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

*

*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 $t\rightarrow\infty$!


*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...
