An integral equals to high order derivative of dirac delta function

Question

I encounter the following equation:

$\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}e^{-ix\theta}\frac{(-1)^m e^{i\mu\theta}\theta^{2m}}{m!2^m}d\theta=\frac{1}{m!2^m}\delta^{(2m)}(x-\mu)$

I wonder whether it means for every "good" function $f(x)$,

$\frac{(-1)^m}{2\pi}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}f(x)\theta^{2m}e^{-i(x-\mu)\theta}d\theta dx=f^{(2m)}(\mu)$.

If so, any hints to prove it? If not, what does the first equation mean? Many thanks.

Answer 1

Yes, this is true. You should study the Fourier transform of distributions. See equation (308) or the unnumbered following it.