I want to calculate Fourier' transformation of $$f(n)=e^{-n^2\pi x}.$$
Using How to calculate the Fourier transform of a Gaussian function. I found Fourier' transformation of $g(n)=e^{-n^2}$, which is $$\hat{g}(\xi)=\sqrt{\pi}e^{-\frac{\xi^2}4}.$$ I know that $g(\pm \sqrt{\pi x}n)=f(n)$. How can I find $\hat{f}$ through $\hat{g}$? I know that the end result should be $$\hat{f}(\xi)=\frac1{\sqrt{\pi}}e^{-\frac{\xi^2 \pi}{x}}.$$
In general the fourier transform of $f(ax)$ is equal to $\frac{1}{|a|}\hat{f}(\frac{\xi}{a}).$ Hence, if the fourier transform of $g(n)=e^{-n^2}$ is $\hat{g}(\xi)=\sqrt{\pi}e^{-\frac{\xi^2}4},$ then the fourier transform of $f(n)=e^{-n^2\pi x} =g(n\sqrt{\pi x})$ is given by $$\hat{f}(\xi) = \frac{1}{\sqrt{\pi x}}\hat{g}\left(\frac{\xi}{\sqrt{\pi x}}\right) =\frac{1}{\sqrt{\pi x}}\sqrt{\pi}e^{-\frac{\xi^2}{4\pi x}}=\frac{1}{\sqrt{x}}e^{-\frac{\xi^2}{4\pi x}}.$$
In general you have the following: if $\mathcal{D}_{\alpha}f(x) = f(\alpha x)$ where $\alpha >0$, then
$$\mathcal{F}\mathcal{D}_{\alpha}f(y) = \frac{1}{\alpha} \mathcal{D}_{\alpha^{-1}}\mathcal{F}f(y).$$
You can see this by doing a change of variable $x' = \alpha x$:
$$\mathcal{F}\mathcal{D}_{\alpha}f(y) = \int_{-\infty}^{\infty} e^{-ixy} f(\alpha x)\,dx = \int_{-\infty}^{\infty} e^{-ix'\frac{y}{\alpha}} f(x')\,\frac{1}{\alpha}dx' = \frac{1}{\alpha} \mathcal{F}f\left(\frac{y}{\alpha}\right)$$
which is of course equal to $\frac{1}{\alpha}\mathcal{D}_{\alpha^{-1}}\mathcal{F}f(y)$. What this says is that if you know $\mathcal{F}f$, then you can easily compute the Fourier transform of a dilation of $f$. In your case, we know what the Fourier transform of $g(n) = e^{-n^2}$ is. To get $f$, note that $f(n) = g(\sqrt{\pi x} n)$ and use the above relationship.
