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.
Best Answer
$$I_2 = i \int_{\infty}^m dy \, \frac{i y \, e^{-r y}}{+i \sqrt{y^2-m^2}} $$
$$I_3 = i \int_m^{\infty} dy \, \frac{i y \, e^{-r y}}{-i \sqrt{y^2-m^2}} $$
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
$$\int_{-\infty}^{\infty} dx \frac{x \, e^{i r x}}{\sqrt{m^2+x^2}} = i 2 \int_m^{\infty} dy \, \frac{y \, e^{-r y}}{\sqrt{y^2-m^2}} = i 2 m K_1(r m)$$