quantum mechanics - Expectation value with plane waves

Hey guys Im a little confused with the concept of plane waves and how to perform an expectation value. Let me show you by an example. Suppose you have a wave function of the form

$\psi_{\boldsymbol{p}_{0}}(x)=f(p_{0})e^{\frac{i}{\hbar}\boldsymbol{p}_{0}\cdot\boldsymbol{x}}$

where $\boldsymbol{p}_{0}=(0,0,p_{0})$ and suppose you want to perform an expectation value of the position of the particle, that is

$=f^{2}(p_{0})\int\,d^{3}\boldsymbol{x}\,x\,e^{\frac{i}{\hbar}\boldsymbol{p}_{0}\cdot\boldsymbol{x}}e^{-\frac{i}{\hbar}\boldsymbol{p}_{0}\cdot\boldsymbol{x}}=f^{2}(p_{0})\int\,d^{3}\boldsymbol{x}\,x$

wich I think is nonsense. But if you define an arbitrary momentum vector $\boldsymbol{p}'=(p_{1}',p_{2}',p_{3}')$, and perform the transition probability

\begin{align} \left<\psi_{\boldsymbol{p}'}(\boldsymbol{x})|\,x\,|\psi_{\boldsymbol{p}_{0}}(\boldsymbol{x})\right>&=f(p')f(p_{0})\int\,d^{3}\boldsymbol{x}\,xe^{-\frac{i}{\hbar}(\boldsymbol{p}'-\boldsymbol{p}_{0})\cdot\boldsymbol{x}}=f(p')f(p_{0})\int\,d^{3}\boldsymbol{x}\left( i\hbar\frac{\partial}{\partial p_{x}'}\right)e^{-\frac{i}{\hbar}(\boldsymbol{p}'-\boldsymbol{p}_{0})\cdot\boldsymbol{x}}= \\ &=i\hbar f(p')f(p_{0})\frac{\partial}{\partial p_{x}'}\delta^{3}(\boldsymbol{p}'-\boldsymbol{p}_{0})=-i\hbar\delta^{3}(\boldsymbol{p}'-\boldsymbol{p}_{0})\frac{\partial}{\partial p_{x}'}\left( f(p')\right)f(p_{0}) \end{align}

where I made use of the property $f(x)\delta'(x)=-f'(x)\delta(x)$. So now with my new expresion I have a meaningful result and I can evaluate for $\boldsymbol{p}'=\boldsymbol{p}_{0}$ and get a result that I wasnt able to get with the first method. What I'm doing wrong or the the second way is the correct way to do it? Thanks!