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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 $$\hat{f}(\xi)=\frac1{\sqrt{\pi}}e^{-\frac{\xi^2 \pi}{x}}.$$

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  • $\begingroup$ Did you mean to put a negative in the exponent for $f$? $\endgroup$ – Cameron Williams Mar 25 '16 at 17:47
  • $\begingroup$ It's fixed now. $\endgroup$ – weqfrr Mar 25 '16 at 17:49
  • $\begingroup$ if the function is integrable, then its Fourier transform is $\int_{-\infty}^\infty f(x) e^{-2 i \pi \xi x} dx$ .... a change of variable might solve your question $\endgroup$ – reuns Mar 25 '16 at 17:54
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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 $$\hat{f}(\xi) = \frac{1}{\sqrt{\pi x}}\hat{g}\left(\frac{\xi}{\sqrt{\pi x}}\right) =\frac{1}{\sqrt{\pi x}}\sqrt{\pi}e^{-\frac{\xi^2}{4\pi x}}=\frac{1}{\sqrt{x}}e^{-\frac{\xi^2}{4\pi x}}.$$

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  • $\begingroup$ How come the Fourier' transformation is different from my result? $\endgroup$ – Rasmus Erlemann Mar 25 '16 at 18:11
  • $\begingroup$ @C.Dubussy Does it hold that the fourier transform of |x| is equal to $-\frac{4 \pi}{\xi^2}$ ? $\endgroup$ – Evinda Mar 29 '16 at 18:21
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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.

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  • $\begingroup$ juste to say that in the context of linear operators, $D$ is the derivative operator, I think $S$ for stretch would be safer $\endgroup$ – reuns Mar 25 '16 at 17:56
  • $\begingroup$ I changed it to $\mathcal{D}$ for this reason (also I don't like $D$ - looks weird next to the Fourier transform). Good point though. Using $S$ runs into the issue of the Schwartz space as well. $\endgroup$ – Cameron Williams Mar 25 '16 at 17:56
  • $\begingroup$ @REr Where did you get the answer from? There are several different (though effectively equivalent) definitions of the Fourier transform. Your answer might vary depending on which convention is chosen. $\endgroup$ – Cameron Williams Mar 25 '16 at 19:06

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