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This question was previously asked here in the Mathematics StackExchange but using a slightly different notation. But I did not find the answer I was looking for or rather got two very different answers.

Consider the following integral $$I=\int\limits_{-\infty}^{\infty}d\omega \frac{e^{-i\omega t}}{\omega^2-c^2k^2}$$ which has poles at $\omega=\pm ck$. When poles exists on the real line, we can either indent the contour to bypass the poles (or equivalently, use an appropriate $i\varepsilon$-prescription).

Let us choose a contour that lies entirely in the upper half of the complex $\omega$ plane. Let the contour consists of two small C1 and C2, each of radius $\varepsilon$, that goes over the poles at $\omega=\pm ck$, and the larger semicircle $C$ also lies in the upper half of the complex plane. Please see the diagram of Tong's Lecture notes on Electromagnetism on page 125, figure 50.

With this choice of contour, I Tong shows that $I=0$ for $t<0$.

But when I am trying to calculate it, for different parts of the contour, piece by piece, I get a nonzero result. Let me discuss how. Using Cauchy's residue theorem and Jordan's lemma, gives (schematically),

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

In the limit, $\varepsilon\to 0$, the sum of the 1st, 3rd and 5th integrals reduce to the required integral $I$. Thus, $$I+\int\limits_{-ck-\varepsilon}^{-ck+\varepsilon}+\int\limits_{ck-\varepsilon}^{ck+\varepsilon}=0.$$ Therefore, in the limit $\varepsilon\to 0$, $$I=i\pi{\rm Res}(z=+ck)+i\pi{\rm Res}(z=+ck)\\ =i\pi\left(\frac{e^{-ickt}}{2ck}\right)+i\pi\left(\frac{e^{ickt}}{-2ck}\right)\\ =\frac{\pi}{ck}\sin(ckt)\neq 0$$

Therefore, this does not agree with Tong's notes I linked. What is wrong with my calculation? Tong's result is correct but I am unable to reproduce it in my way. Please note that I am not directly using the $i\epsilon$-presciption of shifting poles but using the prescription of indenting contours.

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Your calculation of the principal value of the integral along the real line is (I think) correct. The problem is that the retarded Green's function is not defined to be this principal value, rather it is defined to be the integral along the contour including the semicircles.

This integral appears in the context of quantum field theory: there is a good discussion of the different ways to define such integrals in Timo Weigand's QFT lecture notes, section 1.11.3.

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Consider that diagram you have there. The retarded propagator has the contour running over both poles, no matter what. When $t<0$, we must run the integral in the upper half-plane to apply Jordan’s lemma to kill the semicircle portion of the contour. However, when we do that, no poles are within the contour, so by the residue theorem, $$\int_C f(t)=\int_{-\infty}^\infty f(t)=0$$ When $t>0$, we must run the integral over the bottom half-plane; both poles are in this, so then you run the residue theorem as usual.

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  • $\begingroup$ You forgot the contribution from the little semicircles. Actually, I think I have the answer now. Arthur Morris's comment is correct. Also see the answer and comments math.stackexchange.com/questions/4106170/… by user. $\endgroup$ – mithusengupta123 Apr 18 at 16:45
  • $\begingroup$ @mithusengupta123 regardless of the little semicircles, there are still no poles enclosed by the contour in the $t<0$, so it is automatically zero by the Cauchy integral theorem $\endgroup$ – John Dumancic Apr 19 at 1:05
  • $\begingroup$ Yes, but only if you are not talking about principal value integral. $\endgroup$ – mithusengupta123 Apr 19 at 7:18

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