\infty} \frac{1}{k(k+1)^2} = \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1} - \frac{1}{(k+1)^2} \right)$$ $$ = \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right) - \sum_{k=1}^{\infty} \frac{1}{(k+1)^2} = 1 - \left( \frac{\pi^2}{6} - 1\right) = 2-\frac{\pi^2}{6}$$
A: Integrating by parts we find:
$$\begin{align}
\mathcal{I}
&=\int_{0}^{1}\ln{\left(x\right)}\ln{\left(1-x\right)}\,\mathrm{d}x\\
&=\left[x\ln{\left(x\right)}\ln{\left(1-x\right)}\right]_{0}^{1}-\int_{0}^{1}x\cdot\frac{d}{dx}\left[\ln{\left(x\right)}\ln{\left(1-x\right)}\right]\,\mathrm{d}x\\
&=-\int_{0}^{1}x\left[\frac{\ln{\left(1-x\right)}}{x}-\frac{\ln{\left(x\right)}}{1-x}\right]\,\mathrm{d}x\\
&=\int_{0}^{1}\left[\frac{x\ln{\left(x\right)}}{1-x}-\ln{\left(1-x\right)}\right]\,\mathrm{d}x\\
&=\int_{0}^{1}\frac{x\ln{\left(x\right)}}{1-x}\,\mathrm{d}x-\int_{0}^{1}\ln{\left(1-x\right)}\,\mathrm{d}x\\
&=\int_{0}^{1}\frac{\ln{\left(x\right)}}{1-x}\,\mathrm{d}x-\int_{0}^{1}\ln{\left(x\right)}\,\mathrm{d}x-\int_{0}^{1}\ln{\left(1-x\right)}\,\mathrm{d}x\\
&=\int_{0}^{1}\frac{\ln{\left(x\right)}}{1-x}\,\mathrm{d}x-2\int_{0}^{1}\ln{\left(x\right)}\,\mathrm{d}x\\
&=2+\int_{0}^{1}\frac{\ln{\left(x\right)}}{1-x}\,\mathrm{d}x\\
&=2+\int_{0}^{1}\frac{\ln{\left(1-t\right)}}{t}\,\mathrm{d}t;~~~\small{\left[1-x=t\right]}\\
&=2-\operatorname{Li}_{2}{\left(1\right)}\\
&=2-\frac{\pi^2}{6},\\
\end{align}$$
where $\operatorname{Li}_{2}{\left(z\right)}$ is the dilogarithm function.
