The task is to evaluate, using the specific given hint, and only real-analysis methods, no complex analysis is allowed in the problem. $$\int_0^1 \log^2(x)\log^2(1-x)\, dx$$ The hint given is: $$ \sum_{k=1}^{\infty} (H_k)(x^k) = (-)\frac{\log(1-x)}{1-x},$$ where $H_k$ is the harmonic number.
If we integrate both sides we get. $$ (2)\cdot\sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{(k+1)} = \log^2(1-x)$$ Recalling the hint: $$\sum_{k=1}^{\infty} (H_k)(x^k) = (-)\frac{\log(1-x)}{1-x}$$ Substitute $x \to 1-x$ $$\sum_{k=1}^{\infty} (H_k)((1-x)^k) = \frac{-\log(x)}{x}$$ Integrate both sides to get: $$(2)\cdot\sum_{k=1}^{\infty} (H_k)\frac{((1-x)^{k+1})}{k+1} = \log^2(x)$$
I think it is a good idea to evaluate the sums differently.
Or simply, $$\log^2(x)\log^2(1-x) = \left[(2)\cdot\sum_{k=1}^{\infty} (H_k)\frac{((1-x)^{k+1})}{k+1}\right]\left[(2)\cdot\sum_{k=1}^{\infty} \frac{(H_k)(x^{k+1})}{(k+1)}\right]$$
But that would result in quadruple sums... Any ideas?
