Can $ \int_0^{\pi/2} \ln ( \sin(x)) \; dx$ be evaluated with "complex method"? Can the following integral be evaluated using complex method by substituting $\sin(x) = {e^{ix}-e^{-ix} \over 2i}$?
$$ I=\int_0^{\pi/2} \ln ( \sin(x)) \; dx = - {\pi \ln(2) \over 2}$$

Let us change variables $u=\sin^2(x)$, $x=\arcsin(\sqrt{u})$, $\mathrm{d}x = \frac{1}{2} \frac{\mathrm{d}u}{\sqrt{u(1-u)}}$:
$$
   \int_0^{\pi/2} \ln(\sin(x)) \mathrm{d}x = \int_0^1 \frac{1}{2} \ln(u) \cdot \frac{1}{2} \frac{\mathrm{d}u}{\sqrt{u(1-u)}} \stackrel{u=1-v}{=} \frac{1}{4} \int_0^1 \ln(1-v) \cdot \frac{\mathrm{d}v}{\sqrt{v(1-v)}}
$$
Adding two last integrals, and dividing by $2$:
$$
 \int_0^{\pi/2} \ln(\sin(x)) \mathrm{d}x = \frac{1}{4} \int_0^1 \frac{\ln\left(\sqrt{u(1-u)}\right)}{\sqrt{u(1-u)}} \mathrm{d}u  = -\frac{1}{4} \left.\frac{\mathrm{d}}{\mathrm{d}s}\operatorname{B}\left(s,s\right)\right|_{s=\frac{1}{2}}
$$
Since $\operatorname{B}\left(s,s\right) = \frac{\Gamma(s)^2}{\Gamma(2s)}$:
$$
   \frac{\mathrm{d}}{\mathrm{d}s}\operatorname{B}\left(s,s\right) = 2 \operatorname{B}\left(s,s\right) \left(\psi(s) - \psi(2s)\right)
$$
where $\psi(s)$ is the digamma function.  Using $\operatorname{B}\left(\frac{1}{2},\frac{1}{2}\right) = \pi$, and duplication identity for the digamma function:
$$
    2 \psi(2s) = \psi(s) + \psi\left(s+\frac{1}{2}\right) + 2 \log(2) \quad \stackrel{s=\frac{1}{2}}{\implies} \quad \psi\left(\frac{1}{2}\right) - \psi(1) = -2\ln(2)
$$
Combining, we arrive at the result:
$$
  \int_0^{\pi/2} \ln(\sin(x)) \mathrm{d}x = \frac{1}{4} \cdot \pi \cdot \left(-2 \ln(2)\right) = -\frac{\pi}{2} \ln(2)
$$
