# A tricky countour integral (branch cuts)

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):

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

I am transforming it using $$z=e^{i\omega}$$ into
$$\frac{1}{i}\int \frac{dz}{\sqrt{z-1/2}\sqrt{z-3}}$$, 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 $$z-3$$ instead of $$\sqrt{z-3}$$, 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...

Last edited:

Hurkyl
Staff Emeritus