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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?

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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: enter image description here

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  • $\begingroup$ Could you explain $f(x) = g(i\sqrt{\frac\alpha\pi}x)$? $\endgroup$ – Hendrra Mar 16 '20 at 17:51
  • $\begingroup$ @Hendrra I made a mistake there. $\endgroup$ – Luke Collins Mar 16 '20 at 17:51
  • $\begingroup$ Thank you. Your $g$ is something different than the function $g$ in my post, isn't it? $\endgroup$ – Hendrra Mar 16 '20 at 18:14
  • $\begingroup$ No, it's the same: $g(\sqrt\frac\alpha\pi x) = \exp(-\pi(\sqrt\frac\alpha\pi x)^2) = \exp(-\pi\frac{\alpha}{\pi}x^2) = \exp(-\alpha x^2) = f(x)$. $\endgroup$ – Luke Collins Mar 16 '20 at 18:17
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\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.

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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, $\hat{f}(\xi) = \sqrt{\frac{\pi}{\alpha}} e^{-\pi^2\xi^2/\alpha}.$

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