In chapter 2, page 27, eq. 2.51, P&S solves the following integral -

$$ \frac{4\pi}{8\pi^3} \int _0 ^\infty dp \ \frac{p^2 \ \ \ e ^{-it\sqrt{p^2 + m^2}}}{2\sqrt{p^2 + m^2}}.\tag{2.51}$$

My solution to above integral (using Cauchy's Residue theorem) -

$$ \frac{1}{4\pi^2}\ \int _0 ^\infty dp \frac{p^2 \ \ e ^{-it\sqrt{p^2 + m^2}}}{\sqrt{(p+im)(p-im)}} -> 2 \ \ \ poles \ \ -> \pm im \ \ ->\ \ \ closing\ \ \ contour \ \ \ upward \ \ = 2\pi i \frac{(im)^2}{\sqrt{im + im}} $$

$$ = -\sqrt{\frac{im^3}{8}} \ \ \ -> \ \ INCORRECT !!$$

P&S's solution -

$$ = \frac{1}{4\pi ^2} \int _m ^ \infty dE \ \sqrt{E ^2 - m ^2}\ \ e ^{-iEt} \quad \sim \quad e ^{-imt} \quad \text{ for }\quad t \ \to \ \infty .\tag{2.51}$$

As far as the second form of the same integral is concerend, there are no poles, so shouldn't the integral be $0$?

P.S. -I've been out of touch with complex integrals for a while and probably missing something conceptually,would appreciate any help.

  • $\begingroup$ You can't use a semicircular contour like that because $\left|e^{-it\sqrt{p^2+m^2}}\right|=e^{t\Im\sqrt{p^2+m^2}}$. $\endgroup$
    – J.G.
    Aug 2, 2022 at 14:42
  • $\begingroup$ Another way to find the approximation for the final expression is to search it more thoroughly: physics.stackexchange.com/q/105045. $\endgroup$
    – tueda
    Aug 29, 2022 at 11:38

2 Answers 2


Hint: $$\begin{align} \int_{m}^{\infty}\! dE~\sqrt{E ^2 - m ^2} e ^{-iEt} ~=~~~&\left(\int_{m}^{-i\infty} +\int_{-i\infty}^{\infty} \right)\! dE~\sqrt{E ^2 - m ^2} e ^{-iEt}\cr ~\stackrel{E=|E|e^{i\theta}}{=}&\left(\int_{m}^{-i\infty} \! dE+ \lim_{|E|\to\infty}\int_{-\frac{\pi}{2}}^0\! d\theta~iE\right)\sqrt{E ^2 - m ^2} e ^{-iEt}\cr ~=~~~&\int_{m}^{-i\infty}\! dE~\sqrt{E ^2 - m ^2} e ^{-iEt}\cr ~\stackrel{E=-imz }{=}&-m^2\int_{i}^{\infty}\! dz~\sqrt{z^2 + 1} e ^{-z mt} \cr ~\stackrel{E=-imz }{=}&-m^2\int_{i}^{\infty}\! dz~\sqrt{z^2 + 1} e ^{-z mt} \cr ~\stackrel{z=x+i}{=}~&-m^2\int_{0}^{\infty}\! dx~\sqrt{2ix+x^2} e ^{-(x+i)mt} \cr ~\stackrel{mt\gg 1 }{\approx}~~&-m^2\int_{0}^{\infty}\! dx~\sqrt{2ix} e ^{-(x+i)mt} \cr ~\stackrel{y=xmt}{=}~~&-\sqrt{\frac{2i m}{t^3}}e ^{-imt} \underbrace{\int_{0}^{\infty}\! dy~\sqrt{y} e ^{-y}}_{\Gamma(\frac{3}{2})=\frac{\sqrt{\pi}}{2}}\end{align}$$


P&S do not employ the method of residues to reach the final conclusion you cite. Instead, what they do is simply to make a variable substitution of the form $E=\sqrt{p^2+m^2}$.

