This is much more on the mathematical side on the derivation, so I may be barking up the wrong Stack Exchange for this question; however, I have a curiosity about a contour integration performed in deriving Coulomb's law from classical field theory in Matthew Schwartz' QFT book. Through an action upon the time component of the vector potential via a propagator

$$A_0=\frac{e}{\Box}\delta^3(x). \tag{3.61}$$

One eventually obtains, by going into Fourier space, the following integral

$$\int_\infty^\infty dk\frac{e^{ikr}-e^{-ikr}}{k}.$$

To evaluate this integral, the author plays a cute trick by adding a small parameter to the denominator and letting it go to zero. This parameter, $\delta$, is set to be greater than zero (however, it can be set to be less than zero, and I presume that the integration will then just pick the $e^{ikr}$ out of the integration. He therefore lets

$$\int_\infty^\infty dk \frac{e^{ikr}-e^{-ikr}}{k}=\lim_{\delta \rightarrow0}\bigg[\int_\infty^\infty dk \frac{e^{ikr}-e^{-ikr}}{k+i\delta}\bigg]. \tag{3.63}$$

He then states that "for $e^{ikr}$, we must close the contour up to get exponential decay at large k. This missed the pole, so this term gives zero".

I am not necessarily sure what he means by "misses the pole", nor why this implies that $e^{irk}$ does indeed go to zero.


1 Answer 1


We want to integrate $\int_{-\infty}^\infty\frac{e^{ikr}}{k+i\delta}dk$, but we don't know how. But we do know something about complex integrals: a contour integral around a closed loop is equal to $2\pi$ time the sum of the residues enclosed by the loop. So if we can write the integral above as a complex integral around a closed loop, we'll be able to do the integral just by counting the residues inside the loop. Let's pick this loop:

enter image description here

The claim is that in the limit $R\rightarrow 0$,


Why should that be true? Well, it will be true provided the half-circle part of the integral goes to zero as $R\rightarrow \infty$. This is clearly true, since the half-circle has $|k|=R\rightarrow\infty$, and the imaginary part of $k$ is positive, meaning that $e^{-ikr}\propto e^{-\Im(k)r}$, which decays exponentially. Note we could NOT have chosen to close the contour with a half-circle below the real axis, since then $e^{-ikr}$ would grow exponentially in $R$.

Now we can just ask ourselves if there are any residues enclosed by the loop. There aren't, because the only singularity of our function is $k=-i\delta$, which occurs below the real axis. That's what he means by we "missed the pole"--we don't encircle the pole. Thus, the total integral is zero.


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