An integral in Peskin's Quantum Field Theory P. 27 In Peskin's QFT P. 27, there is an integral $$-\frac{i}{2 (2 \pi )^2 r}\int_{-\infty}^{\infty} \mathrm{d}p\frac{p\ e^{ipr}}{\sqrt{p^2+m^2}}.\tag{2.51a}$$
He said that in order to push the contour up to wrap around the upper branch cut. After some manipulation, it gives the following integral
$$\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}.\tag{2.52}$$


*

*I don't understand the phrase "push the contour up to wrap around the upper branch cut". The integral is on the real line and hence there is no singular point along the line. 

*If we define $\rho =-i p$, I still can't get $\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$.
 A: In those paragraphs, P&S want to obtain the behaviour of the amplitude $D(x-y)$ in the limit $r \to \infty$. Starting from the integral
$$ \tag{1} \label{int}
\frac{-i}{2(2\pi)^2 r} \int_{-\infty}^\infty \text{d}p \frac{p e^{ipr}}{\sqrt{p^2
+m^2}},
$$
you can see that a limit for $r \to \infty$ is not well defined, since the exponential has an oscillating behaviour. Also, the final form of the integral can be recognized as a modified Bessel function of the second kind. See this Wikipedia page for the asymptotic expansion of this function. The modification of the integration contour is a way to obtain a more straightforward integral. I hope this answers your question 1.
About point 2., in order to evaluate this integral, P&S apply the Cauchy's theorem defining the integrand as a function of the complex variable $p$. The square root in the denominator causes the branch cut in Figure 2.3. When rewriting the contour integral as
$$ \tag{2} \label{1}
\int_{i \infty}^{i m} \dots + \int_{i m}^{i \infty} \dots
$$
we have to be careful in writing the integrands, since the branch cut generates a discontinuity between the two sides of the cut. In fact the square root in the complex plane is a multi-valued function, and the argument of $p^2+m^2$ shifts of $2\pi$ when passing from the right of the branch cut to the left. Thus
$$
\sqrt{p^2 + m^2} = \cases{|p^2+m^2|^{1/2} \exp\left(i\frac{\arg(p^2+m^2)}{2}\right), \\ |p^2+m^2|^{1/2} \exp\left(i\frac{\arg(p^2+m^2)+ 2\pi}{2}\right) = - |p^2+m^2|^{1/2} \exp\left(i\frac{\arg(p^2+m^2)}{2}\right),}
$$
where the first value is at the right and the second at the left of the branch cut. You see that the difference is a minus sign.
Therefore, making the substitution $p = i \rho$, \eqref{1} becomes
$$
i\int_\infty^m \text{d} \rho \frac{\rho e^{-r \rho}}{- \sqrt{\rho^2 - m^2}} + i\int_m^\infty \text{d} \rho \frac{\rho e^{-r \rho}}{\sqrt{\rho^2 - m^2}},
$$
which finally gives
$$
2i \int_m^\infty \text{d} \rho \frac{\rho e^{-r \rho}}{\sqrt{\rho^2 - m^2}}.
$$
Substituting this in \eqref{int} you find your result. Note that if you had chosen a different contour, for example the analogous of the "pushed" contour of Figure 2.3 but with the U-turn at $p = 0 + i 0$, you would not have the phase difference in the integrands on the lines $(im, 0)$ and $(0, im)$, so these two pieces would have cancelled each other (see \eqref{1}) and you would have obtained the same result with $m$ as lower integration limit.
A: I'll just add a little more information about pushing the contour upward, since this was very confusing to me at first. We may push the contour of integration up, as long as we keep the endpoints fixed and don't pass through any singularities. This is because Cauchy's theorem says that the integral around any closed loop that doesn't enclose any singularities is zero. Therefore, any two contours with the same endpoints and no singularities between them must give the same integral so that they cancel each other when made into a loop contour.
We may create a U shaped contour wrapping around the fixed point $p=im$ and going up to infinity, as long as the endpoints are fixed at $\pm\infty$ on the real axis. This means the full contour is the U contour shown in Peskin and Schroesder plus two giant arcs at infinity, connecting the two top ends of the U contour to $\pm\infty$. However, in the limit as $r\xrightarrow{}\infty$, the imaginary part of $p$ along these arcs will combine with $ir$ in the exponential to make the integrand approach zero everywhere along these arcs everywhere except for a small region near the real axis. However, the portion near the real axis where the integral in non-negligible becomes vanishingly small as $r\xrightarrow{}\infty$. Therefore, the contribution from the arcs is zero in this limit, and we are left with the U contour.
