Residue Theory

• 02-25-2012, 05:02 PM
monomocoso
Residue Theory
I am having trouble understanding how to calculate residues.

I have the function

[TEX]f(z)=\frac {z^a}{1+z^2}[/TEX] and am supposed to find

[TEX]\int^\infty_0 f(z)\,dz[/TEX] using the keyhole contour. So I found singularities at +i and - i and am trying to find the residues there using

[TEX]\lim_{z\to+i}\frac{{z^a}{(z-i)}}{1+z^2}[/TEX] and [TEX]\lim_{z\to-i}\frac{{z^a}{(z+i)}}{1+z^2} [/TEX]

Is ths the right way to find residues? What are these limits?
• 02-25-2012, 05:31 PM
daon2
Quote:

Originally Posted by monomocoso
I am having trouble understanding how to calculate residues.

I have the function

[TEX]f(z)=\frac {z^a}{1+z^2}[/TEX] and am supposed to find

[TEX]\int^\infty_0 f(z)\,dz[/TEX] using the keyhole contour. So I found singularities at +i and - i and am trying to find the residues there using

[TEX]\lim_{z\to+i}\frac{{z^a}{z-i}}{1+z^2}[/TEX] and [TEX]\lim_{z\to-i}\frac{{z^a}{z+i}}{1+z^2} [/TEX]

Is ths the right way to find residues? What are these limits?

That isn't exactly right. Your poles are simple, so $$\text{Res}(i)=\lim_{z\to i}\,\, (z-i)\frac{z^a}{1+z^2}$$.

Also, I assume that $$a\ge 0$$?
• 02-25-2012, 06:17 PM
monomocoso
I forgot to mention

0 < a < 1

Can you explain why the poles are simple?
• 02-25-2012, 06:31 PM
daon2
Quote:

Originally Posted by monomocoso
I forgot to mention

0 < a < 1

Can you explain why the poles are simple?

Simple means that it has multiplicity 1. When multiplicity > 1 you have more calculation to do.
• 02-25-2012, 06:36 PM
monomocoso
So[TEX] res(i) = \frac {i^{a-1}} {2} [/TEX]?
• 02-25-2012, 06:40 PM
HallsofIvy
Do you understand what a "simple pole" is?

A function, f(z), has a "pole" at z= a if, in some neighborhood of z= a, f(z) can be written as a power series in (z- a) with only a finite number of negative powers. We say that the pole "is of order n" if lowest power is -n. In particular, we say that the pole is "simple" if it has order 1.

Now, here, your function is $$\frac{z^a}{1+ z^2}= \frac{z^2}{1- iz}{1+ iz}$$. Using "partial fractions" we can write that as a sum of fractions with denominators z- i and z+ i. Around, say, z= i, we can write that as a power series with non-negative powers of z- i and the single term with $$(z- i)^{-1}$$. That is why it is a simple pole.
• 02-25-2012, 07:01 PM
monomocoso
Now I have found

[TEX] res(i) = \frac {i ^ {a-1}}{ 2}[/TEX] and
[TEX] res(-i) = \frac {-i ^ {a-1}}{ 2}[/TEX]

I am trying to prove that

[TEX]\int^\infty_0 \frac {x^a}{1+x^2}\,dx = \frac {\pi}{2} \sec( {\frac{\pi a }{2}})[/TEX] for [TEX]0<a<1[/TEX]

I have bounded the integrals on the two circular paths and show that they go to zero as the inner radius r goes to zero and the outer radius R goes to infinity. I also know that on the upper line segment,

[TEX]\int^R_r \frac {z^a}{1+z^2}\,dz=\int^R_r \frac {x^a}{1+x^2}\,dx[/TEX] because there is no imaginary part.

For the bottom line segment, I have found

[TEX]\int^r_R \frac {z^a}{1+z^2}\,dz=e^{2 \pi i a} \int^r_R \frac {x^a}{1+x^2}\,dx[/TEX]

and putting it all together, I have

[TEX]\int^\infty_0 \frac {x^a}{1+x^2}\,dx = \frac {2 \pi i^a}{1-e^{2 \pi i^a}}[/TEX]

which is not what I want. Can anyone see where I went wrong?