• Consider the integral $$I=\int\limits_{-\infty}^{\infty}dx \frac{e^{-ixt}}{x^2-a^2}=\int\limits_{-\infty}^{\infty}dx \frac{\cos(xt)}{x^2-a^2}$$ for $t<0$ and $a>0$. The integrand has poles at $x=\pm a$ on the real axis. We can use contour integral to evaluate this.

  • Let us choose a contour that lies entirely in the upper half of the complex plane. Let the contour be so indented that the smaller semicircles $C_1$ and $C_2$, each of radius $\epsilon$, goes over the poles at $x=\pm a$, and the larger semicircle $\Gamma$ also lies in the upper half of the complex plane (so that contribution from it vanishes).

  • With this choice of contour, I want to show that $I=0$. This is a famous integral that arises in solving the wave equation by the method of Green's function. But my physics textbooks do not work out this integral explicitly, by considering different parts of the contour piece by piece. See page 125, figure 50 here.

  • For this contour, using residue theorem and Jordan's lemma, gives (schematically),

$$0=\int\limits_{-\infty}^{-a-\varepsilon} +\int\limits_{-a-\varepsilon}^{-a+\varepsilon}+\int\limits_{-a+\varepsilon}^{a-\varepsilon}+\int\limits_{a-\varepsilon}^{a+\varepsilon} +\int\limits_{a+\varepsilon}^{+\infty}.$$

In the limit, $\varepsilon\to 0$, the sum of the 1st, 3rd and fifth integrals reduce to the required integral I. Thus, $$I+\int\limits_{-a-\varepsilon}^{-a+\varepsilon}+\int\limits_{a-\varepsilon}^{a+\varepsilon}=0.$$ Therefore, in the limit $\varepsilon\to 0$, $$I=i\pi{\rm Res}_f(z=+a)+i\pi{\rm Res}_f(z=+a)\\ =i\pi\left(\frac{e^{-iat}}{2a}\right)+i\pi\left(\frac{e^{iat}}{-2a}\right)\\ =\frac{\pi}{a}\sin(at).$$

This video suggests that the required integral is nonzero. But it does not agree with the notes I linked. What is going on?

  • $\begingroup$ Several solutions are also here math.stackexchange.com/questions/4105166/… $\endgroup$
    – Svyatoslav
    Apr 17, 2021 at 19:56
  • $\begingroup$ What is your question? $\endgroup$
    – user
    Apr 18, 2021 at 7:18
  • $\begingroup$ This integral is supposed to be zero. But I am getting a nonzero value. $\endgroup$ Apr 18, 2021 at 7:20
  • $\begingroup$ Do you mean the integral $I$? Why is it supposed to be zero? Who have supposed this? $\endgroup$
    – user
    Apr 18, 2021 at 7:47
  • $\begingroup$ @user See the pdf I linked. It is a basic integral for getting the retarded potential. $\endgroup$ Apr 18, 2021 at 8:42

2 Answers 2


Use twice the corollary to the lemma given in the most upvoted answer here: Definite integral calculation with poles at $0$ and $\pm i\sqrt{3}$

Thus, we get, on corresponding little semicircles around $\;z=\pm a\;$ , that

$$Res_f(z=\pm a)=\lim_{z\to\pm a}(z\pm a)f(z)=\begin{cases}\cfrac{e^{-iat}}{2a}\\{}\\\cfrac {e^{iat}}{-2a}\end{cases}$$

But we only want the real part of the above (since the integral for the imaignary part vanishes for sure since we get an odd function on a convergent imporper integral on the whole real line), and thus we get that the residues are

$$\frac{\cos at}{\pm2a}$$

and from here the integral vanishes.

  • $\begingroup$ Isn't the contribution from little semicircles are $-\pi i {\rm Res}_f(z=\pm a)$? $\endgroup$ Apr 17, 2021 at 20:08
  • $\begingroup$ @mithusengupta123 Yes (if you cut the little semicircles out of a hole big semicircle in positive direction), but then you have the sum of the residues times $\;-i\pi\;$ , which yields zero by the above... $\endgroup$
    – DonAntonio
    Apr 17, 2021 at 20:10
  • $\begingroup$ The contour I have in mind is that of figure 50 here (page 125) damtp.cam.ac.uk/user/tong/em/el5.pdf $\endgroup$ Apr 17, 2021 at 20:11
  • $\begingroup$ Not sure if I have made a mistake. The sum of the contributions from two semicircles comes out equal to $-\pi\sin(at)/a$ which is real and nonzero. $\endgroup$ Apr 17, 2021 at 20:27
  • $\begingroup$ That's exactly the contour I had, and still have, in mind. The outcome of the integral is thus $$-i\pi\left[\text{Res}_f(z=a)+\text{Res}_f(z=-a)\right]=-i\pi\cdot0=0$$ Observe in my answer, I wrote down the residues of the real part of the integral, which is what we actually want. $\endgroup$
    – DonAntonio
    Apr 17, 2021 at 20:37

$\def\o{\omega}$ I assume you mean the following quotation:

Let’s first look at the case with $t<0$. Here, $e^{-i\o t}\to 0$ when $\o\to+i\infty$. This means that, for $t<0$, we can close the contour $C$ in the upper-half plane as shown in the figure and the extra semi-circle doesn’t give rise to any further contribution. But there are no poles in the upper-half plane. This means that, by the Cauchy residue theorem, $G_\text{ret}(\mathbf {r},t)=0$ when $t<0$.

Observe that the integral in question: $$ I(k)=\int\limits_{-\infty}^{\infty} \frac{e^{-i\o t}}{\o^2-c^2k²}d\o,\tag1 $$ in fact does not exist in the usual sense (for $ck\in\mathbb R$) but only as (for example) a corresponding Cauchy principal value.

I would assume that the claim quoted above is meaningful only if the integral $(1)$ is understood as including the semicircle paths to bypass the poles and fixing the paths to lie in the upper complex half-plane.

This understanding is consistent with the later definition of the advanced Green's function which fixes the paths for bypassing the poles to lie in the lower complex half-plane.

  • $\begingroup$ What is wrong with the way I have calculated it? $\endgroup$ Apr 18, 2021 at 12:25
  • $\begingroup$ @mithusengupta123 You used different definition of the integral which corresponds to the Cauchy principal value (and you computed it correctly). As easy to see the value represents the average of the retarded and advanced Green's functions (provided the definition described in the answer). $\endgroup$
    – user
    Apr 18, 2021 at 14:23
  • $\begingroup$ Given the integral, how does one know whether the integral is defined to be Cauchy Principal value or not? $\endgroup$ Apr 18, 2021 at 15:13
  • $\begingroup$ @mithusengupta123 Usually one writes "p.v." before the integral symbol, but there are alternative notations as well: en.m.wikipedia.org/wiki/Cauchy_principal_value#Notation. $\endgroup$
    – user
    Apr 18, 2021 at 15:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .