Integral of ln(sin(x))

Question

So I've had a few attempts at this this integral, but they all seem to get different answers to wolframalpha. This is my working out:

$I=\int \ln(\sin(x))dx \\ I=\int \ln(\frac{e^{ix}-e^{-ix}}{2i})dx \\ I=\int \ln(e^{ix}-e^{-ix})-\ln(2i)dx \\ I = \int \ln(e^{ix}-e^{-ix})dx-\ln(2i)x$

I split this up so to make things a bit clearer.

$I_{0}=\int\ln(e^{ix}-e^{-ix})dx \\ I_{0}=\int \ln(e^{ix}(1-e^{-2ix}))dx \\ I_0=\int \ln(e^{ix})+\ln(1-e^{-2ix})dx \\ I_{0}=\int \ln(1-e^{-2ix})dx + \frac{ix^{2}}{2}$

It is mainly this part which I think I have done incorrectly.

$I_{1}=\int \ln(1-e^{-2ix})dx \\ I_{1}=\int \ln(1-e^{-2ix})dx \\ \text{Let }e^{-2ix}=u \Leftrightarrow \frac{-2ix}{u}du=dx \\ I_{1} = -2i\int \frac{\ln(1-u)}{u}du \\ I_{1} = 2i\int \sum^{\infty}_{n=1}\frac{u^{n-1}}{n}du \\ I_{1} = 2i \sum^{\infty}_{n=1}\frac{u^{n}}{n^{2}} \\ I_{1}=2i\text{Li}_{2}(u) \\ I_1=2i\text{Li}_{2}(e^{-2ix})$

After substituting everything back into the original integral I got:

$I=2i\text{Li}_{2}(e^{-2ix})-\ln(2i)x+\frac{ix^{2}}{2}+C$

Answer 1

$$I= \int \log\left(-\frac{\mathrm{i} \left(\mathrm{e}^{\mathrm{i}x} - \mathrm{e}^{-\mathrm{i}x}\right)}{2}\right) \, \mathrm{d}x$$ Substitute $u = \mathrm{i}x$ $$I= -\mathrm{i} \int \left(\log\left(\mathrm{e}^{-u} - \mathrm{e}^{u}\right) + \log\left(\frac{\mathrm{i}}{2}\right)\right) \mathrm{d}u= \int \log\left(\mathrm{e}^{-u} - \mathrm{e}^{u}\right) \, \mathrm{d}u + \log\left(\frac{\mathrm{i}}{2}\right) \int \, \mathrm{d}u=I_1+I_2$$ $I_2$ is simply $u=ix$ For $I_1$ :$$\int \log\left(\mathrm{e}^{-u} - \mathrm{e}^{u}\right) \, \mathrm{d}u= u \log\left(\mathrm{e}^{-u} - \mathrm{e}^{u}\right) - \int \frac{u \left(-\mathrm{e}^{u} - \mathrm{e}^{-u}\right)}{\mathrm{e}^{-u} - \mathrm{e}^{u}} \, \mathrm{d}u$$ Solving $\int \frac{u \left(-\mathrm{e}^{u} - \mathrm{e}^{-u}\right)}{\mathrm{e}^{-u} - \mathrm{e}^{u}} \, \mathrm{d}u$ Substitute $v = \mathrm{e}^{2u} - 1$ $$\int \frac{u \left(-\mathrm{e}^{u} - \mathrm{e}^{-u}\right)}{\mathrm{e}^{-u} - \mathrm{e}^{u}} \, \mathrm{d}u= \frac{1}{4} \, \int \frac{\left(v + 2\right) \log\left(v + 1\right)}{v \left(v + 1\right)} \, \mathrm{d}v= 2 \int \frac{\log\left(v + 1\right)}{v} \, \mathrm{d}v - \int \frac{\log\left(v + 1\right)}{v + 1} \, \mathrm{d}v$$ This is just $$ -\frac{\log^{2}\left(v + 1\right)}{2} - 2 \operatorname{Li}_{2}\left(-v\right)$$And $$I= -\frac{\mathrm{i} \operatorname{Li}_{2}\left(1 - \mathrm{e}^{2u}\right)}{2} - \mathrm{i}u \log\left(\mathrm{e}^{-u} - \mathrm{e}^{u}\right) - \frac{\mathrm{i}u^{2}}{2} - \mathrm{i} \log\left(\frac{\mathrm{i}}{2}\right) \, u\\= -\frac{\mathrm{i} \operatorname{Li}_{2}\left(1 - \mathrm{e}^{2\mathrm{i}x}\right)}{2} + x \log\left(\mathrm{e}^{-\mathrm{i}x} - \mathrm{e}^{\mathrm{i}x}\right) + \frac{\mathrm{i}x^{2}}{2} + \log\left(\frac{\mathrm{i}}{2}\right) \, x$$ Simplifying a bit more we get $$\int\log(\sin(x))= \frac{\mathrm{i}x^{2} - \mathrm{i} \operatorname{Li}_{2}\left(2 \sin\left(x\right) \left(\sin\left(x\right) - \mathrm{i} \cos\left(x\right)\right)\right)}{2} + x \log\left(2 \sin\left(x\right)\right) + \log\left(\frac{\mathrm{i}}{2}\right) \, x + C$$

Answer 2

$$ \begin{aligned} I & =\int \log (\sin \theta) d \theta \\ & =\int \log \left(\frac{e^{2 i}-e^{-\theta i}}{2 i}\right) d \theta \\ & =\int \log \left(\frac{e^{\theta i}}{2 i}\right) d \theta+\int \log \left(1-e^{-2 \theta i}\right) d \theta \\ & =\int(\theta i-\log 2-\log i) d \theta+\frac{i}{2} \int \frac{\log \left(1-e^{-2 \theta i}\right)}{e^{-2 \theta i}} d\left(e^{-2\theta i}\right) \\ &= \frac{\theta^2}{2} i-\theta\left[\log 2+i\left(\frac{\pi}{2}+2 n\pi \right)\right]+\frac{i}{2} \operatorname{Li_2}\left(e^{-2 \theta i}\right)+C \quad \textrm{ for some integer }n \\&= - \frac{\pi}{2}\theta-\frac{1}{2} \Im \operatorname{Li_2}\left(e^{-2 \theta i}\right)+C^{\prime} \end{aligned} $$

In particular, $$ \int_0^{\frac \pi 2} \log (\sin \theta) d \theta =-\frac{\pi}{2} \log 2-\frac{1}{2} \Im\left(\operatorname{Li_ 2}(-1)-\operatorname{Li_ 2}(1)\right) =-\frac{\pi}{2} \ln 2 $$