Show that the Fourier transform of $f(x) = \exp(-x^2/2)$ satisfies the differential equation $F'(x) = -x F(x)$

Question

Let $f(x) = \exp(-x^2/2)$ and $$F(x) = \hat{f}(x) = \int_{-\infty}^\infty f(t) \exp(-ixt) \,\Bbb dt$$ be the Fourier transform of $f$. In my exercise, I need to show that $F$ satisfies the differential equation $F'(x) = -x F(x)$.

I used the rule $i (\hat{f})'(x) = \hat{g}(x)$ with $g(t) = t f(t)$ from our lecture notes to derive

$$F'(x) = \frac{1}{i} \hat{g}(x).$$

Using the definition of the Fourier transform, we further obtain

$$\frac{1}{i} \hat{g}(x) = \frac{1}{i} = t \exp\left(-ixt - \frac{t^2}{2}\right) \,\Bbb dt.$$

I cannot see how I can get the desired expression

$$-x F(x) = -x \hat{f}(x) = -x \int_{-\infty}^\infty f(t) \exp(-ixt) \,\Bbb dt$$

from this point. I feel like I need to make use of the fact that $f(x) = \exp(-x^2/2)$ and hence tried integrating by parts, but I cannot simplify my integral with this.

Can anyone help me with this problem, please?

Answer 1

As you’ve noted, it’s a standard rule that $i\hat{f}’=\hat{xf}$ and similarly $ix\hat{f}=\hat{f’}$.

Noting that $f(x)=e^{-x^2/2}$ satisfies $f’=-xf$, we have that $$F’=\hat{f}’=1/i(\hat{xf})=1/i(\hat{-f’})=1/i (-ix\hat{f})$=-xF$$

Answer 2

We have $$F(x) = \int_{\mathbb{R}}\exp(-t^2/2 -ixt) dt$$ Differentiating with respect to $x$, we get $$F'(x) = -i\int_{\mathbb{R}}t\exp(-(t^2/2 +ixt))dt = -i\int_{\mathbb{R}}t\exp(-t^2/2)\exp(-ixt)dt$$

Integrating by parts with $u = \exp(-ixt)$ and $dv = t\exp(-t^2/2)$, we have

$$F'(x) = -i\left(\left.\exp({-x^2/2})\exp(-ixt)\right|_{-\infty}^\infty+ix\int_{\mathbb{R}}\exp\left({-t^2/2}\right)\exp({-ixt})dt\right)$$

Since the Gaussian is asymptotically $0$ on both ends, we have

$$F'(x) = xF(x)$$