# Evaluating $\int_{-\infty}^\infty \frac{pe^{ipr}}{\sqrt{p^2 + m^2}} dp$ using complex contours

Consider the following integral: $$\int_{-\infty}^\infty \frac{pe^{ipr}}{\sqrt{p^2 + m^2}} dp$$ where $$r, m$$ are positive constants.

This integral appears in a quantum field theory textbook by Peskin & Schroeder. I know that this integral can be solved using a change of variables or Bessel functions (as described here: Peskin and Schroeder, Eq 2.52 with a spacelike interval.).

In the text they consider the integrand as a complex function, and thus because of the square root function there are branch points at $$\pm im$$. The two obvious choices for branch cuts here would be the line segment $$(-im, im)$$ on the imaginary axis or $$(-i\infty, -im) \cup (im, i\infty)$$ also through the imaginary axis. Since the original integral is over the real line only the second branch cut makes sense. I understand everything until this point. The authors then say

To evaluate the integral we push the contour up to wrap around the upper branch cut.

and make the substitution $$\rho = -ip$$ to obtain $$2\int_m^\infty \frac{\rho e^{-\rho r}}{\sqrt{\rho^2 - m^2}} d\rho.$$

Here is a figure of what they're doing.

I am very confused on what they're doing. The first path from $$(-\infty, \infty)$$ does not cross a pole or a branch cut so what is the need for this substitution? How do they get the extra factor of 2 after doing the transformation? How did the lower limit of integration go from $$-\infty$$ to $$m$$? The lower limit is obvious from the picture but this doesn't follow from the transformation they gave.

More importantly how is any of this justified? The first contour $$(-\infty, \infty)$$ is not closed so how can we transform it to the U shaped contour they describe and expect the two integrals to be the same?

It's a simple application of Cauchy's integral theorem. Consider the complex contour integral $$\oint\limits_C dz \frac{z\, e^{irz}}{\sqrt{z^2+m^2}}, \qquad r\gt 0,\tag{1} \label{1}$$ where the closed curve $$C$$ (taken counterclockwise) consists of the following pieces: \begin{align} \gamma_1&: \, p\to z(p)=p, \, -R\le p \le R, \\ \gamma_2&: \, \varphi \to z(\varphi) = R e^{i \varphi}\!,\, 0\lt \varphi \le \pi/2-\epsilon, \; (\epsilon >0), \\ \gamma_3&: \, \rho \to z(\rho)= i \rho+\epsilon , \, \text{ from} \;\rho=R \; \text{to} \; \rho=m,\\ \gamma_4&: \, s \to z(s)= im -s+\epsilon, \; 0\le s \le 2 \epsilon, \\ \gamma_5&: \, \rho \to z(\rho)= i \rho -\epsilon, \; m \le \rho \le R,\\ \gamma_6&: \varphi \to z(\varphi)=R e^{i\varphi}, \; \pi/2+\epsilon \le \varphi \lt \pi. \end{align} \tag{2} \label{2} The branch cut of $$w \to \sqrt{w}$$ is chosen along the negative real axis, such that $$\lim\limits_{\epsilon \downarrow 0} \sqrt{-|w|\pm i \epsilon}= \pm i |w|$$.
As $$C$$ does not enclose any poles of the integrand, the integral \eqref{1} vanishes. In the limit $$R\to \infty$$, the contributions from $$\gamma_2$$ and $$\gamma _6$$ vanish because of the exponential damping factor $$e^{-R r \sin{\varphi}}$$ for $$r \sin{\varphi} >0$$. Taking finally also the limit $$\epsilon \downarrow 0$$, we are left with $$\underbrace{\int\limits_{-\infty}^{+\infty}\!\! dp \frac{p\, e^{irp}}{\sqrt{p^2+m^2}}}_{\text{from} \; \gamma_1} \;\underbrace{-i \int\limits_m^\infty \! \! d \rho \frac{\rho e^{-r \rho}}{\sqrt{\rho^2-m^2}}}_{\text{from} \; \gamma_2} \; \underbrace{-i\int\limits_m^\infty \! \! d\rho \frac{\rho e^{-r \rho}}{\sqrt{\rho^2-m^2}}}_{\text{from} \; \gamma_6}=0, \tag{3} \label{3}$$ giving the desired result $$\int\limits_{-\infty}^{+\infty} \! \! dp \frac{p \, e^{irp}}{\sqrt{p^2+m^2}} = 2i \int\limits_m^\infty \! \! d\rho \frac{\rho \, e^{-r \rho}}{\sqrt{\rho^2-m^2}}. \tag{4} \label{4}$$ (Note that a factor $$i$$ is missing in your last formula.)
Peskin and Schroeder use \eqref{4} to demonstrate that naive "relativistic" one-particle quantum mechanics, obtained by simply repacing $$H=\vec{p}^2 /2m$$ by $$H=(\vec{p}^2+m^2)^{1/2}$$, is in conflict with the requirement of causality. At the same time, they show how relativistic quantum field theory (being the correct description) avoids this problem.
• @AlbertusMagnus $w \to \sqrt{w}$ has a branch point at $w=0$, such that $z \to \sqrt{z^2+m^2}$ has branch points at $z = \pm i m$, explaining the choice of the branch cut by Peskin & Schroeder (see figure in the question). Commented May 7 at 4:13