generalisation of Gaussian integral

The integral

$${\int }_0^{\mathrm{\infty }}{e}^{-x^2}\cos txdx:=w(t)$$

is a generalisation of the Gaussian integral  $w(0)=\frac{\sqrt{\pi }}{2}$.  For evaluating it we first form its derivative which may be done by differentiating under the integral sign (http://planetmath.org/DifferentiationUnderIntegralSign):

$$w^{\prime }(t)={\int }_0^{\mathrm{\infty }}{e}^{-x^2}(-x)\sin txdx=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{e}^{-x^2}(-2x)\sin txdx$$

Using integration by parts this yields

$$w^{\prime }(t)=\frac{1}{2}\underset{x=0}{\overset{\mathrm{\infty }}{/}}{e}^{-x^2}\sin tx-\frac{t}{2}{\int }_0^{\mathrm{\infty }}{e}^{-x^2}\cos txdx=\frac{1}{2}(0-0)-\frac{t}{2}{\int }_0^{\mathrm{\infty }}{e}^{-x^2}\cos txdx=-\frac{t}{2}w(t).$$

Thus $w(t)$ satisfies the linear differential equation

$$\frac{dw}{dt}=-\frac{1}{2}tw,$$

where one can separate the variables (http://planetmath.org/SeparationOfVariables) and integrate:

$$\int \frac{dw}{w}=-\frac{1}{2}\int t𝑑t.$$

So,  $\ln w=-\frac{1}{4}{t}^2+\ln C$,  i.e.  $w=w(t)=C{e}^{-\frac{1}{4}{t}^2}$,  and since there is the initial condition  $w(0)=\frac{\sqrt{\pi }}{2}$, we obtain the result

$$w(t)=\frac{\sqrt{\pi }}{2}{e}^{-\frac{1}{4}{t}^2}.$$

Title	generalisation of Gaussian integral
Canonical name	GeneralisationOfGaussianIntegral
Date of creation	2013-03-22 18:43:36
Last modified on	2013-03-22 18:43:36
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	6
Author	pahio (2872)
Entry type	Derivation
Classification	msc 26B15
Classification	msc 26A36
Related topic	SubstitutionNotation