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?
Best Answer
\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.