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