So, then, if you want to explain why do the authors conclude that $$D(x-y)=\frac{1}{4\pi^2}\int_{m}^{\infty}dE\sqrt{E^2-m^2} e^{-iEt} \xrightarrow[\text{}]{t\rightarrow\infty}e^{-imt}$$ at the limit of $t\rightarrow\infty$, you can simply think how the integrand behaves at the near the two limits of integration.

  1. First, we explore what is going on near the upper bound. Let us assign an upper cutoff $\Lambda$, such that we are able to keep track on how the integrand diverges at the upper boundary. Namely, we are making the evaluating the integral $$\lim_{\Lambda\to\infty}\frac{1}{4\pi^2}\int_m^{\Lambda}dE \sqrt{E^2-m^2}e^{-iEt}$$ which is actually the expression we have above. Let us also introduce another energy scale $\lambda$, whose sole purpose is to isolate the areas of integration that characterise the upper limit we wish to study. So, in other words $\lambda<<m$ and the integral above in the high energy limit behaves as $$\lim_{\Lambda\to\infty}\frac{1}{4\pi^2}\int_{\lambda}^{\Lambda}dE \sqrt{E^2-m^2}e^{-iEt}$$ Then, we approximate $\sqrt{E^2-m^2}$ by $E\sqrt{1-m^2/E^2}\approx E$, because $E\in[\lambda,\Lambda]$ and since $\lambda>>m$, then $E>>m$. So, we write $$\lim_{\Lambda\to\infty}\frac{1}{4\pi^2}\int_{\lambda}^{\Lambda}dE \sqrt{E^2-m^2}e^{-iEt}\approx \lim_{\Lambda\to\infty}\frac{1}{4\pi^2}\int_{\lambda}^{\Lambda}dE Ee^{-iEt}$$ which in turn can be written as $$\lim_{\Lambda\to\infty}\frac{i}{4\pi^2}\int_{\lambda}^{\Lambda}dE \frac{\partial}{\partial t}e^{-iEt}= \lim_{\Lambda\to\infty}\frac{i}{4\pi^2}\frac{\partial}{\partial t} \int_{\lambda}^{\Lambda}dE e^{-iEt}= \lim_{\Lambda\to\infty}\frac{i}{4\pi^2}\frac{\partial}{\partial t} \Bigg[\frac{e^{-iEt}|_{\lambda}^{\Lambda}}{-it}\Bigg]$$ whose real part vanishes when we take the limit $t\rightarrow\infty$!

  2. Second, we explore what is going on in the integration regions near the mass of the propagating particle. So, in this case we introduce $E=m+\delta m$, where $\delta m$ is a very small parameter that has units of mass. Then, the integral reduces to $$\frac{1}{4\pi^2}\int_m^{\delta m}dE \sqrt{E^2-m^2}e^{-iEt}\approx \frac{1}{4\pi^2}(\delta m)^2e^{-i(m+\delta m)t}$$ Assuming that $m>>\delta m$, then the latter reduces further to $$\frac{1}{4\pi^2}\int_m^{\delta m}dE \sqrt{E^2-m^2}e^{-iEt}\approx \frac{1}{4\pi^2}(\delta m)^2e^{-imt}$$ which goes like P&S suggest (i.e. $\sim e^{-imt}$ at the $t\rightarrow\infty$ limit)...

I hope what I write do not contain any sort of mistakes in them (calculational or conceptual). Feel free to comment if something doesn't make sense.

P.S. #1: By the way, it turns out that it is a lot more difficult to evaluate than previously thought and this is why I make so many approximations...

P.S. #2: I understand that my arguments are somewhat less rigorous than expected, but P&S themselves do not provide an exact form for the propagator connecting two timelike points in space...

  • $\begingroup$ I understand that but how do we reach their final answer of $ e ^{-imt} $ I would appreciate the steps leading to the solution. Thanks. $\endgroup$
    – user263315
    Aug 2, 2022 at 10:51
  • $\begingroup$ Okay I will edit in a few moments! $\endgroup$
    – schris38
    Aug 2, 2022 at 10:51

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