Prove that $$I:=\int_0^1 \ln(x+1) \ln(x)dx=2-2\ln(2)-\frac{\pi^2}{12}$$
I've tried integration by parts and u substitution and cannot make it work but I have one potential path
I use
$$\ln(x+1)=\sum^{\infty}_{n=1} \frac{(-1)^{n-1}x^n}{n} $$ $$\Rightarrow I=\int_0^1 \ln(x)\sum^{\infty}_{n=1} \frac{(-1)^{n-1}x^n}{n}dx=\sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{n} \int_0^1 x^{n-1} \ln(x)dx= $$
How does one proceed from here? Thank you for your time
Using your route $$\begin{align} \int_0^1 \ln(x)\ln(1+x)\, dx &= \int_0^1\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}x^n\ln x\,\mathrm dx \\ &= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\int_0^1 x^n\ln x\,\mathrm dx \\ &= \sum_{n=1}^\infty \frac{(-1)^n}{n(n+1)^2}\\ &= \sum_{n=1}^\infty \frac{(-1)^n}{n}-\frac{(-1)^n}{n+1}-\frac{(-1)^n}{(n+1)^2} \\ &= 2 -2 \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} -\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2} \qquad \text{re-indexing} \\ &= 2-2\ln 2 -\frac{\pi^2}{12}\end{align}$$
You can use monotone convergence theorem to justify the interchange of summation and integration.
$ \newcommand{\I}{\mathcal{J}} \newcommand{\d}{\mathrm{d}} $First, let's try to find
$$\I_n := \int_0^1 x^n \ln(x) \, \d x$$
(which is what you meant to write in the integral). Integration by parts (differentiating $\ln(x)$ and integrating $x^n$) gives us
$$\I_n = \frac{1}{n+1} x^n \ln(x) \bigg|_0^1- \frac{1}{n+1} \int_0^1 x^{n} \, \d x$$
To evaluate this properly, we need
$$L := \lim_{x \to 0^+} x^n \ln(x)$$
Using L'Hopital's rule suffices:
$$L = \lim_{x \to 0^+} \frac{\ln(x)}{x^{-n}} = \lim_{x \to 0^+} \frac{1/x}{-nx^{-n-1}} = \lim_{x \to 0^+} - \frac 1 n x^n = 0$$
Thus, we readily can see that (by finishing the evaluation and then evaluating the remaining trivial integral),
$$\I_n =- \frac{1}{(n+1)^2}$$
Now we have that your integral is equal to $$I = \sum_{n=1}^\infty \frac{(-1)^{n}}{n(n+1)^2}$$
A simple decomposition of the summands then gives
$$I = \sum_{n=1}^\infty \left( \frac{(-1)^{n}}{n} - \frac{(-1) ^{n}}{n+1} - \frac{(-1) ^{n}}{(n+1)^2} \right)$$
Assuming the necessary convergence properties, we can split this up:
$$I = \sum_{n=1}^\infty \frac{(-1)^{n}}{n} - \sum_{n=1}^\infty\frac{(-1)^{n}}{n+1} - \sum_{n=1}^\infty\frac{(-1)^{n}}{(n+1)^2} $$
We then reindex the right two sums and factor $-1$ from the first:
$$I =- \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} -\sum_{n=2}^\infty\frac{(-1)^{n -1}}{n } - \sum_{n=2}^\infty\frac{(-1)^{n-1 }}{n^2} $$
Then we write this in terms of the Dirichlet eta function $\eta(s)$:
$$I = -\eta(1) - (\eta(1) - 1) - (\eta(2) - 1) =- 2 \eta(1) -\eta(2) + 2$$
We know $\eta(1) = \ln(2)$ and $\eta(2) = \pi^2/12$, letting us conclude.
Integrate by parts
\begin{align} &\int_0^1 \ln(x+1) \ln x \>dx\\ =& -\int_0^1 \ln(1+x)\>dx -\int_0^1 \ln x \> dx+\int_0^1 \frac{\ln x}{1+x}dx\\ =&-(2\ln2-1)-(-1)+(-\frac{\pi^2}{12}) =2-2\ln2 -\frac{\pi^2}{12} \end{align} where $\int_0^1 \frac{\ln x}{1+x}dx= -\frac{\pi^2}{12}$
