Inverse Laplace transform with branch cut For the purpose of my research on persistent random walks I need to compute the inverse Laplace transform of
$$ F(s)=\frac{\mathrm e^{-b\sqrt{s^2-1}}}{s^2-1}.$$
I looked up in tables of integral transforms such as Erdélyi et al., Gradsteyn & Ryzhik, Prudnikov et al. with no success. So I have looked for a proper contour for the Bromwich integral. I came up with this one 

$\color{green}A$ is the integral we want to compute $$f(t)=\frac1{2\pi\mathrm i}\int_{\gamma-\mathrm i\infty}^{\gamma+\mathrm i\infty}\mathrm e^{st}\frac{\mathrm e^{-b\sqrt{s^2-1}}}{s^2-1}\mathrm ds,$$
$\color{blue}B$ and $\color{blue}K$ vanish when the radius of the large circle goes to infinity. The contributions of $\color{brown}F$ and that of $\color{brown}D$ and $\color{brown}H$ are given by simple poles and yield together a contribution of $\sinh t$. 
The contribution of $\color{red}C$ and $\color{red}J$ cancel each other.
To compute the contributions of $\color{red}E$ and $\color{red}G$, I set $s=-\cos\theta+\mathrm i\varepsilon$ for $\color{red}E$ and $s=\cos\theta-\mathrm i \varepsilon$ for $\color{red}G$. I end up with the following integral
$$\color{red}{E+G}\to\frac1{\pi\mathrm i}\;\text{vp.}\int_0^\pi \frac{\mathrm e^{\mathrm ib\sin\theta}}{\sin\theta}\sinh(t\cos\theta)\mathrm d\theta\tag{1}$$
that I didn't manage to reduce to a handier result. 
Can someone tell me if I'm the right track and perharps help me to come up with a better expression of (1) ?
 A: I think you played a little fast and loose with the contributions over $E$ and $G$.  In considering the contour integral over $E$, I let $z=e^{i \pi} x$ and, over $G$, I let $z=e^{-i \pi} x$.  The integrals I get, after taking appropriate limits, is
$$e^{i \pi} \int_1^{-1} dx \frac{e^{i b \sqrt{1-x^2}}}{-(1-x^2)} e^{-x t} + e^{-i \pi} \int_{-1}^1 dx \frac{e^{-i b \sqrt{1-x^2}}}{-(1-x^2)}e^{-x t}$$
Note that the $\pm i$ in the exponentials comes from the assignment of $\sqrt{s-1} = \sqrt{e^{\pm i \pi} (x+1)}$ from the respective branch.  The result is
$$-i 2 \int_{-1}^1 dx \frac{\sin{b \sqrt{1-x^2}}}{1-x^2}e^{-x t}$$
This integral converges, although I have not been able to unwind it into a closed form.  (I suspect it is some Bessel of complex argument.)
Anyway, I get that
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+ i \infty} ds \frac{e^{-b \sqrt{s^2-1}}}{s^2-1} e^{s t} = \sinh{t} + \frac1{\pi} \int_{-1}^1 dx \frac{\sin{b \sqrt{1-x^2}}}{1-x^2}e^{-x t}$$
A: I have a copy of G.E.Roberts and H.Kaufman "Table of Laplace Transforms", 1966.
On page 252, Item 3.2.55 is a more complex example but might apply if $v = 0$.
Inverse of   $(s + (s^2 - a^2)^{1/2})^v  \exp(-b\sqrt{s^2-a^2})/\sqrt{s^2 - a^2}$  is given as:
$0$ for $0 < t < b$     
and   $a^v ((t-b)/(t+b))^{v/2} I_v(a\sqrt{t^2-b^2})$  for $t > b$
With the normal restrictions that $b>0$,   $|\operatorname{Re}(v)| < 1$,  $\operatorname{Re} s > |\operatorname{Re} a|$.
$I_v$ is the modified Bessel function of order $v$.
I assume that the expressions would be valid for $v = 0$.
