Evaluate: $\int_0^{2 \pi} \ln(1+e^{i \theta}) d\theta$

Question

How to evaluate $\displaystyle \int_0^{2 \pi} \ln(1+e^{i \theta}) d\theta$. W|A is giving $0$. How do I get this result if this is correct?

Attempt: let $z = e^{i \theta}$, then we have

$$\frac{1}{i}\oint_{|z|=1} \frac{\ln (1 +z)}{z} dz $$

Expading log at $z=0$ we get analytic function whose contour integration is zero. It this correct?

What you are doing is right. Note that, you have a pole at $z=0$ in your last integral. Calculating the residue gives you $0$. — Mhenni Benghorbal, Apr 08 '13 at 12:27
@MhenniBenghorbal ok ok ... so if i evaluate the contour I would get it zero anyway? — Mula Ko Saag, Apr 08 '13 at 12:28
Note that, using the above residue technique, you can evaluate $\displaystyle \int_0^{2 \pi}\ln(1+re^{i\theta})d\theta$, By putting $z=re^{i\theta}.$ — Mhenni Benghorbal, Apr 08 '13 at 12:33

Ishan Banerjee · Accepted Answer · 2013-04-08T12:46:49.260

3

According to Gauss' mean value theorem $f(z_0)=\dfrac{1}{2\pi}\int_0^{2\pi}f(z_0+re^{i\theta})d\theta$ Using this with $\log (1+z)$ gives us the result.

edited Apr 08 '13 at 12:46

answered Apr 08 '13 at 12:15

Ishan Banerjee

6,459

I changed my mind (looks like this is okay)... I was confused between $z$ and $\theta$. Still i would like to see more opinion. – Mula Ko Saag Apr 08 '13 at 12:25
and Gauss MVT is $f(z_0)=\frac{1}{2 \pi}\int_0^{2\pi}f(z_0+re^{i\theta})d\theta$ – Mula Ko Saag Apr 08 '13 at 12:26
1

The use of the Gauss mean value theorem here is wrong, as $\log(1+z)$ is not analytic at $z=-1$. – Hans Oct 10 '18 at 08:55

Evaluate: $\int_0^{2 \pi} \ln(1+e^{i \theta}) d\theta$

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