Integrals¶

Gaussian Intergral¶

The Gaussian integral is a handy tool for calculating/estimating integrals.

\[\int_{-\infty}^{\infty} e^{-ax^2} \mathrm dx = \sqrt{\frac{\pi}{a}}\]

A comprehensive description of how this is done is here. It is easier to solve the integral using polar coordinates.

In general, we could define a more general form of the integral,

\[I_n(a) = \int_{0}^{\infty} e^{-ax^2} \mathrm dx.\]

A Practice of Symmetries

The integral we just solved is

\[\begin{split}&\int_{-\infty}^{\infty} e^{-ax^2} \mathrm dx \\ =& \int_{0}^\infty e^{- a x^2} \mathrm dx + \int_{-\infty}^0 e^{-a x^2} \mathrm dx \\ =& I_0(a) - \int_{-x=\infty}^{-x=0} e^{-a (-x)^2} \mathrm d (-x) \\ =& I_0(a) + \int_{-x=0}^{-x=\infty} e^{-a (-x)^2} \mathrm d (-x) \\ =& I_0(a) + I_0(a) \\ =& 2 I_0(a)\end{split}\]

Another view of this is that we could investigate the symmetry of the expression \(I_n(a)\). By changing the sign of \(x\), we know that

\[\begin{split}&\int_{x=0}^{x=\infty} e^{-a(-x)^2} \mathrm d(-x) \\ =& \int_{-x=0}^{-x=-\infty} e^{-a (-x)^2} \mathrm d(-x) \\ =& -\int_{x=0}^{x=\infty} e^{-a x^2} \mathrm dx \\ = - I_n(a).\end{split}\]

Using this we could easily calculate the first integral.