Derivative of function with Fourier transform

Question

I want to calculate below derivative with Fourier transport: $\partial_{x}p(x,t)= \partial_{x} \int_{-\infty}^{+\infty} p(k,t) e^{-ikx} dk$$ = -\int_{-\infty}^{+\infty} p(k,t) i k e^{-ikx} dk$ in my case, p(k,t) is a $ e^{-|k|^{\alpha}t} $. So, above integral is zero! But in the case of $\alpha=2$, $p(x,t)$ is a guasian distribution which its derivative is not zero!! Could anyone underestand what is the problem in this case?

Answer 1

Assume $\alpha>1$. One may observe that $$ \begin{align} \int_{-\infty}^{+\infty} e^{-|k|^{\alpha}} i k e^{-ikx} dk&=\int_{-\infty}^{0} e^{-|k|^{\alpha}} i k e^{-ikx} dk+\int_{0}^{+\infty} e^{-|k|^{\alpha}} i k e^{-ikx} dk \\\\&=-\int_{0}^{\infty} e^{-|k|^{\alpha}} i k e^{+ikx} dk+\int_{0}^{+\infty} e^{-|k|^{\alpha}} i k e^{-ikx} dk \\\\&=2\int_{0}^{\infty} e^{-|k|^{\alpha}} k\: \frac{e^{+ikx}-e^{-ikx}}{2i} dk \\\\&=2\int_{0}^{+\infty} e^{-|k|^{\alpha}} k \sin(kx) dk \\\\&\neq 0. \end{align} $$

Answer 2

Why would the integral always be zero? Note that $p(k, t)ik$ is an odd function in $k$, but $e^{-ikx} = \cos(kx) - i\sin(kx)$ is neither even nor odd.