I am trying to find the inverse Laplace transform for following function and it seems almost impossible for me to find the answer. Can anyone help me please with final answer and also the way to get to final answer if possible at all :)
Here is the function: $$ \frac{1}{s} e^{-a\sqrt{s} + b/s} $$
This is not a complete answer, but too long for a comment.
I think it will be difficult to find such a general expression, but I write some formulas that might help someone else to finish.
One has the Laplace transforms (here erfc denotes the complementary error function) $$ \newcommand{\erfc}{\mathop{\rm erfc}\nolimits} \mathcal L\Bigl(\erfc\bigl(a/(2\sqrt{t})\bigr)\Bigr)(s)=\frac{1}{s}e^{-a\sqrt{s}} $$ and (here $I_1$ denotes a modified Bessel function of the first kind) $$ \mathcal L\Bigl(\frac{\sqrt{b}}{\sqrt{t}}I_1(2\sqrt{bt})\Bigr)(s)=e^{b/s}-1. $$ Moreover $\mathcal L(\delta)=1$. By the convolution rule of Laplace transforms, you could try to calculate $$ (u_1*u_2)(t)=\int_0^t u_1(\tau)u_2(t-\tau)\,d\tau $$ with $$ u_1(t)=\erfc\bigl(a/(2\sqrt{t})\bigr)\quad\text{and}\quad u_2(t)=\frac{\sqrt{b}}{\sqrt{t}}I_1(2\sqrt{bt})+\delta(t). $$ I honestly don't see how that should be done, though.
Consider the following contour integral:
$$\oint_C \frac{dz}{z} e^{-a \sqrt{z} + b/z} e^{t z} $$
where $C$ is the contour defined here. We apply Cauchy's theorem and set the integral to zero. As the outer radius $\to \infty$, we find that
$$\int_{c-i \infty}^{c+i \infty} \frac{ds}{s} e^{-a \sqrt{s} + b/s} e^{s t} + \int_{-\infty}^0 \frac{dx}{x} e^{-i a \sqrt{x}-b/x} e^{-t x} + i \int_{\pi}^{-\pi} d\phi \ \, e^{-a \sqrt{\epsilon} e^{i \phi/2}} e^{b/\epsilon e^{-i \phi}} e^{t \epsilon e^{i \phi}} \\ + \int_0^{\infty} \frac{dx}{x} e^{i a \sqrt{x}-b/x} e^{-t x} = 0$$
The third term on the LHS is actually quite tricky. Through a substitution, I get
$$-i 2 \int_{-\pi/2}^{\pi/2} d\phi \, e^{-a \sqrt{\epsilon} e^{i \phi}} e^{b/\epsilon e^{i 2 \phi}} e^{t \epsilon e^{i 2 \phi}} $$
which we can express as a contour integral over a half-circle connected by a line segment about the imaginary axis with a small detour about the origin. As $\epsilon \to 0$, the integral about that line segment goes to zero and the integral is simply $i \pi$ times the sum of the residues at the origin of
$$(2/z) e^{-a \sqrt{\epsilon} z} e^{b/(\epsilon z^2)} e^{-t \epsilon z^2}$$
Thus, leaving out all of the fun details of expanding the above in a Laurent series, I get that
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} \frac{ds}{s} e^{-a \sqrt{s} + b/s} e^{s t} = I_0\left ( 2 \sqrt{b t} \right ) + \sum_{k=1}^{\infty} \frac{(a^2 b)^k}{k! (2 k)!}-\frac1{\pi} \int_{-\infty}^{\infty} dx \, \frac{\sin{a x}}{x} e^{-b/x^2} e^{-t x^2}$$
You would think that the sum in the middle would be some well-known special function like a Bessel or something, but no dice. (Mathematica returns $\, _0F_2\left(;\frac{1}{2},1;\frac{a^2 b}{4}\right)$ which does not ring a bell.) I imagine I could sit around and play with the properties of this function but I would prefer not to reinvent the wheel if I do not need to.
The integral looks like a train wreck, i.e., I cannot think of a way to evaluate it yet. Of course you all will be the first to know if I get any flashes of brilliance.
