I'm trying to evaluate $$\int_0^{2\pi} \ln(2 R \sin(\frac{\theta}{2})) d\theta$$ using a contour integral, where $R > 0$. Letting $z = \exp(i \frac{\theta}{2})$ I have
$$ = \oint \frac{2}{i z} \ln(i R (\frac{1}{z} – z)) dz$$
$$ = \frac{2}{i} \oint \frac{1}{z} \ln(i R (\frac{1}{z} – z)) dz$$
I think I need to integrate over a keyhole contour now to get rid of the discontinuity at $z = 0$, then using the residue theorem sum up the residues in the unit circle and multiply by $2 \pi i$. I think I see potential poles at $z = \pm 1$ (since it would give $\log(0)$) but it's unclear to me how to determine their residues from here (I keep getting $0$ which I don't think is correct) or if this is even the correct approach.
Any help would be appreciated.
Best Answer
$|1-e^{i\theta}|=2\sin{\frac{\theta}{2}}$, so we need to evaluate $I=\int_0^{2\pi}\log |1-e^{i\theta}|d\theta$ as then the answer is $2\pi\log R+I$.
Now $\log|1-z|=\Re{\log(1-z)}$ is harmonic inside the unit disc, so by the mean value theorem for harmonic function, $\int_0^{2\pi}\log |1-re^{i\theta}|d\theta=0$ for any $r<1$.
But $\log|1-e^{i\theta}|$ is obviously integrable on the unit circle (since near $0$, $\log 2\sin{\frac{\theta}{2}}-\log \theta$ is continuos and the latter is clearly integrable, while near $2\pi$ we can use $\sin{\frac{\theta}{2}}=\sin{\frac{2\pi-\theta}{2}}$ and the result at zero) and $\log|1-re^{i\theta}| \to \log|1-e^{i\theta}|$ a.e.
But now if say $0 \le \theta \le \frac{\pi}{100}$ or $0 \le 2\pi-\theta \le \frac{\pi}{100}$, by drawing the perpendicular from $1$ to the $\theta$ ray which has length $|\sin \theta|$ it follows from elementary geometry that $|1-re^{i\theta}| \ge |\sin \theta|$, while for the rest $|1-re^{i\theta}| \ge c >0$ and since $|1-re^{i\theta}| \le 2$, we get that $|\log |1-re^{i\theta}|| \le \max {(\log 2, \log^- c-\log |\sin \theta|)}$ and that is integrable on the unit circle as before, hence we can apply the Lebesgue dominated convergence and conclude that $0=\int_0^{2\pi}\log |1-re^{i\theta}|d\theta \to I$, so $I=0$ and the final answer is $2\pi\log R$
As an aside, there is also a classic real variables proof that $I=0$ using the doubling formula for the $\sin$ and various changes of variable while the above can be expressed in terms of contour integrals if one wishes using $f(z)=\frac {\log (1-z)}{z}$ which is analytic inside the unit disc, or treat the original integral using $f(z)=\frac {\log R(1-z)}{z}$ which has $\log R$ as residue at $0$ etc, but of course it may not quite be what OP had in mind.
(Edit) As asked let's quickly use Cauchy rather than Poisson:
$I_1=\int_0^{2\pi}\log (2R\sin{\frac{\theta}{2}})d\theta=\int_0^{2\pi}\log R|1-e^{i\theta}|d\theta= \int_0^{2\pi} \Re {\log R(1-e^{i\theta})}d\theta=\Re {\int_0^{2\pi} \log R(1-e^{i\theta})d\theta}$
with the last equality holding because $d\theta$ is a real (positive) measure.
But now with the usual $e^{i\theta}=z, d\theta=\frac{1}{iz}dz$ we have;
$I_1=\Re {\int_{|z|=1} \frac{\log R(1-z)}{iz}dz}$
Now by Cauchy ${\int_{|z|=1} \frac{\log R(1-rz)}{iz}dz}= 2\pi \log R$ as the residue at zero of the integrand is $\frac{\log R}{i}$, while it is analytic anywhere else on the closed unit disc when $0 < r <1$. So we need to be able to pass to the limit $ r \to 1$ to conclude that the above unit circle integral (and hence its real part) is $2\pi \log R$ and the same argument as above works since the only problem comes from $\log |1-rz|$ near the boundary as everything else is obviously bounded so the same estimates work to show that we can use the Lebesgue dominated convergence and conclude that $I_1= 2\pi \log R$
(for $|z|=1$ we have $|\frac{\log R(1-rz)}{iz}|\le |\log R|+ |\log (1-rz)| \le |\log R|+ |\Re \log (1-rz)|+ |\Im \log (1-rz)| $ and $|\Re \log (1-rz)|=|\log |1-rz||$ as above, while $|\Im \log (1-rz)|=|\arg (1-rz)| \le \frac{\pi}{2}$ since $\Re (1-rz) >0, |z| =1$)