Propagator of the Klein-Gorden equation

Question

Does this integral converge?

My question is related to this one: Free particle propagation amplitude calculation

I am reading the book of Peskin and Schroesder. In the second page of their chapter 2, they estimated the propagator for the Klein-Gorden equation. The final step of the calculation is to calculate an integral of the form

$$\int_0^\infty \mathrm{d}p\;p\;\sin(px)e^{-\mathrm{i}t\sqrt{p^2+m^2}}.$$

My question is very simple: does this integral converge in the first place? Note that the integrand is an oscillating function whose amplitude grows without bound. If we take $m=0$, then for sure the integral does not converge.

In fact, I can reproduce their result very easily (namely, the $e^{-m\sqrt{x^2-t^2}}$ behavior). But I am not sure whether my calculation makes sense because I don't know whether the integral has a finite value.

I know this integral can be converted into a loop integral in the complex plane, and then be estimated using stationary-phase method. But since the amplitude of the integrand does not go to zero uniformly at infinity, I feel such a procedure may not be valid.

We may multiply a factor $e^{-\epsilon p}$ ($\epsilon>0$) to the integrand to make it converge, but then we are not calculating what we wanted to calculate in the beginning, right?

Then, if the integral does not converge by itself, how should I make sense of the result in the Peskin and Schroesder book? I am quite confused by this. Thanks for any hints!

Answer 1

The integral does not converge by itself, for exactly the reason you mentioned. But it doesn't need to! The reason being, when you use this integral in such a physical context you're not actually referring to the mathematical definition of the integral of a real-valued function. You're rather referring to exactly the construct that you get when introducing the limiting factor: in actuality, even an "infinite" system is always bound in some way (e.g. finite amount of total energy $\Rightarrow$ upper bound for the momentum to be considered). It's just that you're interested in the case where these boundaries are so far off that their exact values don't matter anymore, and that's exactly what you calculate when introducing some $e^{-\epsilon|p|}$ factor and outside of the integral taking the limit $\epsilon\to 0$.