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Complex integration

  1. Nov 19, 2005 #1
    I have the function


    and the loop

    [tex]\gamma_{r_1,r_2}=I_{r_1,r_2}+II_{r_2}+III_{r_1,r_2}+IV_{r_1},\quad r_1,r_2>0,[/tex]

    which bounds the domain

    [tex]A_{r_1,r_2}=\{z\in\mathbb{C}\mid -r_1<\Re(z)< r_2\wedge0<\Im(z)<\pi\}.[/tex]

    Now I have to show that


    I know that the integral along [itex]\gamma_{r_1,r_2}[/itex] is the sum of the integrals along the four countours [itex]I_{r1,r_2},\dots,IV_{r_1}[/itex]. I also know that if [itex]f[/itex] is continuous in a domain D and has an antiderivative [itex]F[/itex] throughout D, then the integral along a contour lying in D is [itex]F(z_T)-F(z_I)[/itex], where [itex]z_T[/itex] is the terminal point, and [itex]z_I[/itex] the initial point, of the countour.
    However, using this, I am not able to show the aforementioned result, i.e, I get [itex]\int_{\gamma_{r_1,r_2}}f(z)\,dz=0[/itex].

    Could anyone try to calculate the integral, and then report back what they've found.
  2. jcsd
  3. Nov 19, 2005 #2


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    Ok, your contour as stated is tough for me to follow but looks like you're going around the origin in such a was as to encompass poles of the integrand. Right? When is the denominator zero in the complex plane? Thus can't rely on Cauchy's Theorem but rather the Residue Theorem.

    Edit: Alright, suppose Cauchy's Theorem is a special case of the Residue Theorem but you know what I mean.
    Last edited: Nov 19, 2005
  4. Nov 20, 2005 #3
    The function f has singularities in [itex]z=\frac{i}{4}\left(\pi+k2\pi\right),~k=0,1,2,\dots[/itex], and the contour is a rectangle with vertices at [itex]z=-r_1[/itex], [itex]z=r_2[/itex], [itex]z=r_2+i\pi[/itex], and [itex]z=-r_1+i\pi[/itex], where [itex]r_1,r_2>0[/itex]. It is also oriented counter-clockwise.
    Moreover, in the first question of the exercise, I am asked to calculate the residues at the singularities. Comparing Cauchy's Integral Theorem and Cauchy's Residue Theorem, I clearly see that the Residue Theorem can be used here, because the Integral Theorem requires the domain, which contains the contour, to be simply connected, while the Residue Theorem allows singularities inside the contour.
    Thus the Residue Theorem would be the right one to use here. I have not tried to apply this yet, but thanks for pointing me towards this.

    EDIT: I have just used Maple to quickly calculate the integral, using the Residue Theorem, and I got the right result.
    Last edited: Nov 20, 2005
  5. Nov 20, 2005 #4


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    Hey Sigmund, you do know how to calculate the residues by hand right? That is, how to show:

    [tex]\mathop\lim\limits_{z\to \pi i/4} (z-\pi i/4) \frac{e^z}{1+e^{4z}}=-\frac{1/ \sqrt{2}+i/ \sqrt{2}}{4}[/tex]
  6. Nov 20, 2005 #5
    Just factor the denominator by determining the poles : [tex]z = \frac{i( \pi + 2k \pi)}{4}[/tex] and k = 0,1,2,3

    So you get 4 factors, one for each k. Then you will see that the [tex](z-\pi i/4)[/tex] term will also appear in the denominator.

    Be sure that this only goes for first order poles

  7. Nov 21, 2005 #6
    I am not sure how to do this. The denominator of f has a zero of order 1 at [itex]z=i\pi/4[/itex], whence it can be written as [itex](z-i\pi/4)g(z)[/itex], where g(z) is analytic at [itex]i\pi/4[/itex] and [itex]g(i\pi/4)\neq0[/itex]. I do not know how to do this. Couldn't you give me a hint to how to solve this?
  8. Nov 21, 2005 #7


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    Yea, I'd like to know too. I was referring to the relation:

    [tex]\text{Res}_{z=z_0} \left(\frac{p(z)}{q(z)}\right)=\frac{p(z_0)}{q^{'}(z_0)}[/tex]

    Note: Only for a simple pole.
    Last edited: Nov 21, 2005
  9. Nov 21, 2005 #8
    The function f has a simple pole at [itex]z=z_0[/itex] (z_0 has been stated several times in this thread, so I omit it here). Hence the residue at that point is [itex]\text{Res}(f;z_0)=\lim_{z\to z_0}(z-z_0)f(z)[/itex].
    If we let [itex]f(z)=p(z)/q(z)[/itex], where p and q are both analytic at z_0, and q has a simple zero at z_0, while [itex]p(z_0)\neq0[/itex], we can do the following simplification:
    [tex]\text{Res}(f;z_0)=\lim_{z\to z_0}(z-z_0)f(z)=\lim_{z\to z_0}(z-z_0)\frac{p(z)}{q(z)}=\lim_{z\to z_0}\frac{p(z)}{\frac{q(z)-q(z_0)}{z-z_0}}=\frac{\lim_{z\to z_0}p(z)}{\lim_{z\to z_0}\frac{q(z)-q(z_0)}{z-z_0}}=\frac{p(z_0)}{q'(z_0)}[/tex]

    This can be used in the actual problem, because the requirements about analyticity at z_0, and that p(z) is nonzero at z_0, are met. Using this, it is easy to calculate the residue at z_0, while factoring the denominator undoubtedly could be done, although difficult, i presume.
    As Albert Einstein said: "Everything should be made as simple as possible, but no simpler."
    Last edited: Nov 21, 2005
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