Say I am asked to solve $$\int_0^\infty \frac{sin(x)}{x} dx$$

I examine $$\int_C\frac{e^{iz}}{z} dz$$ where $z$ is complex and C is the contour below.


As part of my solution, I show that $$\lim_{R \to \infty} \int_{\Gamma_R}\frac{e^{iz}}{z} dz = 0 $$

Although I have not done an extensive amount of these problems, I have yet to see a case where the integral over this outer contour goes to anything but $0$. Is this a necessary condition of the residue theorem, or is it purely a coincidence of the problems I have been attempting?

For some context, I am studying for qualifying exams in physics (we like to hand-wave). A common problem is "Use the residue theorem to show [integral = finite value]". A lot (all) of the solutions I am using to study say this integral goes to $0$ without any additional work shown. I don't know if this is hand-waving, or referencing something I haven't found.

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    $\begingroup$ Typically the contour is constructed in such a way that the portion of the contour that does not coincide with the relevant integral tends to $0$. In many cases it is enough to choose just a circle or semicircle in the complex plane that does this. If it does not tend to $0$, you should probably pick another contour (or perhaps the integral does not converge). $\endgroup$ – Rellek Sep 2 '16 at 21:10
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    $\begingroup$ You've been studying examples where it works out. It doesn't always work out. $\endgroup$ – zhw. Sep 2 '16 at 21:12
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    $\begingroup$ It is probably important to mention that there are some very typical contours that are chosen. Notice that in your example they intentionally choose a contour in the upper half of the complex plane. You should try reflecting this contour in the bottom half -- it will not work out so nicely. $\endgroup$ – Rellek Sep 2 '16 at 21:14

It is a necessary step to show that that the contour goes to zero.


$\int_{-\infty}^{\infty} \frac {z}{z^2 +1}$ does not converge.

You can use the same contour. You don't need to take the detour around 0, but close enough to the same contour.

You can evaluate the residual. But you need to check to see what happens as the radius of the contour gets to be very large. And in this case we see that it does not converge.

In all of the "nice" problems the path integral contour goes away as R gets big. But that doesn't mean it always happens and that you don't need to check it.

  • $\begingroup$ Out of curiosity, do you know of any instances where this portion of the contour converges to a non-zero value? As has been pointed out, if I chose to take the contour (from my original problem) in the bottom half of the complex plane, I get $2 i \pi$. So, I would think something like this exists in the upper half as well. $\endgroup$ – user363165 Sep 2 '16 at 22:43

You can use Jordan's Lemma, or directly with the M-L estimmation theorem:

$$\left|\int_{\Gamma_R}\frac{e^{iz}}zdz\right|\le \pi R\max_{z\in\Gamma_R}\left|\frac{e^{iz}}z\right|=\pi R\max_{z\in\Gamma_R}\frac{e^{-\text{Im}\,z}}R=\pi e^{-\text{Im}\,z}\xrightarrow[R\to\infty]{}0$$

since Im$\,z>0\;$ and thus Im$\,z\to\infty\;$ when $\;R\to\infty\;$

  • $\begingroup$ I am not so much asking how to show it, but rather if it needs to be shown (within my specific context). The comments on the question do a good job of addressing this in the manner that would be most helpful to me. $\endgroup$ – user363165 Sep 2 '16 at 21:46
  • $\begingroup$ Of course it needs to be shown. I've taught physics and engineers and they must give a complete explanation. I can't say what they demmand from you where you study, but I'd say this is a must. $\endgroup$ – DonAntonio Sep 2 '16 at 21:50

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