# 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?

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:

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

\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.

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