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Homework Help: Using Contour Integration and the Residue Theorem

  1. Feb 14, 2013 #1
    Using contour integration and the residue theorem, evaluate the following
    "Fourier" integral:
    [tex]F_1(t) := \int_{-\infty}^\infty \frac{\Gamma sin(\omega t)}{\Gamma^2 + (\omega +\Omega )^2} dw[/tex]

    with real-valued constants [itex]\Gamma > 0[/itex] and [itex]\Omega[/itex]. Express your answers in terms of [itex]t, \Gamma[/itex] and [itex]\Omega[/itex].

    Hints: Before starting the contour integral formulation:
    (1) Since [itex]cos(\omega t) = Re(e^{i \omega t})[/itex] and [itex]sin(\omega t) = Im(e^{i \omega t})[/itex], write F1 as either the real part or the imaginary part of complex-valued integrals where the cos- and sin-functions have been replaced by the eiwt function, using e.g., [itex] \int_{-\infty}^{+\infty} Re[f(\omega)]d\omega = Re[\int_{-\infty}^{+\infty} f(\omega)d\omega][/itex].
    (2) Get rid of the [itex]\Gamma[/itex] and [itex]\Omega[/itex] parameters insde the integrals, by the following substitutions:
    express [itex]\omega[/itex] by [itex]x := (\omega - \Omega)/\Gamma;[/itex]
    [itex]t[/itex] by [itex]s := \Gamma t,[/itex]
    and [itex]\Omega[/itex] by [itex]\nu := \Omega/\Gamma.[/itex]
    (3) Depending on the sign of t, then "close" the real-axis contour with a semi-circle of radius R either in the upper half or in the lower half of the complex plane such that contribution from the semi-circle vanishes in the limit of [itex]R \rightarrow \infty[/itex].
    (5) Use your Residue Recipes to find the residues and complete the calculation.

    [tex]F_1(t) := \int_{-\infty}^\infty \frac{\Gamma sin(\omega t)}{\Gamma^2 + (\omega +\Omega )^2} dw[/tex]
    [tex]= \int_{-\infty}^\infty \frac{\Gamma Im(e^{i \omega t})}{\Gamma^2 + (\omega +\Omega )^2} dw[/tex]
    [tex]= Im(\int_{-\infty}^\infty \frac{\Gamma e^{i \omega t}}{\Gamma^2 + (\omega +\Omega )^2} dw)[/tex]

    [itex]\omega \rightarrow x:= (\omega - \Omega)/\Gamma \rightarrow \omega = \Gamma x + \Omega \rightarrow d\omega = \Gamma dx[/itex]

    [itex]t \rightarrow s := \Gamma t \rightarrow t= s/\Gamma [/itex]

    [itex]\Omega \rightarrow \nu := \Omega/\Gamma[/itex]

    [itex]\omega t = \frac{\Gamma x + \Omega)s}{\Gamma} = (x + \frac{\Omega}{\Gamma})s = (x+\nu)s[/itex]

    where: [itex] I = \int_{-\infty}^\infty \frac{\Gamma e^{i \omega t}}{\Gamma^2 + (\omega +\Omega )^2} dw[/itex]

    [tex] I = \int_{-\infty}^\infty \frac{\Gamma^2 e^{i(x+\nu)s}}{\Gamma^2 (x^2 + 1)}dx = \int_{-\infty}^\infty \frac{e^{i(x+\nu)s}}{(x^2 + 1)}dx = e^{i \nu s}\int_{-\infty}^\infty \frac{e^{ixs}}{(x+i)(x-i)}dx[/tex]

    Now, I know that the next step is concatenating the line with with the semicircle, but I am not quite sure how to go about setting it up:
    [tex] I = \int_{I_R^+} + \int_{C_R^+} - \int_{C_R^-}[/tex]
    [tex] I = \oint_{P_R^+} -\int_{C_R^-}[/tex]
    Last edited: Feb 14, 2013
  2. jcsd
  3. Feb 14, 2013 #2


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    You haven't defined what ##I_R^+##, etc., are, so I'm not sure what exactly you mean by them. However, let me point out that your poles don't lie on the real axis, so you only need two pieces to your semicircle contour: the straight-line piece along the real axis and the semicircular piece.

