# Solve $\int_0^{2\pi} \ln(2 R \sin(\frac{\theta}{2})) d\theta$ using a contour integral

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.

$$|1-e^{i\theta}|=2\sin{\frac{\theta}{2}}$$, 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 $$\Re (1-rz) >0, |z| =1$$)

• Thanks there was a lot to learn here. Can you expand on the last paragraph? I'm specifically interested in how a contour integral would work here. I don't see a nice way to handle the absolute value. Mar 21 '20 at 20:35
• edited to show how one proceeds Mar 21 '20 at 22:19
• Can you explain why $\Re \log(1-z) = \log(|1-z|)$? Otherwise I think I follow your reasoning. Mar 22 '20 at 10:58
• Ah, I got it. Since $\log(z) = \ln(|z|) + i * \arg(z)$, $\Re \log(z) = \ln(|z|)$ Mar 22 '20 at 15:13
• yes that's right and the fact that $\Re{(1-z)}>0$ in the unit disk ensures that we do not move across branches of the logarithm - as the unit disc is simply connected any continuous function that doesn't have zeroes has a continuos logarithm, but the imaginary part of the logarithm which is a continuous choice of arguments for the function may jump across branches so may be quite unbounded as you can see with say $f(z)=e^{\frac{1}{z-1}}$ - so the argument is bounded in absolute value (by $\frac{\pi}{2}$ if we use the standard $[-\pi, \pi]$ range when dealing with logarithms) Mar 22 '20 at 15:43