1. ## Residue Theory

I am having trouble understanding how to calculate residues.

I have the function

$f(z)=\frac {z^a}{1+z^2}$ and am supposed to find

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

$\lim_{z\to+i}\frac{{z^a}{(z-i)}}{1+z^2}$ and $\lim_{z\to-i}\frac{{z^a}{(z+i)}}{1+z^2}$

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

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

I have the function

$f(z)=\frac {z^a}{1+z^2}$ and am supposed to find

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

$\lim_{z\to+i}\frac{{z^a}{z-i}}{1+z^2}$ and $\lim_{z\to-i}\frac{{z^a}{z+i}}{1+z^2}$

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 $a\ge 0$?

3. I forgot to mention

0 < a < 1

Can you explain why the poles are simple?

4. 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.

5. So$res(i) = \frac {i^{a-1}} {2}$?

6. 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.

7. Now I have found

$res(i) = \frac {i ^ {a-1}}{ 2}$ and
$res(-i) = \frac {-i ^ {a-1}}{ 2}$

I am trying to prove that

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

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,

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

For the bottom line segment, I have found

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

and putting it all together, I have

$\int^\infty_0 \frac {x^a}{1+x^2}\,dx = \frac {2 \pi i^a}{1-e^{2 \pi i^a}}$

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

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