## Gaussian Blur

One dimension:

$G(x) = \frac{1}{\sqrt{2\pi \sigma^2}} e^{-\frac{x^2}{2\sigma^2}}$

Two dimensions:

$G(x, y) = \frac{1}{2\pi \sigma^2} e^{-\frac{x^2 + y^2}{2\sigma^2}}$

This is because:

$I(t) = \int_{-t}^{t} e^{-x^2}dx \\ I^2(t) = (\int_{-t}^{t} e^{-x^2}dx) \cdot (\int_{-t}^{t} e^{-y^2}dy) = \int_{-t}^{t} \int_{-t}^{t} e^{-(x^2+y^2)}dx dy \\ \int_{0}^{t} \int_{0}^{2\pi}e^{-r^2}r dr d\theta < I^2(t) < \int_{0}^{t\sqrt 2} \int_{0}^{2\pi}e^{-r^2}r dr d\theta \\ \int_{0}^{t} 2\pi e^{-r^2} r dr d\theta < I^2(t) < \int_{0}^{t\sqrt 2} 2\pi e^{-r^2}r dr d\theta \\ \pi(1-e^{-t^2}) < I^2(t) < \pi(1-e^{-2t^2}) \\ \lim_{t \to +\infty} \pi(1-e^{-t^2}) < \lim_{t \to +\infty} I^2(t) < \lim_{t \to +\infty} \pi(1-e^{-2t^2}) \\ \pi < \lim_{t \to +\infty} I^2(t) < \pi$

so we obtain:

$\lim_{t \to +\infty} I^2(t) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)}dx dy = \pi \\ \lim_{t \to +\infty} I(t) = \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt \pi \\ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{2\pi \sigma^2} e^{-\frac{x^2+y^2}{2\sigma^2 }} dx dy = 1$