Calculating Integral Using Residue Theorem & Complex Variables

[BOAS]

Homework Statement

I have never formally studied complex analysis, but I am reading this paper: http://adsabs.harvard.edu/abs/1996MNRAS.283..837S

wherein section 2.2 they make use of the residue theorem. I am trying to follow along with this (and have looked up contour integration, cauchy's formula etc..).

I want to compute the integral

$I_1 = \int_{0}^{2\pi} \frac{d \phi}{(X-Y)^2}$ where X and Y are complex variables

Homework Equations

The Attempt at a Solution

[/B]
Using the substitution $X = |\vec x|e^{i\phi}$, $dX = i X d\phi$ we can write this as

$I_1 = -\oint \frac{i dX}{X(X-Y)^2}$ which has two poles, at X=0 and X=Y.

The residue theorem allows me to say that the result of this integral is $2 \pi i$ times the sum of the residues. I have been able to compute the residue by expanding at $X=0$ as follows:

$f(X) = - \frac{1}{X(X-Y)^2} = - \frac{1}{X}(\frac{1}{(X-Y)^2})$

I then taylor expand $\frac{1}{(X-Y)^2}$ around X=0 and find:

$f(X) = -\frac{1}{X}(\frac{1}{Y^2} + \frac{2X}{Y^3} + ...)$

and so by identifying the residue with the term proportional to $\frac{1}{X}$, I see that the residue of the function expanded at that pole is $- \frac{1}{Y^2}$

I am stuck with how to find the residue of the $X=Y$ expansion as it can't be done in the same way as above.

Thank you for any hints you can give!

 

[member 587159]

Hint: Replace all the '$' with '##'. Then your post will become readable :)

 

[BOAS]

Hint: Replace all the '$' with '##'. Then your post will become readable :)

woops - I had a latex plugin in my browser, so everything looked normal to me. Thanks for pointing it out.

(I have excluded physics forums from the plugin now)

 

[vela]

The residue theorem allows me to say that the result of this integral is $2 \pi i$ times the sum of the residues. I have been able to compute the residue by expanding at $X=0$ as follows:

$f(X) = - \frac{1}{X(X-Y)^2} = - \frac{1}{X}(\frac{1}{(X-Y)^2})$

I then taylor expand $\frac{1}{(X-Y)^2}$ around X=0 and find:

$f(X) = -\frac{1}{X}(\frac{1}{Y^2} + \frac{2X}{Y^3} + ...)$

and so by identifying the residue with the term proportional to $\frac{1}{X}$, I see that the residue of the function expanded at that pole is $- \frac{1}{Y^2}$

I am stuck with how to find the residue of the $X=Y$ expansion as it can't be done in the same way as above.

Thank you for any hints you can give!

You can do pretty much the same thing, expanding the $1/X$ part as a series in powers of $(X-Y)$. Start by rewriting $1/X$ as follows:
$$\frac{1}{X} = \frac{1}{X-Y+Y} = \frac{1}{Y}\cdot\frac{1}{1+\frac{X-Y}{Y}}.$$ Now expand the second factor as a series.

 

[BOAS]

You can do pretty much the same thing, expanding the $1/X$ part as a series in powers of $(X-Y)$. Start by rewriting $1/X$ as follows:
$$\frac{1}{X} = \frac{1}{X-Y+Y} = \frac{1}{Y}\cdot\frac{1}{1+\frac{X-Y}{Y}}.$$ Now expand the second factor as a series.

Thank you, I have managed to find that the second residue is indeed $\frac{1}{Y^2}$ with the help of your hint