Does anybody knows how to show that the position expectation value $$\langle x\rangle = \int_{-\infty}^{\infty} x|\psi(x)|^2dx$$
can be expressed in terms of momentum?
$$\langle x\rangle = i\hbar\int_{-\infty}^{\infty} \tilde{\psi}^*(p)\frac{\partial \tilde{\psi}(p)}{\partial p}dp$$
My attempts:
\begin{eqnarray} \langle x \rangle &=&\int \langle \psi |\hat{x}| \psi \rangle \int dp |p\rangle \langle p | \\ &=& \int \langle \psi |\hat{p}|p\rangle\langle p |\psi \rangle dp \\ &=& \int \psi^*(p) x \psi(p) dp \end{eqnarray}
I am stuck at this part ... any help will be appreciated.