A proof of $\int_{0}^{1}\left( \frac{\ln t}{1-t}\right)^2\,\mathrm{d}t=\frac{\pi^2}{3}$

What is the proof of the following: $$\int_{0}^{1} \left(\frac{\ln t}{1-t}\right)^2 \,\mathrm{d}t=\frac{\pi^2}{3} \>?$$

Solution 1:

$$\int_0^1\frac{\log t}{1-t}dt=\int_0^1\log(1-u)\frac{du}{u}=\int_0^\infty \log(1-e^{-v})dv =-\frac{\pi^2}{6}.$$ For the last part see an answer of mine here.

For the revised question, substitute $u=1-t$ and expand into a product of Taylor series, then use some of partial fraction decomposition, sum splitting, reindexing, and telescoping properties: $$\int_0^1\left(\frac{\log t}{1-t}\right)^2dt=\int_0^1\left(\frac{\log(1-u)}{u}\right)^2du=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{nm}\int_0^1 u^{n+m-2}du$$ $$=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{nm(n+m-1)}=\sum_{m=1}^\infty \frac{1}{m^2}+\sum_{n=2}^\infty\frac{1}{n}\frac{1}{n-1}\sum_{m=1}^\infty\left(\frac{1}{m}-\frac{1}{n+m-1}\right)$$ $$=\frac{\pi^2}{6}+\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+1}\right)\sum_{m=1}^n\frac{1}{m}=\frac{\pi^2}{6}+\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{3}. $$

Solution 2:

This was intended to be a comment on Mike Spivey's answer, but alas, it was too long.

For $n>0$,

$$ \begin{align} \int_0^1\left(\frac{\log(t)}{1-t}\right)^n\mathrm{d}t &=\int_0^\infty\left(\frac{-s}{1-e^{-s}}\right)^ne^{-s}\mathrm{d}s\\ &=(-1)^n\int_0^\infty s^ne^{-s}\sum_{k=0}^\infty\binom{-n}{k}(-1)^ke^{-ks}\;\mathrm{d}s\\ &=(-1)^n\Gamma(n+1)\sum_{k=0}^\infty\binom{-n}{k}(-1)^k(k+1)^{-n-1}\\ &=(-1)^n\Gamma(n+1)\sum_{k=0}^\infty\binom{k+n-1}{n-1}(k+1)^{-n-1}\\ &=(-1)^n\Gamma(n+1)\sum_{k=1}^\infty\binom{k+n-2}{n-1}k^{-n-1}\\ &=(-1)^n\Gamma(n+1)\sum_{k=1}^\infty\frac{1}{(n-1)!}\sum_{j=0}^{n-1}\genfrac{[}{]}{0}{0}{n-1}{j}k^jk^{-n-1}\\ &=(-1)^nn\sum_{j=0}^{n-1}\genfrac{[}{]}{0}{0}{n-1}{j}\zeta(n-j+1) \end{align} $$ where $\genfrac{[}{]}{0}{0}{n}{k}$ is a Stirling number of the first kind.