    Now, you need to decide based on the sign of t which half-plane to close your contour in. Do you know how to decide that? Also, you (presumably) know that the straight-line segment will be the integral you want to evaluate, and the straight-line segment plus the semi-circular segment should equal the contour integral around the entire contour.

    It sounds like you're not sure what to write for the overall contour integral - is that right? Well, you know that along the straight-line segment you want the contour integral to reduce to the integral you want to calculate, so usually the easiest thing to try is just replacing x with the complex variable ##z## to get the contour integral integrand. Then you only need think about a) the change of variables from z for the semi-circle integral and b) how to evaluate the overall contour integral.

    Does this help you go further?
  4. Feb 14, 2013 #3
    I meant that notation as the segments that I was concatenating, where I is the line segment on the real axis, and the C was the semi-circle on the positive, imaginary axis ( I meant to write the integral about C(+) minus the same integral, not the C(-) )

    Because t is positive, that means I need to follow the counter-clockwise direction, therefore I need to close my contour within the positive imaginary plane. Would that be about right?

    It helps a bit, but I'm not sure how to right out the semi-circle integral. I'm just having trouble understanding how to put the semi-circle in the equation.
  5. Feb 14, 2013 #4


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    I think you've got the right general idea about things. If you close the contour in the correct half plane then the integral around the semicircle will go to zero as the radius of the semicircle goes to infinity. Which semicircle you close around will depend on whether Re(ω) is positive or negative. You need to break into two cases. And one of your initial steps is quite wrong. You can say sin(ωt)=Im(exp(iωt) but you can't factor the Im out of the integral. ω is a complex variable. You need to write sin(ωt)=(exp(iωt)-exp(-iωt))/(2i) and split it into two integrals. You may want to close the semicircles in opposite directions for each one.
  6. Feb 15, 2013 #5


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    The ##\omega## in sikrut's original integral is real, so s/he can pull the Im outside of the integral, and then generalize the integral to a complex integral.

    If you had a purely circular contour of radius R and were trying to perform some contour integral ##\oint_{|z| = R} dz~f(z)##, do you recall what change of variables you should make? You will basically make the same change of variables for the semicircle arc you're dealing with in your problem.
  7. Feb 15, 2013 #6
    So like you said, I should substitute z for x, leaving me with:[tex]e^{ivs}\int_{-\infty}^\infty \frac{e^{izs}}{(z-i)(z+i)}[/tex]

    I could call that I(z). Then, I could define the semi-circle as: z = re, and plug this into I(z), where I would both add, and subtract that integral?
  8. Feb 15, 2013 #7


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    Close! When you replace x with z, the full contour integral is

    $$e^{ivs}\oint_C \frac{e^{izs}}{(z-i)(z+i)},$$

    where C is your straight line contour with the semi-circle. You can then write this contour integral as the sum of the integral along the straight-line contour, on which you parameterize z = x, and the integral around the semi-circle, on which you parameterize z = re, as you said.

    Now, you want to find the value of the integral along the real-axis, so you'll need to evaluate the contour integral and the semi-circular integral (in the limit that the semi-circle becomes large). Can you take it from here?
  9. Feb 18, 2013 #8
    Sorry for taking so long to reply. I've been busy working on this and other work. Just wanted to say thank you for your help.

