<s>[INST] 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 $\log x =\log_e x$.
 [/INST] 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 $$\sum_{k=1}^{\infty} \frac{1}{k(k+1)^2} = \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1} - \frac{1}{(k+1)^2} \right)$$ $$ = \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right) - \sum_{k=1}^{\infty} \frac{1}{(k+1)^2} = 1 - \left( \frac{\pi^2}{6} - 1\right) = 2-\frac{\pi^2}{6}$$
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