Evaluate complex integral $\int_0^{\pi}\frac{1}{\ln{(e^{ix}\sin{x})}}dx$ I want to find the integral:
$$\int_0^{\pi}\frac{1}{\ln{(e^{ix}\sin{x})}}dx$$
From $\frac{1}{\ln{(e^{ix}\sin{x})}} = \frac{\ln(\sin{x})}{x^2+\ln^2(\sin{x})}-\frac{x}{x^2+\ln^2(\sin{x})}i$,
therefore 
$$\int_0^{\pi}\frac{1}{\ln{(e^{ix}\sin{x})}}dx= \int_0^{\pi}\frac{\ln(\sin{x})}{x^2+\ln^2(\sin{x})}dx-i\int_0^{\pi}\frac{x}{x^2+\ln^2(\sin{x})}dx$$.
But it is hard to evaluate these integral.
Are there another method? Thank you.

Notice that the integral can be written as
$$ I = \int_{0}^{\pi} \frac{dx}{f(e^{2ix})} = \frac{1}{2} \int_{0}^{2\pi} \frac{dx}{f(e^{ix})}, $$
where $f(z) = \log\left(\frac{i}{2}(1-z)\right)$. Using the standard branch cut of the complex logarithm, $f$ has the branch cut $\{ 1 - it : t \geq 0 \}$ and the unique zero at $1+2i$. Thus $\frac{1}{f(z)}$ is a well-defined holomorphic function on $\Bbb{D} = \{z \in \Bbb{C} : |z| < 1\}$ and by the mean-value property we have
$$ \frac{1}{2} \int_{0}^{2\pi} \frac{dx}{f(e^{ix})} = \frac{\pi}{f(0)} = \frac{\pi}{\log(i/2)}. $$