    I think I got it, but I'm not certain. Still a bit confused about what I did, but i found some stuff in my notes and combined it with your advice. Hopefully I did it correctly.
  10. Feb 18, 2013 #9
    i'm going to throw some more of my work in here and let you check it if thats ok:
    [tex] e^{ivs}\oint_C \frac{e^{izs}}{(z-i)(z+i)} = e^{ivs}\int_{-\infty}^\infty \frac{e^{ixs}}{(x^2 + 1)} + e^{ivs}\int_{-R}^R \frac{e^{i(Re^{i\theta})s}}{R^2e^{2i\theta}} - e^{ivs}\int_{-R}^R \frac{e^{i(Re^{i\theta})s}}{R^2e^{2i\theta}}[/tex]
    where:[tex] z = Re^{i\theta} = Rcos(\theta) + iRsin(\theta)[/tex]
    [tex]|e^{izt}| = e^{Re(izt)} = e^{-tRsin(\theta)} \rightarrow because \rightarrow iz = icos(\theta) - sin(\theta)[/tex]

    Therefore:[tex]e^{ivs}\oint_C \frac{e^{izs}}{(z-i)(z+i)}dz = e^{ivs}\oint_C \frac{e^{izs}}{(z-i)(z+i)}dz - e^{ivs}\int_{-R}^R \frac{e^{-Rtsin(\theta)}}{(R^2e^{2i\theta}} dt[/tex]

    So the numerator of the semicircle is a negative exponent, throwing it to the denominator. Now all the R values are in the denominator, so as R ----> infinity, the who equation goes to 0, as long as 0 < [itex]\theta[/itex] < [itex]pi[/itex] .

    How does that look?

    Also, I wasn't too sure how to show the limit as R goes to infinity. Do I simply disregard the integral and take the limit of the semicircle equation?
  11. Feb 18, 2013 #10


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    I'm not sure what your last two terms here are. On the right hand side you should have 1) the integral along the real line, which you have, although the limits should be from -R to +R at the moment, and 2) the integral over the semicircle. Instead of (2), you have two integrals which appear to be exactly the same and would cancel out to zero.

    Going back a step for a second, you have

    [tex] e^{ivs}\oint_C \frac{e^{izs}}{(z-i)(z+i)} = e^{ivs}\oint_{C_1} \frac{e^{izs}}{(z-i)(z+i)} + e^{ivs}\oint_{C_2} \frac{e^{izs}}{(z-i)(z+i)},[/tex]
    where ##C = C_1 + C_2## is the full, closed contour, C_1 is the piece along the real axis and C_2 is the semi-circle.

    You now want to parameterize z along each of these contour pieces. As we discussed previously, on the straight-line piece we can just set z = x. This gives you the first term on the right hand side that you wrote down.

    On the semicircular piece, we previously discussed that you should set ##z = \exp(Ri\theta)##. This is basically a change of variables from z to ##\theta##, so you need to account for that. i.e., ##dz## needs to be replaced with ##d\theta## time an appropriate factor. It looks like you wrote the factor down correctly, but you neither wrote down the dz, dx or ##d\theta## at all in the quoted equation. You also need to change the limits of integration to the appropriate limits for theta. So, the second term you wrote down is almost what term (2) should be, aside from those few errors. The third term you added appears to just be the subtraction of the second term - you can't do that. The third term doesn't exist.

    So, try that again and show us what you get. Once you have it, the next step will be to solve for (1), the integral of interest. To do that, you can evaluate the closed contour integral with the residue, but you still need to figure out (2) - again, you can't just subtract it away. Typically, one shows that the integral (2) will go to zero as R tends to infinity. This argument goes along the lines of what you started showing in the rest of your post.

    Again, the equation is wrong. There's an additional mistake beyond what I mentioned above, and that's that you replaced ##\exp(iRe^{i\theta})## with its real part only - the imaginary part is still there. The reason I'm pointing this out is that even though the theta integral is wrong, it almost has the form you need to show the integral vanishes. You've noted that the real part is an exponential of ##-Rt \sin\theta##, which means that the sign of t determines which half-plane you need to close the contour in. The reason that matters is that as you take R to infinity, this exponential will basically kill this integral. (You need to show it rigorously, though).

    To rigorously show that the integral tends to zero, the typical approach is to show that your integral satisfies Jordan's Lemma. (I'm linking to wikipedia for a quick reference, but you should look it up in your textbook to be sure there are no errors on the wikipedia article). You may need to make some minor modifications to the argument for the case in which you close the circle in the negative half-plane, but I don't think it should be too hard.

    Let us know if you get stuck again.
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