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Show that if $|a|\le1$, then $\int^{2\pi}_0 {\log|1-ae^{i\theta}|} \,d\theta =0$

This is from Stein's Complex Analysis.

I think that if $|a|<1$ then the function $\log|1-ae^{i\theta}|$ is holomorphic in the unit disk so the integral is clearly zero.

Am I making a mistake? What is the solution when $|a|=1$?

Help me please.

Thanks.

user115608
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  • when $a=1$, $\lim_{\theta\to \pi}\ln|1-ae^{i\theta}|=-\infty$ – Ellya May 31 '14 at 08:43
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    $\log|1-az|$ is real valued on the unit disc, so certainly not holomorphic. Also, that is beside the point: you are integrating over a real interval. Try transforming the integral to a complex curve integral first. – mrf May 31 '14 at 08:46

1 Answers1

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Using Cauchy formula, you can show that $$\int_0^{2\pi} \log (1 - ae^{i\theta}) d\theta = 2\pi\log (1) = 0 \tag{1}$$

Similarly (going counter clockwise direction) you can show that $$\int_0^{2\pi} \log (1 - ae^{-i\theta}) d\theta = 2\pi\log (1) = 0 \tag{2}$$ Adding $(1)$ and $(2)$ you get the required result. $$ \int_0^{2\pi} \log |1 - ae^{i\theta}|^2 d\theta = 0 $$

ADDED:: looks like I need strict inequality for above :S for case $a=1$, the integral you need to evaluate is $$\int_0^{2\pi} \log \sqrt{(1-\cos x)^2 + \sin^2x}dx = \int_0^{2\pi} \frac 1 2 \log (2- 2\cos(x) )dx \\ = \pi \log 2 + \frac 1 2 \int_{0}^{2\pi} \log (1 - \cos(x))dx \tag{3}$$

Then integral converges because we know $\int_0^1 \log(x) dx $ converges and there is only two finite points where $\log(x)$ tends to singularity.

The integral on the right is $-\pi \log(2)$ making the whole expression zero.

$$\int_0^{2\pi} \log(1 - \cos(x)) dx = \int_0^{2\pi} \log(2 \sin^2 (x/2) )dx = 2 \pi \log (2) + 4 \int_0^{\pi} \log(\sin(y)) dy \tag{4}$$

The last integral is pretty famous, you can find it here and use it's value and show that the whole expression on $(3)$ is zero.

Community
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S L
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  • :in the case $|a|=1$,can we assume $a=1$ WLOG? – user115608 May 31 '14 at 11:55
  • @user115608 yes you can ... since $a = 1 e^{i\phi}$ and constant argument is not going to make any difference since we are going around the circles – S L May 31 '14 at 11:57