Gaussian Blur
As an example, investigate the Laplace transform of $exp(-\mu t^2)$:
$$ \int_{-\infty}^{+\infty} e^{\, -pt} e^{\,-\mu t^2} \, dt =
\int_{-\infty}^{+\infty} e^{\,-\mu x^2} \, dx . e^{\, p^2/4\mu }
= \sqrt{ \frac{\pi}{\mu} } e^{\, p^2 / 4\mu } $$
with help of: $ x = t + p/2\mu $.
Laplace transform $H$ and inverse Laplace transform $h$ are mutually related
as follows, after replacing $1/4\mu$ by $1/2\sigma^2$:
$$ H(p) = e^{\, \frac{1}{2} \sigma^2 p^2 } \quad \Longleftrightarrow \quad
h(t) = \frac{1}{ \sigma \sqrt{2\pi} } e^{\, -t^2 / 2\sigma^2 } $$
A convolution integral with the normal distribution $h(t)$ as a kernel can thus
be written, with the help of Operator Calculus, as:
$$ \int_{- \infty}^{+ \infty} \! h(\xi) \phi(x-\xi) \, d\xi =
e^{\, \frac{1}{2} \sigma^2 \frac{d^2}{dx^2} } \phi(x) $$
The physical meaning of this is that the operator
$ exp(\frac{1}{2} \sigma^2 \frac{d^2}{dx^2}) $ "smears out" the function
$\phi(x)$ over domains with size of the order
$ \sigma $.
"Moment generating functions" play a role in Statistics. They are
the expectation values of the exponential function $\exp(pt)$; with other words:
Laplace transforms of probability densities. In a handbook about Statistics it
is read as follows, again:
$$ M(t) = \frac{1}{\sigma \sqrt{2 \pi}} \, \int_{- \infty}^{+ \infty}
exp \left[ \frac{ - (x - \mu)^2 }{ 2 \sigma^2 } \right] e^{tx} \, dx = $$
$$ \frac{ e^{\mu t + \frac12 \sigma^2 t^2 } }{ \sigma \sqrt{ 2 \pi }}
\int_{- \infty}^{+ \infty}
exp \left[ \frac{ - (x-\mu-\sigma^2 t)^2}{ 2 \sigma^2} \right] \, dx =
e^{\mu t + \frac12 \sigma^2 t^2 } $$
The corresponding Normal "smear"-operator is in general given by:
$$ M(\frac{d}{dx}) = e^{\mu \frac{d}{dx} + \frac12 \sigma^2 \frac{d^2}{dx^2} }
= e^{\mu \frac{d}{dx}} \, e^{\frac12 \sigma^2 \frac{d^2}{dx^2} } $$
The outcome is immediately applicable to the following problem. Consider
the (partial differential)equation for diffusion of heat in space and time:
$$ \frac{\partial T}{\partial t} = a \frac{\partial^2 T}{\partial x^2} $$
Rewrite in the first place:
$$ \lambda \frac{\partial}{\partial t} T = \lambda a \frac{\partial^2}{\partial x^2} T $$
As a next step we exponentiate at both sides the operator in place:
$$ e^{\lambda \partial / \partial t} \, T = e^{\lambda a \partial^2/\partial x^2} \, T $$
At both sides are operator-expressions which can be converted into classical
mathematics with the acquired knowledge:
$$ T(x,t+\lambda) = \int_{- \infty}^{+ \infty} \! h(\xi) T(x-\xi,t) \, d\xi $$
Where $ \frac12 \sigma^2 = \lambda a $. Therefore:
$$ h(t) = \frac{1}{ \sigma \sqrt{2\pi} } \, e^{\, -t^2 / 2\sigma^2 } $$
Now exchange $t$ and $\lambda$, and substitute $\lambda = 0$. Then we find,
while travelling via a very short route, the solution of the equation for
thermal diffusion:
$$ T(x,t) = \int_{- \infty}^{+ \infty} \!
\frac{1}{\sqrt{4\pi a t}}\, e^{\, - \xi^2/(4 a t) }\, T(x-\xi,0) \, d\xi $$