I am trying to understand how to get general expression for fourier transform for $$\large f(x)=e^{-\frac{ax^2}{2}}$$
if $a=1$ I understand that I can calculate it through the fact that $$ f'(x)=-xf(x)$$ $$ w \hat f(w) = -\hat f'(w)$$ and because $f$ and $\hat f$ solve the same differential equation they are same function that might different in scalar multiply. so we can get $$\hat f(w)=\frac{1}{\sqrt{2\pi}}e^{-\frac{w^2}{2}}$$ how do I get general form for $a>0$?
$\int e^{-itx} e^{-ax^{2}/2}dx=\frac1 {\sqrt a}\int e^{-ity/\sqrt a} e^{-y^{2}/2}dy$ by the substitution $y=\sqrt a x$. The integral here is the value of the FT in the case $a=1$ with $t$ changed to $\frac t {\sqrt a}$.