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?
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.