Contour Integral

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kreil
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Homework Statement


Calculate the following integral along three different circular contours,

[tex]\int_{C_j}\frac{dz}{z(3z-1)^2(z+2)}[/tex]

where
[tex]C_1:0<r_1<1/3[/tex]
[tex]C_2:1/3<r_2<2[/tex]
[tex]C_3: r_3>2[/tex]

The Attempt at a Solution



The function has singularities at z=0, z=1/3 and z=-2. Thus all three contours enclose singularities and Cauchy's integral theorem doesn't hold (none of the integrals are immediately zero).

Along each circular contour,

[tex]z=re^{i \theta}\implies dz=ire^{i \theta}d \theta[/tex]

Am I going to need to use partial fractions for this? What is the best way to get started?
 
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Answers and Replies

  • #2
vela
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Perhaps calculate the residue for each pole.
 
  • #3
kreil
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we haven't covered that..i think i need to it a different (probably harder) way
 
  • #4
owlpride
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Partial fractions might be the easiest way. You could also expand the function into its Maclaurin series and apply Cauchy's integral formula to the individual terms (it will apply to most terms, and you can do the exceptions by hand).
 
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  • #5
kreil
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After expanding I got the following:

[tex]\frac{1}{z(3z-1)^2(z+2)}=\frac{1}{2z}+\frac{9}{7(3z-1)^2}-\frac{98}{z+2}[/tex]

When I put this back into the integral and substitute for z and dz (defined in OP) I can't get the wolfram integrator to integrate the second term:

[tex]\frac{9ir}{7}\int_0^{2 \pi}\frac{e^{i \theta}}{(3re^{i \theta}-1)^2}d \theta[/tex]

did I do something incorrectly, or does this integral need further manipulation?
 
  • #6
owlpride
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Do you know that the Fundamental Theorem of Calculus holds true for complex-valued functions? How would you usually evaluate

[tex]\int_a^b \frac{f'(x)}{(3 f(x)-1)^2}dx[/tex]
 
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  • #7
kreil
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u = f(x)
du = f'(x) dx

doesnt this essentially go backwards though? I already subbed in [itex]z=re^{i \theta}[/itex]
 
  • #8
owlpride
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Not exactly. The initial substitution reduced your integral from an integral of a complex variable to an integral in terms of a real variable. The second substitution computes an antiderivative for the function of a real variable. Your integrand assumes complex values, but it is a function of a real variable.
 
  • #9
kreil
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This will give the last two integrals equal limits (since [itex]e^{2i \pi}=e^0=1[/itex]) so am I correct to say that the second and third terms drop off and the total integral is [itex]i \pi[/itex]? Also, this would mean that all three integrals are equal to this (despite the fact that they enclose differing numbers of singularities)...?
Just to be clear the whole expression is:

[tex]\frac{i}{2}\int_0^{2\pi}d \theta+\frac{9ir}{7}\int_0^{2 \pi}\frac{e^{i \theta}}{(3re^{i \theta}-1)^2}d \theta-98\int_0^{2 \pi}\frac{ire^{i \theta}}{re^{i \theta}+2}d \theta[/tex]
 
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  • #10
owlpride
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If r < 1/3, the total integral is i*Pi indeed. You should get a different answer for r > 2 though.

What happens to the third fraction when r > 2?
 
  • #11
vela
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No. You have to be careful. When you integrate the last term, you'll get log(z+2), and you'll have to worry about a branch cut.
 
  • #12
kreil
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If r < 1/3, the total integral is i*Pi indeed. You should get a different answer for r > 2 though.

What happens to the third fraction when r > 2?


when you say r>2 do you mean r--> inf?

also what about 1/3 < r < 2?
 
  • #13
kreil
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No. You have to be careful. When you integrate the last term, you'll get log(z+2), and you'll have to worry about a branch cut.

but when you use the u-sub the last two integrals are essentially

[tex]\int_a^af(x)dx=0[/tex]...right?
 
  • #14
owlpride
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I mean any r>2, not necessarily the limit as r -> inf (though that wouldn't make a difference).

Try this: what does Euler's integral theorem let you conclude about the difference between the integral of 1/(z+2) over the circle |z| = 3 vs the circle |z+2|=1?
 
  • #15
kreil
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sorry I am not sure what euler's integral theorem is (we haven't covered it yet)

if i was to let r--> inf though, the last term would tend to 1
 
  • #16
owlpride
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Sorry, I meant Cauchy's integral theorem. My bad!
 
  • #17
owlpride
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The problem with the complex log is that it is not continuous: you will jump from [itex]2 \pi[/tex] to 0 at some point. That often causes trouble.
 
  • #18
kreil
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Haha ok. In that case I believe the integrals would be equal? (they enclose the same singularity so one could deform the contours to be equal without crossing a singularity or changing the value of the integral)
 
  • #19
owlpride
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Yes, they are equal. And the integral over |z+2|=1 is a bit easier to compute.
 
  • #20
kreil
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Deforming the z>2 contour in that way in this problem would cross a singularity (z=1/3) wouldn't it?

Also, the integral expression I have produced is general and not part a-c specific. I'm confused as to how the last two terms are not zero after the u-sub, since they have equal limits...should i be integrating wrt u and then subbing back in [itex]re^{i \theta}[/itex]? Why do the last two terms drop for r<1/3 but not for parts (b) and (c)?
 
  • #21
owlpride
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The middle term always drops (you will learn why later). The last term does not always drop because it does not have an antiderivative. log|z+2| looks like an antiderivative, but it is not - the log function is not defined continuously on a closed loop around the origin.

When you do the r > 2 contour, I suggest you integrate the first two fractions over the given contour and only modify the circle for the last fraction. This way your contours don't cross additional singularities.
 
  • #22
kreil
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modifying the circle amounts to changing the value of r, a constant, correct? I guess this is what I don't understand..how is it that changing r can cause terms to drop when it is a constant?
 
  • #23
kreil
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after the u-sub i have

[tex]
\frac{i}{2}\int_0^{2\pi}d \theta+\frac{9}{7}\int \frac{du}{(3u-1)^2}-98\int \frac{du}{u+2}
[/tex]

Integrating,

[tex]i\pi + \frac{9}{7(3-9u)}-98ln(u+2)[/tex]

now what are the evaluation limits on the second two terms? or should I keep 0-2pi and resub re^itheta?
 
  • #24
owlpride
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You are doing fine on the middle term. The problem is that log(z+2) is NOT an antiderivative for 1/(z+2) when r > 2. Why? Because the log function is not continuous (or differentiable!) on a closed loop around the origin (which means that log(z+2) is not differentiable on a closed loop around z=2). Google "branch cut" for an explanation.
 
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  • #25
kreil
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so how is it that when r<1/3 the second and third terms drop? is it because this contour does not enclose any part of these functions? I'm sorry that I'm not getting this but it's all quite new for me.
 
  • #26
owlpride
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Yes, that's the reason. As long as r < 2, 1/(z+2) has a well-defined antiderivative: the log-function is a perfectly happy differentiable function as long as you don't need it on all sectors of the plane. 1/(3z -1)^2 *always* has an antiderivative, and that's why its contour integral will always come out to be zero.
 
  • #27
kreil
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ok, that makes sense. so a and b have the same answer of [itex]i \pi [/tex] correct?

how should i go about solving part c? I read the article on branch cuts and also found a section in my book on it, but since we haven't covered the residue theorem yet I am having trouble employing it
 

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