A tricky countour integral (branch cuts)

[dnquark]

Hi all. I need to take the following integral by contour integration (it's actually a simplified version of another integral that I'm working on):

[tex]\int_{-\pi}^\pi \frac{e^{i\omega}}{\sqrt{e^{i\omega}-1/2}\sqrt{e^{i\omega}-3}}[/tex].

I am transforming it using [tex]z=e^{i\omega}[/tex] into
[tex]\frac{1}{i}\int \frac{dz}{\sqrt{z-1/2}\sqrt{z-3}}[/tex], where the integral is around the unit contour around the origin.

Now, it looks to me that there are two branch points: z=1/2 and z=3, and I'm choosing a branch cut from 1/2 to -infinity and from 3 to infinity. This is illustrated in the picture here
http://img509.imageshack.us/my.php?image=clipboardqx7.jpg

Along the branch cut from 1/2 to -infinity, the sqrt(z-1/2) is multivalued. The illustration that I linked shows how this can be used to solve the integral. However, I don't know how to account for the second branch cut.

To clarify: if, for instance, I use [tex]z-3[/tex] instead of [tex]\sqrt{z-3}[/tex], the integral works out. However, with the square root it doesn't work -- I don't think the second branch cut should matter because I'm not integrating anywhere in its vicinity, but apparently it does. I don't have anyone to ask about it... so I'm hoping there's a contour integration maven that reads this and explains things to me...

Edit: I wanted to clarify the linked diagram (link doesn't click for some reason so I have to copy/paste the URL).
The green line is the branch cut, the yellow contour is the integral that I seek (i.e. int dw from -pi to pi is the same as int of dz over the yellow contour); the red contour does not enclose singularities, so it's zero by the residue theorem...

 

[Hurkyl]

Your original integral: is it supposed to be w.r.t. omega?

Anyways, I note that your original integrand is multivalued; what branch cut is it supposed to be using? I'm going to assume the branch cut required for the original integrand is the same one you decided to use for the integrand to your contour integral.

Could you clarify in what way your approach doesn't work?

 

[tacman]

Hi there,

Thank you for sharing your question and diagram. Contour integration can definitely be tricky, especially when dealing with branch cuts. Let me try to provide some insight to help you understand why the second branch cut matters in this integral.

First, let's review the concept of branch cuts. A branch cut is a line or curve on the complex plane where a function is not single-valued. In your case, the function \sqrt{z-1/2}\sqrt{z-3} is not single-valued along the branch cut from 1/2 to -infinity. This means that when you take the square root of a complex number along that branch cut, you will get two different values depending on which side of the branch cut you are on.

Now, let's focus on the integral itself. The reason why the second branch cut matters is because when you transform the integral using z=e^{i\omega}, you are essentially integrating over a contour that encloses both branch points, z=1/2 and z=3. This means that you need to take into account both branch cuts in order to properly evaluate the integral.

To see why this is important, let's consider the integral over a small circular contour centered at z=1/2. Along this contour, the function \sqrt{z-1/2} will only take one value, since we are not crossing the branch cut. However, when we consider the integral over a contour that encloses both branch points, we are essentially integrating over a larger region where the function \sqrt{z-1/2} will take two different values. This is why the second branch cut matters in this case.

I hope this helps to clarify things for you. Contour integration can be challenging, but keep practicing and seeking help when needed. Good luck with your integral!