# Thread: The exponential and the Fourier transform.

1. ## The exponential and the Fourier transform.

Hi all,

For what reason is exp(-1/2 x^2) a rapidly decreasing function? I cannot see it. I know that |exp(-1/2 x^2)| <= 1, but I don't think this helps me.

Also, we know that it satisfies the differential equation y' + xy = 0. I have to show that it's Fourier transform also satisfies it (this is in a proof of showing that it equals its own FT, so we cannot use this fact).

I have tried to use the definition and differentiation under the integral sign, but I can't get it to 0. I end up with an 'x' term inside one integral and an 'x' term on the outside of another.

Thanks.

2. Originally Posted by measureman
For what reason is exp(-1/2 x^2) a rapidly decreasing function? I cannot see it. I know that |exp(-1/2 x^2)| <= 1, but I don't think this helps me.
By an easy induction proof, the n'th derivative of $\exp(-\tfrac12x^2)$ is of the form $p_n(x) \exp(-\tfrac12x^2)$ for some polynomial $p_n(x).$ This goes to 0 at $\pm\infty$ because of the fact that exponential growth always exceeds polynomial growth.

Originally Posted by measureman
Also, we know that it satisfies the differential equation y' + xy = 0. I have to show that its Fourier transform also satisfies it (this is in a proof of showing that it equals its own FT, so we cannot use this fact).

I have tried to use the definition and differentiation under the integral sign, but I can't get it to 0. I end up with an 'x' term inside one integral and an 'x' term on the outside of another.
Differentiation under the integral sign is certainly the right way to go about this, combined with integration by parts.

If $f(x) = \exp(-\tfrac12x^2)$ then its Fourier transform is given by $\hat{f}(\omega) = \int_{-\infty}^\infty e^{-\frac12x^2}e^{-i\omega x}dx.$ Differentiate under the integral sign (with respect to $\omega$) to get $\hat{f}'(\omega) = \int_{-\infty}^\infty e^{-\frac12x^2}\bigl(-ix e^{-i\omega x}\bigr)dx = i\int_{-\infty}^\infty \bigl(-xe^{-\frac12x^2})e^{-i\omega x}dx.$ Now integrate that by parts to get

$\Bigl[ie^{-\frac12x^2}e^{-i\omega x}\Bigr]_{-\infty}^\infty - i\int_{-\infty}^\infty e^{-\frac12x^2}\bigl(-i\omega e^{-i\omega x}\bigr)dx = -\omega\hat{f}(\omega).$

3. Brilliant, thank you. The first one seems really obvious now you mention it. The second one, I was differentiating wrt x, which caused me problems.