Question:
What is the ILT of $$H(s)=\frac{\sqrt{s^2-2c^2s}}{2sc} e^{-(a/c)\sqrt{s^2-2c^2s}},$$ where $a \geq 0$ and $c>0$. That is, how can we compute $$\frac{1}{2\pi i} \int_{\gamma -i\infty}^{\gamma +i\infty} H(s) e^{st} ds?$$
Attempt
I am a bit rusty at dealing with contour integration involving branch-cuts but here we go. Let $R>0$ be big, and $\epsilon >0$ be small and consider the vertical line intersecting the real axis at $\gamma$ from $\gamma-i\sqrt{R^2-\gamma^2}$ to $\gamma+i\sqrt{R^2-\gamma^2}$, call it's curve in the plane $A$, the portion of a circle centered at zero with radius $R$ running from $\theta = \arctan(\sqrt{R^2-\gamma^2}/\gamma)$ to $\theta = \arctan(-\sqrt{R^2-\gamma^2}/\gamma)$, call this $C_R$, and a dogbone contour that captures the branch cut $[0, 2c^2]$, call this $C_i$, and let its orientation run opposite of the outer contour, which we shall call $C_0$: the union of the verticle line and the semicircle. Then $H(z)$ is holomorphic on $\operatorname{int} C_0\setminus \operatorname{int}C_i$, and between the outer and the inner contour no poles or branch points are contained. Thus the contour integral over the closed region $C_0-C_i$ should be $0$ by Cauchy's residue theorem.
The contour integral over the semicircle ought to vanish as $R\to \infty$, the small one about zero either vanishes or goes to 1/2 the residue at $z=0$ (again, I am rusty here), which is zero anyway, and the small one about the branch cut $z=2c^2$ should vanish too. All that is left is the contribution from the line segment parallel and above the interval $[0, 2c^2]$ and the one below it in reverse direction. Ultimately, I did some trial and error how to handle these two segments and eventually got $$\frac{1}{2\pi i} \int_{\gamma -i\infty}^{\gamma +i\infty} \frac{\sqrt{s^2-2c^2s}}{2sc} e^{-(a/c)\sqrt{s^2-2c^2s}} e^{st} ds=\frac{1}\pi \int_0^{2c^2} \frac{\sqrt{x^2+2c^2s}}{2xc}\sin((a/c)\sqrt{x^2+2c^2 x}) e^{xt},$$ and then this remains to be computed.
But of course I am not very happy with playing fast and loose. I have consulted Ahlfors' book for review but unfortunately his only mention of a keyhole contour (which is an alternative here, I believe) is for computing $\int_0^\infty x^\alpha R(x) dx$ and writes "we have to use the branch of $z^\alpha$ [here $0<\alpha<1$ is assumed] whose argument lies in between $0$ and $2\pi \alpha$. This method needs some justification, for it does not conform to the hypotheses of the residue theorem. The justification is trivial." If anyone, in the process of computing this ILT or verifying mine (and possibly filling in the fast and loose parts), could clarify these justifications in either the keyhole or dogbone-Bromwich case and for $\sqrt{z^2-2c^2}$, I would be very grateful.