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Disclaimer: the knowledge I have about contour integration is solely from the book "Mathematical Methods in the Physical Sciences" by Mary L. Boas.

I am trying to understand how the following function is derived: \begin{equation} \displaystyle\lim_{\epsilon \to 0^+} \int\limits^\infty_{-\infty} \frac{e^{-ixt}}{x+i \epsilon} \; \mathrm{d} x = - 2 \pi i \Theta(t) \end{equation} where $\Theta(t)$ denotes the unit step function.

In order to do this, I start by considering the integral: \begin{equation} \int\limits_{-\infty}^\infty \frac{e^{-i x t}}{x + i \epsilon} \; \mathrm{d} x \end{equation} where $t>0$. In order to evaluate this integral I believe we can use the ``contour integration'' technique and thus I consider: \begin{equation} \oint_C \frac{e^{-i z t}}{z + i \epsilon} \; \mathrm{d} z \tag{1} \end{equation} where $C$ is the (clockwise) contour as shown in the figure:

enter image description here

Clearly, there is a simple pole at $z= -i \epsilon$ and the residue can be easily found: \begin{equation} \mathrm{Res}(-i \epsilon) = \displaystyle\lim_{z \to -i \epsilon}\left\{ (z+i \epsilon) \frac{e^{-i z t}}{z + i \epsilon} \right\}= e^{-\epsilon t} \end{equation} Therefore, by the residue theorem, we can evaluate equation $(1)$ as: \begin{equation} \oint_C \frac{e^{-i z t}}{z + i \epsilon} \; \mathrm{d} z = - 2 \pi i e^{-\epsilon t} \end{equation} where the minus sign is due to the fact that the contour is clockwise. Subsequently, by taking $\epsilon$ to be a very small positive constant this becomes: \begin{equation} \oint_C \frac{e^{-i z t}}{z + i \epsilon} \; \mathrm{d} z = - 2 \pi i \end{equation} Letting $z=\rho e^{i \theta}$, we can write: \begin{equation} \oint_C \frac{e^{-i z t}}{z + i \epsilon} \; \mathrm{d} z = \int\limits^\rho_{-\rho} \frac{e^{-i x t}}{x + i \epsilon} \; \mathrm{d} x + \int\limits^{-\pi}_{0} \frac{e^{-i z t}}{\rho e^{i \theta} + i \epsilon} \rho i e^{i \theta} \; \mathrm{d} \theta \tag{2} \end{equation} At this point I am stuck. I believe the second term on the right-hand side of the above equation should tend to zero as $\rho \rightarrow \infty$, however I do not see why this would be true. Any help would be much appreciated.

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  • $\begingroup$ I take it that $\rho$ is the radius of your circle? What is z in your second integral -- should it not be in terms of $\theta$?. I do not see that integral going to 0 because if you ignore the i$\epsilon$ it is essentially the integral of $e^{-izt}$. But do you want it to be 0? Shouldn't it add your step function? $\endgroup$ – Betty Mock Jan 15 '14 at 21:48
  • $\begingroup$ @BettyMock yes $\rho$ is the radius of the circle. And $z=x+iy=\rho e^{i \theta}$ is a complex number to take the first integral to the complex plane. I believe that the second term on the right-hand side of equation $(2)$ should go to zero (because this is the usual way I have learnt how to use contour integration). If it does not go to zero, then I really do not how to use the residue theorem in this case. $\endgroup$ – Hunter Jan 15 '14 at 21:53
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    $\begingroup$ $\epsilon \to 0^{+}$. $\endgroup$ – Felix Marin Jan 20 '14 at 9:40
  • $\begingroup$ @FelixMarin thanks; I have edited it. $\endgroup$ – Hunter Jan 20 '14 at 12:01
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The second integral on the RHS is

$$i \rho \int_0^{-\pi} d\theta \, e^{i \theta} \frac{e^{-i t \rho e^{i \theta}}}{\rho e^{i \theta}+i\epsilon} = i \rho \int_0^{-\pi} d\theta \, e^{i \theta} \frac{e^{-i t \rho \cos{\theta}} e^{t \rho \sin{\theta}}}{\rho e^{i \theta}+i \epsilon}$$

The magnitude of the above integral is bounded by

$$\int_0^{-\pi} d\theta \, e^{t \rho \sin{\theta}} = \int_0^{\pi} d\theta \, e^{-t \rho \sin{\theta}}$$

Use symmetry and $\sin{\theta} \ge 2 \theta/\pi$, and the integral is further bounded by

$$ 2\int_0^{\pi} d\theta \, e^{-2 t \rho \theta/\pi} \le \frac{\pi}{t \rho}$$

As $\rho \to \infty$, the integral therefore vanishes as $\pi/(t \rho)$ when $t \gt 0$.

When $t \lt 0$, however, the above bounds do not apply; rather, we must close above the real axis. because there are no poles there, the integral is zero. Thus, the step function.

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  • $\begingroup$ Thanks for your reply. I apologize if this is a stupid question, but could you explain what you mean by: "The magnitude of the above integral is bounded by $$\int_0^{-\pi} d\theta \, e^{t \rho \sin{\theta}} = \int_0^{\pi} d\theta \, e^{-t \rho \sin{\theta}}$$ I have never seen this before, so if it is too much to explain, then could you provide a reference. $\endgroup$ – Hunter Jan 15 '14 at 22:46
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    $\begingroup$ @Hunter: This is nothing more than an application of the Cauchy-Schwartz lemma, in that the magnitude of the integral is bounded by the integral of the magnitude of the integrand. You may also refer to Jordan's lemma. $\endgroup$ – Ron Gordon Jan 15 '14 at 23:17
  • $\begingroup$ I will have to read up on the Cauchy-Schwartz lemma (probably tomorrow, because I am too tired now) and see how I can apply it to this problem. As for now, I will accept your answer and thank you again for your help. $\endgroup$ – Hunter Jan 15 '14 at 23:21
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If $\,t<0\,$, because there are no singularities in the upper half plain, the integral (the principal value of it, same applies for further down) is zero by Jordan's lemma (see any textbook on complex analysis). For $\,t>0\,$ the integral is equal to $\,-2\pi ie^{-\varepsilon t}\,$ (by Jordan's lemma again), what tends to $\,-2\pi i\,$, when $\,\varepsilon\to0$. When $\,t=0\,$ the integral is $\,-\pi i\,$. The equality holds also at that point if the value of the Heaviside step function at zero is assumed to be $\,1/2$.

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