Evaluar el cerrado de:
$$I=\int_{0}^{1}\ln[\sin(\pi x)]\ln^2[2\sin(\pi x)]\mathrm dx$$
Tenemos lo siguiente:
$$\int_{0}^{1}\ln^2[2\sin(\pi x)]\mathrm dx={\zeta(2)\over 2}\tag2$$
$$\int_{0}^{1}\ln[\sin(\pi x)]\mathrm dx=-\ln(2)\tag3$$
Aproximadamente
$$I\approx-{\zeta(2)\over \ln (2)}$$
Integral indefinida:
$u=\sin(\pi x)\implies du=\pi\cos(\pi x) dx$
$${1\over \pi}\int{\ln(u)\ln^2(2u)\over \sqrt{1-u^2}}\mathrm du\tag4$$
$$(1-x)^{-1/2}=\sum_{k=0}^{\infty}{(2k-1)!!\over (2k)!!}x^k$$
$${1\over \pi}\sum_{k=0}^{\infty}{(2k-1)!!\over (2k)!!}\int{ u^{2k} \ln(u)\ln^2(2u)}\mathrm du\tag5$$