$$ \int _0^ \frac{ \pi }{2} \frac{ln(sinx)ln( \sqrt{1- sin ^{2}x } )}{tanx} dx \wedge sin x=u \Longleftrightarrow ?= \int _0^1 \frac{ln(u)ln( \sqrt{1- u^{2} } )}{u} du \wedge \ast \ast ln(1- u^{2} )=- \sum _ {k=1} ^ \infty \frac{ ( u^{2} )^{k} }{k} \Longrightarrow ?=- \frac{1}{2} \int _0^1lnu \sum _ {k=1} ^ \infty \frac{ u^{2k-1} }{k} du =- \frac{1}{2} \sum _ {k=1} ^ \infty \frac{1}{k} \int _0^1 u^{2k-1} lnudu=- \frac{1}{2} \sum _ {k=1} ^ \infty \frac{(-1)}{k (2k)^{2} } = \frac{1}{8} \sum _ {k=1} ^ \infty \frac{1}{ k^{3} } = \frac{1}{8} \zeta (3) \wedge ( \ast \int _0^1 x^{ \eta -1} ln^{z-1} (x)dx= \frac{ (-1)^{z-1} \Gamma (z)}{ \eta ^{z} } ) $$
$$ \zeta (3)=ثابت آپری$$
