Given that:
\displaystyle \sum_{n = 1}^\infty \frac 1 {n^2} = \frac {\pi^2} 6
Show that:
\displaystyle \sum_{n = 1}^\infty \frac {(-1)^{n - 1}} {n^2} = \frac {\pi^2} {12}
(i) Show that:
\displaystyle \int_0^1 \log x \log (1 - x) \mathrm dx = 2 - \frac {\pi^2} 6
(ii) Evaluate:
\displaystyle \int_0^1 \log x \log (1 + x) \mathrm dx
You can ignore issues of convergence and swapping sums/integrals.
Not too hard - just neat.
