TITLE: inverse Laplace transform of $e^{-\tau s\sqrt{\frac{s+q}{s+p}}}$
QUESTION [8 upvotes]: I'm trying to compute the inverse Laplace transform of a function:
$$
g(s)=e^{-\tau s\sqrt{\frac{s+q}{s+p}}}
$$
where $\tau$, $p$ and $q$ are all positive real numbers, and $q>p$. The ILT is given by:
$$
\mathcal{L}^{-1}\left[g(s)\right] = \frac{1}{2\pi i}\int_{\epsilon-i\infty}^{{\epsilon-i\infty}}dse^{ts}e^{-\tau s\sqrt{\frac{s+q}{s+p}}}
$$
I chose a deformed integral contour avoiding the branch cut of $g(s)$ to evaluate the Bromwich integration. 
integral contour
The integrals along $C_2$, $C_{10}$,  $C_{4}$,  $C_{8}$ and $C_{6}$ all vanish as the radius of the large circle goes to infinity and that of the small circle goes to 0. The integrals over $C_3$ and $C_9$ cancel each other. Letting $s=-p+(x-p)e^{i\pi}$ over $C_5$ and $s=-p+(x-p)e^{-i\pi}$ over $C_7$, I get this result:
$$
\mathcal{L}^{-1}\left[g(s)\right]=-\frac{1}{2\pi i}\int_q^pdx e^{-tx}e^{i\tau x\sqrt{1+\frac{p-q}{x-p}}}+\frac{1}{2\pi i}\int_q^pdx e^{-tx}e^{-i\tau x\sqrt{1+\frac{p-q}{x-p}}}
=-\frac{1}{\pi i}Im\left(\int_q^pdx e^{-tx}e^{i\tau x\sqrt{1+\frac{p-q}{x-p}}}\right)
$$
Is my derivation correct? And does anyone know how to evaluate this integration?

REPLY [4 votes]: When working with distributions, the integral transforms are, in fact, not defined as integrals. They can be formally written as integrals, but that's just a symbolic notation for the transforms. To separate the regular and singular parts, notice that
$$ \lim_{A \to \pm \infty} \left(
 g(\gamma + i A) - e^{-\tau (\gamma + i A) + \tau (p - q)/2} \right) =
0,$$
therefore
$$\mathcal L^{-1}[g](t) =
e^{\tau (p - q)/2} \delta(t - \tau) +
 \mathcal L^{-1}[s \mapsto e^{\tau s} g(s) - e^{\tau (p - q)/2}](t - \tau).$$
The inverse transform on the rhs exists as an ordinary function (the Bromwich integral converges by Dirichlet's test) but is unlikely to be evaluable in closed form.