Is this document using laplace transforms incorrectly? (in particular, laplace transform of 1/r) http://folk.uio.no/helgaker/talks/SostrupIntegrals_10.pdf
If you look at page 16/34, you can see this bit:
$
\frac{1}{r_c} = \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} \exp{(-r_c^2 t^2 )dt}
$
The page notes that it is a laplace transform. However, I always thought the laplace transform of 1/r doesn't exist, for reasons that are explaind well here:
Laplace transform of $1/t$
I am wondering how they came up with the above formula. I tried plugging the integral into mathematica to see what popped out, and the result was:
$
\frac{1}{\sqrt{r_c^2}}
$
aka just 1/r. I suppose it formats it in that way to force the sign?
I also found another document (which is also talking about the same thing; molecular integrals) which has a similar expression which it describes at a laplace transform, which evaluates the same in mathematica:
$
\frac{1}{r_c} = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \exp{(-sr_c^2 )s^\frac{-1}{2}ds}
$
Note that in the context of these two documents, rc is shorthand for:
$
r_c = \sqrt{(x_c^2 + y_c^2 + z_c^2)}
$
where xc, etc are themselves shorthand for:
$
xc = (x-c_x)
$
Where $ c_x $ is the x coordinate of atom c, etc. 
Maybe that's why the transform exists, because it's actually 3 dimensional...? I'm sure the transform (from both documents) are correct, as I have played around with them a bit, but I still don't understand where they came from. 
The motivation for these transforms is to be able to evaluate the following types of integrals:
$
\int_{-\infty}^{\infty} dr \frac{\exp{(-\gamma r_c^2)}}{r_c}
$
r again being shorthand as above, and dr meaning dxdydz 
 A: 
I am uncertain as to the specific question in the OP.  It seems that the request is to show the rationale for calling the integral $\frac1{\sqrt \pi}\int_{-\infty}^\infty e^{-r_c^2t^2}\,dt$ a Laplace Transform.  It is to that end that we now proceed.


First, exploiting even symmetry reveals
$$\frac1{\sqrt \pi}\int_{-\infty}^\infty e^{-r_c^2t^2}\,dt=\frac{2}{\sqrt \pi}\int_{0}^\infty e^{-r_c^2t^2}\,dt$$
Then, letting $t=\frac{\sqrt{u}}{r_c}$, we obtain
$$\frac1{\sqrt \pi}\int_{0}^\infty e^{-r_c^2t^2}\,dt=\frac{1}{r_c\sqrt \pi}\int_{0}^\infty u^{-1/2}e^{-u}\,dt \tag 1$$
is in the form of a Laplace Transform of $u^{-1/2}$ evaluated at $s=1$.

Note that nowhere is there the claim that the result is the Laplace Transform of $\frac1{r_c}$ or the inverse Laplace Transform of $\frac{1}{r_c}$. In fact, $(1)$ has the explicit appearance of $1/r_c$.  Hence, given that the expression in $(1)$ is equal to $1/r_c$, we find that
$$\int_0^\infty u^{-1/2}e^{-u}\,du=\sqrt \pi$$


The authors motivation for using the representation $\frac1{r_c}=\frac1{\sqrt{\pi}}\int_{-\infty}^\infty e^{-r_c^2t^2}\,dt$ is to work with a $4$-dimensional integral with $r_c$ appearing as part of the argument in an exponential term rather than work with the $3$-dimensional integral with the appearance of $\frac1{r_c}$.

A: Using symmetry, the right side of the first equation is
$$ \frac{2}{\sqrt{\pi}} \int_0^\infty \exp(-r_c^2 t^2)\; dt$$
The change of variables $t = \sqrt{u}$ makes this into
$$ \frac{1}{\sqrt{\pi}} \int_0^\infty \frac{\exp(-r_c^2 u)}{\sqrt{u}}\; du $$
which is essentially the Laplace transform of $1/\sqrt{u}$ at $r_c^2$.  This one does exist, because $1/\sqrt{u}$ is integrable at $0$.
The value is indeed $1/\sqrt{r^2} = 1/|r|$, assuming $r$ is real.
Actually, this is slightly backwards.  To do this Laplace transform we need to know that $\Gamma(1/2) = \sqrt{\pi}$, and the usual way to show $\Gamma(1/2) = \sqrt{\pi}$ is to reverse the change of variables and apply the well-known fact that $$\int_{-\infty}^\infty \exp(-t^2)\; dt = \sqrt{\pi}$$
