Contour Integral

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Homework Statement


I have to evaluate the following integral by means of the Cauchy Integral Theorem:

[tex]\int_{- \infty}^{\infty}\frac{e^{-ikx}}{(x+i)(x+2i)}dx[/tex]


Homework Equations


[tex]f(z_0)=\frac{1}{2 \pi i} \oint_C \frac{f(z)}{z-z_0}dz[/tex]


The Attempt at a Solution


The idea I had was to consider the complex contour integral,

[tex]\oint_C\frac{e^{-ikz}}{(z+i)(z+2i)}dz= \oint_C f(z)dz[/tex]

If the contour C is a semi circular contour in the lower plane with radius > 2 (so that both poles are enclosed), we can break this integral up into two parts:

[tex]\oint_C f(z)dz = \int_a^{-a}f(z)dz + \int_{arc}f(z)dz[/tex]

[tex]\implies \int_{-a}^a f(z)dz = \int_{arc}f(z)dz - \oint_C f(z)dz[/tex]

At this point I can evaluate the contour integral by splitting into two integrals using partial fractions and then directly applying Cauchy's Integral Theorem. My problem is with this arc integral, which I think goes to zero in the limit [itex]a,R \rightarrow \infty[/itex]? I know Jordan's Lemma States, for a semi circular contour in the upper plane:

[tex]\lim_{R \rightarrow \infty} \int_{C_R} e^{i \alpha z} f(z) dz = 0 [/tex]

but this formula is for alpha greater than or equal to zero. In this case, alpha is less than zero and the contour is in the lower plane.

So, does this arc term vanish or need to be evaluated, and in either case how can I proceed?
 

Answers and Replies

  • #2
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I know Jordan's Lemma States, for a semi circular contour in the upper plane:

[tex]\lim_{R \rightarrow \infty} \int_{C_R} e^{i \alpha z} f(z) dz [/tex]

but this formula is for alpha greater than or equal to zero. In this case, alpha is less than zero and the contour is in the lower plane.


I do not believe [itex]\alpha[/itex] can be less than zero. Assuming that's correct, then carefully go through the derivation of Jordan's Lemma as you have written it above, then re-do the derivation with a contour in the lower half-plane ([itex]-C_R[/itex]) with the expression:

[tex]\lim_{R \rightarrow \infty} \int_{-C_R} e^{-i \alpha z} f(z) dz [/tex]
 
  • #3
kreil
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I guess my wording was a little awkward. I'm aware that alpha can't be less than zero in Jordan's lemma, and was trying to say that's why I can't apply the lemma.

Is it OK to just move the poles into the upper plane by taking z = -z? Then the contour would be in the upper plane and the form of the integral would be suitable for the lemma. This integral would differ from the one I have by only a sign (I think..), so if it is equal to zero by jordan's lemma then the other would be as well.
 
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kreil said:
Is it OK to just move the poles into the upper plane by taking z = -z?

Why would it not be?
 
  • #5
kreil
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excellent. then the arc term goes to zero and i get, letting a and R diverge:

[tex]\int_{- \infty}^{\infty}\frac{e^{-ikz}}{(z+i)(z+2i)}dz=- \oint_C \frac{e^{-ikz}}{(z+i)(z+2i)}dz = i \oint \frac{e^{-ikz}}{z+i}dz - i \oint \frac{e^{-ikz}}{z+2i}dz = 2 \pi i^2 \left ( e^{-k} - e^{-2k} \right ) = \frac{2 \pi}{e^k} \left ( e^{-k}-1 \right ) [/tex]

[tex]\implies \bar f (k) = \int_{- \infty}^{\infty} \frac{e^{-ikx}}{(x+i)(x+2i)}dx = \frac{2 \pi}{e^k} \left ( e^{-k}-1 \right ) [/tex]
 
  • #6
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I was wrong with my last post and removed it. I noticed you incorporated the -1 into the expression. Sorry.

Your answer is correct.
 

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