I'm interested in this integral: $\int_{{\frac {\pi}{8}}}^{{\frac {7\,\pi}{8}}}\!{\frac {\ln \left( 1- \cos \left( t \right) \right) }{\sin \left( t \right) }}\,{\rm d}t$
I found this particular closed form with Maple and finally : $-{\frac { \left( \ln \left( 2-\sqrt {2+\sqrt {2}} \right) \right) ^{
2}}{2}}-{\frac {{\pi}^{2}}{12}}+{\frac {11\, \left( \ln \left( 2
\right) \right) ^{2}}{8}}-{\frac { \left( \ln \left( 1+\sqrt {2}
\right) \right) ^{2}}{2}}+{\frac {3\, \left( \ln \left( 2+\sqrt {2}
\right) \right) ^{2}}{4}}+{\it {Li_2}} \left( -{\frac {\sqrt {2+
\sqrt {2}}}{4}}+{\frac{1}{2}} \right)
$, where \Li_2 is the dilogarithm function.
Please, can someone prove it ?
Best Answer
$$I=\int{\frac {\log \left( 1- \cos \left( t \right) \right) }{\sin \left( t \right) }}\,dt=\int \sin(t){\frac {\log \left( 1- \cos \left( t \right) \right) }{1-\cos^2\left( t \right) }}\,dt$$
Let $$x=\cos(t)\implies I=\int \frac{\log(1-x)}{x^2-1} \,dx=-\frac12\int\frac{\log(1-x)}{ 1-x}\,dx-\frac12\int\frac{\log(1-x)}{ x+1}\,dx$$
$$\int\frac{\log(1-x)}{ 1-x}\,dx=-\frac{1}{2} \log ^2(1-x)$$ For the second integral, integration by parts gives $$\int\frac{\log(1-x)}{x+1}\,dx=\text{Li}_2\left(\frac{1-x}{2}\right)+\log (1-x) \log \left(\frac{x+1}{2}\right)$$ Combining the results
$$I=\int \frac{\log(1-x)}{x^2-1} \,dx=\frac{1}{4} \left(\log (1-x) (\log (4(1-x))-2 \log (x+1))-2 \text{Li}_2\left(\frac{1-x}{2}\right)\right)$$ Go back to $t$ if you wish and use bounds.