I'm currently reading Griffiths' book about Quantum Mechanics but I cannot understand how he derives the formula for the time derivative of the expected value of position in 1 dimension.
He writes:
$$\frac{\mathrm d\langle x \rangle}{ \mathrm dt} = \int x\,\frac{\partial}{\partial t}\left(|\psi|^2\right) \mathrm dx =\frac{i\hbar}{2m}\int x\,\frac{\partial}{\partial x}\left(\psi^*\frac{\partial\psi}{\partial x}+\psi \frac{\partial\psi^*}{\partial x}\right)\mathrm dx \quad . \tag 1$$
Now he integrates by parts: He integrates $$\frac{\partial}{\partial x}\left(\psi^*\frac{\partial\psi}{\partial x}+\psi \frac{\partial\psi^*}{\partial x}\right) \tag 2 $$
and he differentiates $x$. Explicitly the integral above becomes:
$$x\left(\psi^*\frac{\partial\psi}{\partial x}+\psi\frac{\partial\psi^*}{\partial x}\right)\big\rvert_\mathbb{R} + \int\psi^*\frac{\partial\psi}{\partial x}+\psi\frac{\partial \psi^*}{\partial x} \mathrm d x\quad . \tag 3$$
The author concludes that the boundary term evaluates to $0$, but I can't see how. $x$ goes to $\infty$ and both $\psi$ and $\psi^*$ go to $0$ when $x$ approaches $\infty$, so in principle we should have an indeterminate form of the type $\infty * 0$. How can I make some progress here?
Reference: David J. Griffiths - Introduction to Quantum Mechanics 2nd Ed.
Chapter 1, Section 1.5, Pages 15-16