Evaluating the Integral \int_0^{2\pi}log|e^{i\theta}-1|d\theta

[Poopsilon]

Homework Statement

Evaluate the integral
\int_0^{2\pi}log|e^{i\theta}-1|d\theta

Homework Equations

The Attempt at a Solution

So I'm essentially integrating log|z| around a circle of radius 1 centered at -1. Evaluating at the endpoints gives a singularity, but I feel like that shouldn't matter since they are only the endpoints. I guess my main issue is I don't know how to integrate log|z|. I was inclined to say that:

\int_0^{2\pi}log|e^{i\theta}-1|d\theta = \int_0^{2\pi}log(1 - cos(\theta))d\theta.

Is this equality correct?

I also tried u-du substitution with u = e^{i\theta} - 1 and d\theta = \frac{du}{i(u - 1)}, but then when I update the limits of integration I get that I'm integrating from zero to zero, I know there's a conceptual issue which explains why that happens and how rectify it, but I still don't understand it.

Can anyone help me with this? Thanks.

 

[Dustinsfl]

Homework Statement

Evaluate the integral
\int_0^{2\pi}log|e^{i\theta}-1|d\theta

Homework Equations

The Attempt at a Solution

So I'm essentially integrating log|z| around a circle of radius 1 centered at -1. Evaluating at the endpoints gives a singularity, but I feel like that shouldn't matter since they are only the endpoints. I guess my main issue is I don't know how to integrate log|z|. I was inclined to say that:

\int_0^{2\pi}log|e^{i\theta}-1|d\theta = \int_0^{2\pi}log(1 - cos(\theta))d\theta.

Is this equality correct?

I also tried u-du substitution with u = e^{i\theta} - 1 and d\theta = \frac{du}{i(u - 1)}, but then when I update the limits of integration I get that I'm integrating from zero to zero, I know there's a conceptual issue which explains why that happens and how rectify it, but I still don't understand it.

Can anyone help me with this? Thanks.

In the complex plane, log is a multivalued function. You can try defining a branch cut. I checked the solution on Mathematica, and it returned 0 as well.

 

[jackmell]

[
\int_0^{2\pi}log|e^{i\theta}-1|d\theta = \int_0^{2\pi}log(1 - cos(\theta))d\theta.

Is this equality correct?

I don't think so. I get:

\int_0^{2\pi} \log\left|e^{it}-1\right|dt=\int_0^{2\pi}\left[\log \sqrt{2}+1/2 \log \left(1-\cos(t)\right)\right] dt

 

[Dickfore]

<br /> \vert e^{i \theta} - 1 \vert = \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta} = \sqrt{2 \, (1 - \cos \theta)} = \sqrt{2 \, 2 \, \sin^{2} \frac{\theta}{2}} = 2 \, \left\vert \sin \frac{\theta}{2} \right\vert<br />

But, if 0 \le \theta \le 2 \pi, then \sin \frac{\theta}{2} \ge 0, and the absolute sign is unimportant.