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Consider integrating the following function of a complex variable, $z$,

$$ f(z)=\frac{z e^{irz}}{\sqrt{z^2+m^2}}, $$

around the contour shown here. It seems straightforward to show that the infinite radius arc segments, $C_1$ and $C_2$, vanish and we are left with

$$ 0=I_1+I_2+I_3, $$

via the residue theorem. Now, I expect that the integrals along each side of the branch cut which runs from $im$ to $i\infty$ will differ "by a phase" so that they will add together rather than cancel. However, I don't see how to show this explicitly:

\begin{align} I_2 & = \int^m_\infty \frac{Re^{i\pi/2}}{\sqrt{R^2e^{i\pi}+m^2}}e^{irRe^{i\pi/2}}e^{i\pi/2}dR \\ & = \int^\infty_m \frac{R}{\sqrt{-R^2+m^2}}e^{-rR}dR \\ \end{align}

But

\begin{align} I_3 &= \int^\infty_m \frac{Re^{i3\pi/2}}{\sqrt{R^2e^{i3\pi}+m^2}}e^{irRe^{i3\pi/2}}e^{i3\pi/2}\\ &= -\int^\infty_m \frac{R}{\sqrt{-R^2+m^2}}e^{rR}dR, \end{align} so it appears $I_2\ne I_3$ which I believe to be wrong.

1 Answers 1

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$$I_2 = i \int_{\infty}^m dy \, \frac{i y \, e^{-r y}}{+i \sqrt{y^2-m^2}} $$

$$I_3 = i \int_m^{\infty} dy \, \frac{i y \, e^{-r y}}{-i \sqrt{y^2-m^2}} $$

The difference in phase comes from how we define the behavior of the square root in the vicinity of the branch cut $[i m,i \infty)$. So the contributions from $I_2$ and $I_3$ add.

One should not necessarily ignore the contribution around the branch point where $z=i m + \epsilon e^{i \phi}$, $\phi \in [\pi/2,-3 \pi/2]$:

$$I_4 = i \epsilon \int_{\pi/2}^{-3 \pi/2} d\phi \, e^{i \phi} \frac{\left (i m +\epsilon e^{i \phi} \right ) e^{i r \left (i m +\epsilon e^{i \phi} \right )} }{\sqrt{m^2+\left (i m +\epsilon e^{i \phi} \right )^2}}$$

which indeed vanishes as $\epsilon \to 0$. Thus, by Cauchy's theorem we have the relation

$$\int_{-\infty}^{\infty} dx \frac{x \, e^{i r x}}{\sqrt{m^2+x^2}} = i 2 \int_m^{\infty} dy \, \frac{y \, e^{-r y}}{\sqrt{y^2-m^2}} = i 2 m K_1(r m)$$

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    Indeed, I should have mentioned that the branch point does not contribute. So, am I to understand that the minus sign comes from our _definition_ of the square root on each side of the branch cut (i.e., we put it in "by hand") as opposed to coming directly from the argument of z as I have attempted here (where, in fact, I took the wrong phase angle, anyway, it would seem) ? – 2017-01-28
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    btw, I already had a question related to the very last equality you wrote which relates this to the Bessel function. It's submitted [here](http://physics.stackexchange.com/questions/307856/bessel-function-representation-of-spacelike-kg-propagator) – 2017-01-28
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    @user143410: yes, that's right. There is a $2 \pi$ phase jump across a branch cut; in this case, in the opposite direction as we are traversing clockwise around the cut. This explains why we go from a $+$ to a $-$. To get the Bessel, just sub $y=m \cosh{t}$. – 2017-01-28
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    Yeah, I think that substitution would relate it to a different integral representation of the Bessel function than I want to use. I was trying to see how to put it in the same form as one finds in the "timelike" case, where the $r$ is being interpreted as a spacetime distance for the Klein Gordon propagator. – 2017-01-28