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