Computing Fourier transform for $\exp(- \alpha t^2)$ I use the Fourier Transform for the function $f(x)$ (let's say $f$ is from Schwartz class) in the following form:
$$\widehat{f}(\xi) = \int f(\eta) e^{-2 \pi i \xi \eta} \, d\eta,$$
My task is to compute Fourier Transform the the function $f(x) = e^{-\alpha x^2}$ for $\alpha > 0$ knowing that Fourier Transform for the function $g(x) = e^{- \pi x^2}$ is equal to $\widehat{g}(\xi) = e^{-\pi \xi^2}$.
I tried to solve that problem in the following way.
I know that $f(ax) \mapsto \frac{1}{a} \widehat{f}(\frac{\xi}{a})$. Using that fact I was managed to rewrite $f(x) = \exp(-\alpha \frac{1}{\sqrt{\pi}} (\sqrt{\pi x}))$. Thus we have $\widehat{f}(\xi) = \frac{\sqrt{\pi}}{\alpha} \exp (-\frac{\sqrt{\pi}}{\alpha} \xi^2)$. 
Is my attempt correct?
 A: \begin{align}
\widehat{f}(\xi) & = \int f(\eta) e^{-2 \pi i \xi \eta} \, d\eta. \\[8pt]
g(\eta) & = f(k\eta) \qquad \text{for all values of } \eta. \\[8pt]
\widehat{g\,}(\xi) & = \int g(\eta) e^{-2\pi i\xi\eta} \, d\eta \\[8pt]
& = \int f(k\eta) e^{-2\pi i\xi\eta} \, d\eta \\[8pt]
& = \frac 1 k \int f(k\eta) e^{-2\pi i\big(\xi/k\big)\big(k\eta\big)} \big( k\, d\eta\big) \\[8pt]
& = \frac 1 k \int f(\theta) e^{-2\pi i \big(\xi/k\big) \theta} \, d\theta \\[8pt]
& = \frac 1 k \widehat f\left( \frac \xi k \right)
\end{align}
Find the right value of $k$ for the occasion and then use this.
A: You have $f(x) = g(\sqrt{\frac\alpha\pi}x)$. Thus $$\hat f(\xi) = \frac{1}{\left|\sqrt{\frac\alpha\pi}\right|}\hat g\left(\frac\xi{\sqrt\frac\alpha\pi}\right)=\sqrt\frac{\pi}{\alpha}\,\hat g\Big(-\frac{\xi\sqrt\pi}{\sqrt\alpha}\Big)=\sqrt\frac\pi\alpha\exp\Big(-\frac{\pi^2\xi^2}{\alpha}\Big),$$
using the property that $\widehat{f(a{}\cdot{})}(\xi) = \frac1{|a|}\hat f(\xi/a)$.

In your calcuation, you didn't square the $\sqrt\pi$ and you didn't get the extra $\pi$ factor which is already in the exponent of $\hat g$. Here is what Mathematica gives, for verification:

A: The function $f(x) = e^{-\alpha x^2}$ satisfies the differential equation $f'(x) = -2\alpha x f(x).$ 
Taking the Fourier transform of both sides of the differential equation gives $2\pi i\xi \hat{f}(\xi) = -2\alpha (\frac{i}{2\pi}\frac{d}{d\xi}) \hat{f}(\xi),$ i.e. $\hat{f}$ satisfies the differential equation $\hat{f}'(\xi) = -\frac{4\pi^2}{2\alpha}\xi \hat{f}(\xi).$ The solutions to this are $\hat{f}(\xi) = C e^{-\pi^2\xi^2/\alpha},$ where
$C = \hat{f}(0) = \int f(x) \, e^{-2\pi i0 x} dx = \int e^{-\alpha x^2} dx = \sqrt{\pi/\alpha}.$
Thus, $\hat{f}(\xi) = \sqrt{\frac{\pi}{\alpha}} e^{-\pi^2\xi^2/\alpha}.$
