In 1D wave mechanics, is there a counterexample to the relation $m \frac{d \langle x\rangle}{dt} = \langle p \rangle$? The standard physicists' proof of the identity $m \frac{d\langle x\rangle}{dt} = \langle p \rangle$ involves integration by parts. For example, in Griffiths's "Introduction to Quantum Mechanics", the derivation goes as follows:
\begin{equation}
\begin{split}
m\frac{d\langle x \rangle}{dt} &= m\int x \frac{\partial|\psi|^2}{\partial t} dx\\ &= \frac{i\hbar}{2}\int x\frac{\partial}{\partial x}\left(\psi^\ast\frac{\partial\psi}{\partial x}-\frac{\partial\psi^\ast}{\partial x}\psi\right) dx\\
&= -\frac{i\hbar}{2}\int \left(\psi^\ast\frac{\partial\psi}{\partial x}-\frac{\partial\psi^\ast}{\partial x}\psi\right) dx\\
&= -i\hbar \int \psi^\ast\frac{\partial\psi}{\partial x} dx\\
& = \langle p \rangle,
\end{split}
\tag{1}
\label{a}
\end{equation}
Here, (among other things) one should integrate by parts to obtain the third line, where the associated boundary term is assumed to vanish, i.e.,
\begin{equation}
x\left(\psi^\ast\frac{\partial\psi}{\partial x}-\frac{\partial\psi^\ast}{\partial x}\psi\right) \Bigg|_{x=-\infty}^{\infty} = 0.
\tag{2}
\label{b}
\end{equation}
But is it really OK to make such an assumption? In fact, for the normalizable wave function
\begin{equation}
\psi_1(x) = \frac{e^{ix^4}}{x^2 + 1},
\tag{3}
\end{equation}
the boundary term [Eq. $(\ref{b})$] does not vanish, making the whole derivation in Eq. $(\ref{a})$ invalid. Still, it is easy to see that $\langle p \rangle$ itself is ill-defined for the above wave function (i.e., the integral $\langle\psi_1|p|\psi_1\rangle$ is not convergent), so this counterexample is not very interesting.
Hence, my question is the following:

Is it possible to construct a counterexample to the relation $m \frac{d\langle x\rangle}{dt} = \langle p \rangle$, where both $\langle x\rangle$ and $\langle p\rangle$ are well-defined?

 A: I would say the usual proof of this statement comes from Ehrenfest's theorem:
$\frac{d<Q>}{dt} = -\frac{i}{\hbar} [Q,H]$
Then with the usual single particle Hamiltonian one has $H=\frac{p^2}{2m} +V(x)$ and so  $\frac{d<X>}{dt}=\frac{1}{2m}[p^2,x]$ This evaluate via standard commutation rules to your identity $m\dot{<x>}=<p>$.
At no point here did we invoke integration by parts. The proof of Ehrenfest's theorem, which does involves inner products (and so carries the risk of IBP), simply requires that all the inner products exist and this is equivalent to the statement that our wavefunctions are normalisable.
A: The derivation in equation (1) assumes the following: in order that the boundary at infinity term to be 0 in the limit, it is enough/sufficient to consider that the wavefunctions are Schwartz test functions (i.e. they go to 0 to infinity quicker than any polynomial function). It is when this set of functions is used that the operators $x$ and $p_x$ are essentially self-adjoint, their matrix elements $\langle \psi, x\psi\rangle$ and $\langle \psi, p_x\psi\rangle$ are properly defined, well-behaved functions of the parameter $t$ by means of $\psi(t)$ and the space $\mathcal{S}(\mathbb R)$ is invariant under the uniparameter group $e^{itH}$ by the theorem of Hunziker. These are the sufficient conditions to make (1) into a valid derivation.
So the final question of the original post has this answer: "NO, because making the matrix elements as well-behaved functions of $t$ makes all manipulations in (1) mathematically valid".
Ehrenfest's theorem has been properly formulated and proven only as late as 2009 (i.e. more than 80 years after the original work by P. Ehrenfest). This can be found here: "On the Ehrenfest theorem of quantum mechanics", Gero Friesecke and Mario Koppen, Journal of Mathematical Physics 50, 082102 (2009); doi: 10.1063/1.3191679
