generalisation of Gaussian integral

The integral

 $\int_{0}^{\infty}\!e^{-x^{2}}\cos{tx}\,dx\;:=\;w(t)$

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):

 $w^{\prime}(t)\;=\;\int_{0}^{\infty}\!e^{-x^{2}}(-x)\sin{tx}\,dx\;=\;\frac{1}{2% }\int_{0}^{\infty}\!e^{-x^{2}}(-2x)\sin{tx}\,dx$

Using integration by parts this yields

 $w^{\prime}(t)\;=\;\frac{1}{2}\!\operatornamewithlimits{\Big{/}}_{\!\!\!x=0}^{% \,\quad\infty}\!e^{-x^{2}}\sin{tx}-\frac{t}{2}\int_{0}^{\infty}\!e^{-x^{2}}% \cos{tx}\,dx\,=\,\frac{1}{2}(0-0)-\frac{t}{2}\int_{0}^{\infty}\!e^{-x^{2}}\cos% {tx}\,dx\,=\,-\frac{t}{2}w(t).$

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:

 $\int\!\frac{dw}{w}\;=\;-\frac{1}{2}\int\!t\,dt.$

So,  $\ln{w}\,=\,-\frac{1}{4}t^{2}+\ln{C}$,  i.e.  $w=w(t)=Ce^{-\frac{1}{4}t^{2}}$,  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 GeneralisationOfGaussianIntegral 2013-03-22 18:43:36 2013-03-22 18:43:36 pahio (2872) pahio (2872) 6 pahio (2872) Derivation msc 26B15 msc 26A36 SubstitutionNotation