Is this document using laplace transforms incorrectly? (in particular, laplace transform of 1/r)

Question

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

Answer 1

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.

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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 $s=1$.

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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

$$\int_0^\infty u^{-1/2}e^{-u}\,du=\sqrt \pi$$

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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 $\frac1{r_c}$.

Answer 2

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 $$\int_{-\infty}^\infty \exp(-t^2)\; dt = \sqrt{\pi}$$