Would like help with a contour integral.

Question

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:

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.

Answer 1

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.

Answer 2

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 $\,1/2$.

Answer 3

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\begin{align} &\bbox[5px,#ffd]{\int_{-\infty}^{\infty} {\expo{-\ic xt} \over x + \ic 0^{+}}\,\dd x} = \mrm{P.V.}\int_{-\infty}^{\infty} {\expo{-\ic xt} \over x}\,\dd x + \int_{-\infty}^{\infty} \expo{-\ic xt}\bracks{-\ic\pi\,\delta\pars{x}}\,\dd x \\[5mm] = &\ \int_{0}^{\infty} {\expo{-\ic xt} - \expo{\ic xt} \over x}\,\dd x - \ic\pi = -2\ic\int_{0}^{\infty} {\sin\pars{xt} \over x}\,\dd x - \ic\pi \\[5mm] = &\ -2\ic\on{sgn}\pars{t}\int_{0}^{\infty} {\sin\pars{x} \over x}\,\dd x - \ic\pi = -\pi\ic\on{sgn}\pars{t} - \ic\pi \\[5mm] = &\ \bbx{-2\pi\,\ic\,\Theta\pars{t}} \\ & \end{align}