I found in Peskin-Schroeder, while reading Quantum Field Theory. the following integration.
$$\frac{1}{4\pi^2 r}\int_m^\infty \frac{se^{-sr}}{\sqrt{s^2-m^2}} = e^{-mr}$$ at the limit $r \rightarrow \infty$.
So, what is going on here? If this is the case that, at large r and large m, there is drastic exponential decay. That's why we are keeping only the first term, which is $e^{-mr}$. Then the question is: in that case why $\frac{1}{r}$ does not have any contribution to the answer? Thank you
Ok, this is the physicists quick and dirty route. For a more formal analysis one could apply Laplace method or calculate the integral exactly in terms of Bessel function's (the solution is $\frac{m}{4\pi r} K_1(m r)$) and use their well known asymptotics.
Due to the exponetial decay, the integral will be dominated by a small region around $s=m$.
We put $s=m$ and $\sqrt{s+m}=\sqrt{2m}$ in this region,because their change will be very small compared to the changes in the singular term $\sqrt{s-m}$ and the exponential $e^{-rs}$
We get
$$ I(m,r)\sim\frac{\sqrt{m}}{4 \pi^2 r\sqrt{2}} \int_m^{m+\delta}\frac{e^{-rs}}{\sqrt{s-m}}ds $$
where $\delta$ is some small number at first. The trick now is that even if we choose $\delta=\infty$ the error we introduce is exponentially small. We might therefore use with very good accuracy
$$ I(m,r)\sim\frac{\sqrt{m}}{4 \pi^2 r \sqrt{2}} \int_m^{\infty}\frac{e^{-rs}}{\sqrt{s-m}}ds $$
this integral can be solved by setting $s-m=p^2$ and using the standard Gaussian integral. We obtain
$$ I(m,r)\sim\frac{\sqrt{m\pi}}{4 \pi^2 \sqrt{2}}\frac{e^{-mr}}{r^{3/2}} $$
So there is another prefactor given by $r^{-3/2}$ which PS somehow surpresses most likely because it is not related to an important physical quantity like the correlation length in the exponent.