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: $$ T(x,t) = \int_{- \infty}^{+ \infty} \! \frac{1}{\sqrt{4\pi a t}}\, e^{\, - \xi^2/(4 a t) }\, T(x-\xi,0) \, d\xi $$