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Thread: Residue Theory

  1. #1

    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?
    Last edited by monomocoso; 02-25-2012 at 04:28 PM.

  2. #2
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    Quote Originally Posted by monomocoso View Post
    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 [tex]\text{Res}(i)=\lim_{z\to i}\,\, (z-i)\frac{z^a}{1+z^2}[/tex].

    Also, I assume that [tex]a\ge 0[/tex]?

  3. #3
    I forgot to mention

    0 < a < 1

    Can you explain why the poles are simple?
    Last edited by monomocoso; 02-25-2012 at 05:23 PM.

  4. #4
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    Quote Originally Posted by monomocoso View Post
    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. #5
    So[TEX] res(i) = \frac {i^{a-1}} {2} [/TEX]?
    Last edited by monomocoso; 02-25-2012 at 05:42 PM.

  6. #6
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    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 [tex]\frac{z^a}{1+ z^2}= \frac{z^2}{1- iz}{1+ iz}[/tex]. 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 [tex](z- i)^{-1}[/tex]. That is why it is a simple pole.

  7. #7
    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?
    Last edited by monomocoso; 02-27-2012 at 04:48 PM.

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