The following integral absolutely converges for $Re(z)<0$ and is analytic in this domain: $$F(z):=\int_{0}^{+\infty}\frac{\sin r}{r}e^{\sqrt{r^2+m^2}z}dr ,$$ where $m>0$ is fixed.
Question. To what maximal domain of $\mathbb{C}$ does $F$ extend as an analytic (possibly multi-valued) function of $z$? Does it extend to the whole of $\mathbb{C}$ without a discrete subset? I am particularly interested whether it extends to the whole real line without 0.
Gradshteyn and Ryzhik give the formula (3.914) $$\int_0^\infty \exp(\sqrt{r^2+m^2}z)\cos(br)\,dr=-{zm\over\sqrt{z^2+b^2}}K_1(m\sqrt{z^2+b^2}).$$ Let us denote the right hand side by $g(b,m,z)$. We find that $$\int_0^\infty r\sin r\exp(\sqrt{r^2+m^2}z)\,dr=-{\partial\over\partial b}g(b,m,z)|_{b=1}.$$ Finally, this last integral equals $F''(z)-m^2F(z)$. So you can get an analytic continuation by solving this differential equation.
The computation below does not give a complete answer, but gives a family of extensions to real positive $z$.
Let $\gamma_\theta:=\{re^{i\theta}:r>0\}$ and $ F_\theta(z):=\int_{\gamma_\theta}\frac{\sin r}{r}e^{\sqrt{r^2+m^2}z}dr. $ This integral converges in the half-plane $$H_\theta:=\{\ \Re\mathfrak{e} z\cos\theta -\Im\mathfrak{m}z\sin\theta<\sin\theta\}.$$ By considering pizza-slice shaped contours, we see that for $0\leq \theta\leq \frac\pi2$ these functions actually coincide on their common domain of definition.
To continue this procedure beyond $\theta=\frac\pi2$, we need to be precise about the branch of the square root in the exponential. Namely, le us assume that the branch of $\sqrt{r^2+m^2}$ in the definition of $F_\theta$ is analytic in $\mathbb{C}\setminus [-im;im]$ and positive at positive $r$. Thus defined, $F_\theta$ is discontinuous at $\theta=\frac\pi2$, but we can write $$ F_{\frac\pi2-0}(z)=F_{\frac\pi2+0}(z)+\int_0^{im+0}\frac{\sin r}{r}e^{\sqrt{r^2+m^2}z}dr+\int_{im-0}^{0}\frac{\sin r}{r}e^{\sqrt{r^2+m^2}z}dr, $$ where the first integral runs over the right side of the cut, and the second one runs over the left side, with the same convention about the branch.
This way, we can continue rotating the contour all the way up to $\theta=2\pi$. The domain $H_\theta$ will sweep $\mathbb{C}\setminus [-i;i]$, and, collecting the integrals, we find the as $z$ goes around the cut $[-i;i]$, the analytic continuation picks up the additive term $$ -4i\int_{0}^{m}\frac{\sinh r\sinh(z\sqrt{m^2-r^2})}{r}dr. $$
So, at least $F$ can be continued to the universal cover of $\mathbb{C}\setminus [-i;i]$, with this deck transformation.
