Fourier Transform of Partial Derivative w.r.t x of [ x*f(x) ] Can someone please help with the Fourier Transform of : !enter image description here Thank you in advance! ::Edit:: This is what I am trying to solve: $\frac{\partial(p(x, t))}{\partial t} = -A\frac{\partial(xp(x, t))}{\partial x}+\frac{B}{2}\frac{\partial^{2}(xp(x, t))}{\partial x^{2}}$ Where:[{A, B} = Constants] Define: $FT\\{p(x, t)\\}(\omega) = \int_{-\infty }^{\infty }p(x, t)e^{-2\pi ix\omega }\,dx$ and $FT^{-1}\\{\bar{p}(\omega , t)\\}(x) = \int_{-\infty }^{\infty }p(\omega , t)e^{2\pi ix\omega }\,d\omega$ The next step is to convert each term so I can reduce the order but I started reading about Fourier Transforms two days ago, so I do not know all the tricks. I did use the properties below to get rid of the derivative - but I have no idea how to convolve x with p(x,t) - since p(x,t) is unknown. p(x,t) is a density function - so it goes to zero in the infinities (if this is important) Thank you again!

So let's try and tackle the first order derivative, $\partial_x(xp)=p+x\partial_x p$, let's look at the Fourier of the second term:

$F(x\partial_x p)=\int_{\mathbb R}x\partial_xpe^{-2\pi i xw}\,dx$

Notice that $\partial_we^{-2\pi i xw}=-2\pi ixe^{-2\pi i xw}$ so

$F(x\partial_x p)=\frac{i}{2\pi}\partial_wF(\partial_xp)=-w\partial_wF(p)$. By the way I may be out by some constants, but use the magic $2\pi=1$ formula and its all good.