0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$
    – Jay Lemmon
    Mar 21 '20 at 20:35
  • $\begingroup$ edited to show how one proceeds $\endgroup$
    – Conrad
    Mar 21 '20 at 22:19
  • $\begingroup$ Can you explain why $\Re \log(1-z) = \log(|1-z|)$? Otherwise I think I follow your reasoning. $\endgroup$
    – Jay Lemmon
    Mar 22 '20 at 10:58
  • $\begingroup$ Ah, I got it. Since $\log(z) = \ln(|z|) + i * \arg(z)$, $\Re \log(z) = \ln(|z|)$ $\endgroup$
    – Jay Lemmon
    Mar 22 '20 at 15:13
  • $\begingroup$ 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) $\endgroup$
    – Conrad
    Mar 22 '20 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.