Does the inverse Laplace transform of the square root exist? Does the inverse Laplace transform, defined by the integral,
\begin{equation}
F(t) = \mathscr L_s^{-1}\left[\sqrt s\right](t) =  \int_{c - i\infty}^{c + i\infty} \sqrt s ~e^{-st} ds
\end{equation}
exist? 
Here are a few things I've considered in trying to answer this question myself:


*

*$\mathscr{L}[t^p] = \frac{\Gamma (p+1)}{s^{p+1}}$, provided that $p>-1$.  The latter condition precludes the case of interest, which would correspond to $p=-3/2$.

*The square root function can be written as $ \sqrt s = e^{\frac{1}{2}\log s}$.  So $\sqrt s$ is not holomorphic at the origin.  

*This topic has been discussed previously here without clear resolution.

*Mathematica throws the transform back at me without a solution.
 A: I have encountered the same problem! But let me emphasize, I am a geophysicist, not a mathematician!
My observations:


*

*I do not find these (seemingly simple) transforms in most tables of Laplace transforms (not in Abramowitz and Stegun, for example). So, this is, to me, a "red flag".

*I can do the complex-s-plane contour integral to obtain inverse LT of $\sqrt{s}$. It seems simple, there is a branch cut but no pole. The result is $\frac{-1}{2\sqrt{\pi}}t^{-3/2}$

*Another way to evaluate the inverse LT of $\sqrt{s}$ is to use the convolution theorem and multiply by the LT of a Heaviside step function (1/s). The product is 1/$\sqrt{s}$ and the inverse LT of this IS found in standard tables, 1/$\sqrt{\pi t}$. If this is, then, differentiated with respect to t, we obtain $\frac{-1}{2\sqrt{\pi}}t^{-3/2}$

*However, the forward Laplace transform of $t^{-3/2}$ does not converge! The problem, as might be suspected, is at t=0. This can be seen by calculating the indefinite integral and noting that the equation "blows up" at t = 0.

*So this leaves us with what seems to me to be a paradox. The LT does not exist, but its inverse does! To me, this is also a red flag, but I don't know what it means.

*Subsequent comment: My thinking, now, is that the forward LT of $t^{-3/2}$ diverges because the LT is not exactly designed for simple transformation of quantities that are singular at t=0. One must properly accommodate the "pre-zero 0-" boundary condition with the "post-zero 0+" function. This might be addressed with "generalized functions" that can be discontinuous (But I don't, for the moment, know the details). Furthermore, I note that Abramowitz and Stegun do not layout mathematics that is explicitly compatible with this, see, for example section 29.1.1, where the lower limit of the LT is not very specifically identified as coming from the positive or negative direction. This might explain why they don't list the transforms of relevance to this discussion. Once the boundary condition at t=0 is handled, then the term that diverges in the LT of $t^{-3/2}$ might cancel, the remaining term is $\sqrt{s}$ -- thus there is the possibility of both forward LT compatibility with inverse LT. ;-)
A: As you understand, in the cases you are interested in, the direct and inverse Laplace transforms do not exist in a straightforward way. However both make sense within theory of distributions. See
A. H. Zemanian,  Generalized Integral Transformations, Interscience, New York, 1968;
Yu. A. Brychkov and A. P. Prudnikov, Integral transforms of generalized functions. Gordon and Breach, New York, 1989.
A: Let us work first formally. You want to "calculate''
$
f(s)=\int_0^{+\infty}t^{1/2} e^{-st} dt.
$
It is indeed possible to give a meaning to
$
F(\tau)=\int_0^{+\infty}t^{1/2} e^{-i\tau t} dt
$
as the Fourier transform of the homogeneous distribution $t_+^{1/2}$. $F$ is an homogeneous distribution of degree $-\frac 32$ and in fact
$$
F(\tau)=C(\tau-i0)^{-3/2}.
$$
As a result,
$
f(s)=F(s/i)=C(-is-i0)^{-3/2}=C_1\frac{d}{ds}(s^{-1/2}).
$
It was obvious from the beginning since for $s> 0$
$$
f(s)=\int_0^{+\infty}x e^{-sx^2} 2xdx=-2\frac{d}{ds}\int_0^{+\infty} e^{-sx^2} dx
=-\sqrt{π}\frac{d}{ds}(s^{-1/2}).
$$
A: According to Wolfram Alpha, it is $\frac{1}{\Gamma(-1/2)x^{3/2}}$. This, of  course, involves regularization, so to get the full answer, we have to subtract the divergent part so that its Laplace transform to converge.
So, the answer is $ \frac{\sqrt{i}}2\delta ^{(1/2)}(x)+\text{f.p.}\frac{1}{\Gamma(-1/2)x^{3/2}}$, following Urs Graf's "Introduction to Hyperfunctions and Their Integral Transforms", page 36:

More explanation, using the theory of divergent integrals.
The Fourier transform of $x^n$ is $2\pi i^{n}\delta ^{(n)}(s )$.
This means that at $s=0$ we have
$\int_{-\infty}^\infty x^ndx=2\pi i^n\delta ^{(n)}(0 )$.
If the integral is taken from zero, we have half of that value:
$\int_{0}^\infty x^ndx=\pi i^n\delta ^{(n)}(0 )$.
But there is an equality of divergent integrals
$\int_0^\infty \frac1{x^n} dx=\frac1{(n-1)!}\int_0^\infty x^{n-2} dx$
it is based on the fact that the following Laplace-based transform preserves the area under the integral, defining the equivalence class of divergent integrals:
$\int_0^\infty f(x)dx=\int_0^\infty\mathcal{L}_t[t f(t)]\left(x\right)dx=\int_0^\infty\frac1x\mathcal{L}^{-1}_t[ f(t)]\left(x\right)dx$
So, we have:
$\int_0^\infty \frac1{x^{n+1}} dx=\frac1{n!}\int_0^\infty x^{n-1} dx=\frac\pi{n!}i^{n-1}\delta ^{(n-1)}(0)$
For $n=3/2$, and adding coefficient $\frac{3/2}{\Gamma(-1/2)}$ to the both sides we get
$\frac{3/2}{\Gamma(-1/2)}\int_0^\infty \frac1{x^{5/2}} dx=\frac1{\Gamma(-1/2)\Gamma(1/2)}\int_0^\infty \sqrt{x} dx=\frac{\sqrt{i}\delta ^{(1/2)}(0)}{\Gamma(-1/2)\Gamma(1/2)}=-\frac{\sqrt{i}}2\delta ^{(1/2)}(0)$
So to normalize our equation we should take this term with an opposite sign.
A: The inverse transform of $\sqrt{s}$ doesn't exist, because the integral representing it,
$$
\int\limits_{\gamma -i\infty }^{\gamma +i\infty }\sqrt{s}e^{st}\mathrm{d}s,
$$
diverges as $s \rightarrow \gamma \pm i\infty$. People typically don't notice this, and just go ahead with "evaluating" this integral (by closing the contour, etc.) -- and end up with a result which doesn't make sense (as the resulting function doesn't have a forward transform).
