Using Contour Integration and the Residue Theoremby sikrut Tags: contour, integration, residue, theorem 

#1
Feb1413, 07:32 PM

P: 37

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 realvalued 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 F_{1} as either the real part or the imaginary part of complexvalued integrals where the cos and sinfunctions have been replaced by the e^{i}^{w}^{t} function, using e.g., [itex] \int_{\infty}^{+\infty} Re[f(\omega)]d\omega = Re[\int_{\infty}^{+\infty} f(\omega)d\omega][/itex].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 realaxis contour with a semicircle of radius R either in the upper half or in the lower half of the complex plane such that contribution from the semicircle vanishes in the limit of [itex]R \rightarrow \infty[/itex]. (5) Use your Residue Recipes to find the residues and complete the calculation. Attempt: [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)(xi)}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] 



#2
Feb1413, 07:57 PM

HW Helper
P: 1,391

Now, you need to decide based on the sign of t which halfplane to close your contour in. Do you know how to decide that? Also, you (presumably) know that the straightline segment will be the integral you want to evaluate, and the straightline segment plus the semicircular 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 straightline 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 semicircle integral and b) how to evaluate the overall contour integral. Does this help you go further? 



#3
Feb1413, 08:42 PM

P: 37





#4
Feb1413, 09:50 PM

Sci Advisor
HW Helper
Thanks
P: 25,170

Using Contour Integration and the Residue Theorem 



#5
Feb1513, 12:13 AM

HW Helper
P: 1,391





#6
Feb1513, 05:41 PM

P: 37

I could call that I(z). Then, I could define the semicircle as: z = re^{iθ}, and plug this into I(z), where I would both add, and subtract that integral? 



#7
Feb1513, 08:02 PM

HW Helper
P: 1,391

$$e^{ivs}\oint_C \frac{e^{izs}}{(zi)(z+i)},$$ where C is your straight line contour with the semicircle. You can then write this contour integral as the sum of the integral along the straightline contour, on which you parameterize z = x, and the integral around the semicircle, on which you parameterize z = re^{iθ}, as you said. Now, you want to find the value of the integral along the realaxis, so you'll need to evaluate the contour integral and the semicircular integral (in the limit that the semicircle becomes large). Can you take it from here? 



#8
Feb1813, 01:52 PM

P: 37

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. 



#9
Feb1813, 02:32 PM

P: 37

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}}{(zi)(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}}{(zi)(z+i)}dz = e^{ivs}\oint_C \frac{e^{izs}}{(zi)(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? 



#10
Feb1813, 03:36 PM

HW Helper
P: 1,391

Going back a step for a second, you have [tex] e^{ivs}\oint_C \frac{e^{izs}}{(zi)(z+i)} = e^{ivs}\oint_{C_1} \frac{e^{izs}}{(zi)(z+i)} + e^{ivs}\oint_{C_2} \frac{e^{izs}}{(zi)(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 semicircle. You now want to parameterize z along each of these contour pieces. As we discussed previously, on the straightline 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. 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 halfplane, but I don't think it should be too hard. Let us know if you get stuck again. 


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