This question is from Quantum Physics, but my problem is essentially with the Mathematics.
Let $\Psi(x,t)$ be a complex square-integrable function that satisfies the Schrödinger equation:
$$\frac{\partial \Psi(x,t)}{\partial t} = \frac{i\hbar}{2m} \frac{{\partial}^2 \Psi(x,t)}{\partial x^2} - \frac{i}{\hbar} V(x) \Psi(x,t)$$
where $\hbar$ and $m$ are physical constants that neither depend on $x$ nor $t$ and $V(x)$ is a real function. $i$ is the imaginary unit, the complex number whose square is -1.
I want to prove the Ehrenfest theorem:
$$i\hbar \frac{d}{dt}\int_{-\infty}^{+\infty} \Psi^\star(x,t) \frac{\partial \Psi(x,t)}{\partial x} dx = \int_{-\infty}^{+\infty} |{\Psi(x,t)}|^2 \frac{\partial V(x)}{\partial x} dx$$
This is how I went about doing it:
$$i\hbar \frac{d}{dt}\int_{-\infty}^{+\infty} \Psi^\star \frac{\partial \Psi}{\partial x} dx = i\hbar \int_{-\infty}^{+\infty} \frac{\partial \Psi^\star}{\partial t} \frac{\partial \Psi}{\partial x} dx + i\hbar \int_{-\infty}^{+\infty} \Psi^\star \frac{\partial^2 \Psi}{\partial t \partial x} dx$$
Now, using integration by parts:
$$\int_{-\infty}^{+\infty} \frac{\partial \Psi^\star}{\partial t} \frac{\partial \Psi}{\partial x} dx = [\frac{\partial \Psi^\star}{\partial t} \Psi]|_{-\infty}^{+\infty} - \int_{-\infty}^{+\infty} \frac{\partial^2 \Psi^\star}{\partial t \partial x} \Psi dx$$
which using the square integrability of $\Psi$ becomes:
$$\int_{-\infty}^{+\infty} \frac{\partial \Psi^\star}{\partial t} \frac{\partial \Psi}{\partial x} dx = - \int_{-\infty}^{+\infty} \frac{\partial^2 \Psi^\star}{\partial t \partial x} \Psi dx$$
Thus the left hand side of the proof statement becomes:
$$i\hbar \int_{-\infty}^{+\infty} \Psi^\star \frac{\partial^2 \Psi}{\partial x \partial t} dx - i\hbar \int_{-\infty}^{+\infty} \frac{\partial^2 \Psi^\star}{\partial x \partial t} \Psi dx$$
Let this expression be labelled (1).
Further simplification requires the use of the Schrödinger equation. Differentiating w.r.t $x$, the Schrödinger equation and taking complex conjugate, we get two equations:
$$\frac{\partial^2 \Psi}{\partial x \partial t} = \frac{i\hbar}{2m} \frac{\partial^3 \Psi}{\partial x^3} - \frac{i}{\hbar} \frac{\partial V}{\partial x} \Psi - \frac{i}{\hbar} V \frac{\partial \Psi}{\partial x}$$
$$\frac{\partial^2 \Psi^\star}{\partial x \partial t} = - \frac{i\hbar}{2m} \frac{\partial^3 \Psi^\star}{\partial x^3} + \frac{i}{\hbar} \frac{\partial V}{\partial x} \Psi^\star + \frac{i}{\hbar} V \frac{\partial \Psi^\star}{\partial x}$$
From which, we get:
$$\Psi^\star \frac{\partial^2 \Psi}{\partial x \partial t} = \frac{i\hbar}{2m} \Psi^\star \frac{\partial^3 \Psi}{\partial x^3} - \frac{i}{\hbar} \frac{\partial V}{\partial x} |\Psi|^2 - \frac{i}{\hbar} V \Psi^\star \frac{\partial \Psi}{\partial x}$$
$$\frac{\partial^2 \Psi^\star}{\partial x \partial t} \Psi = - \frac{i\hbar}{2m} \Psi \frac{\partial^3 \Psi^\star}{\partial x^3} + \frac{i}{\hbar} \frac{\partial V}{\partial x} |\Psi|^2 + \frac{i}{\hbar} V \Psi \frac{\partial \Psi^\star}{\partial x}$$
And thus,
$$\Psi^\star \frac{\partial^2 \Psi}{\partial x \partial t} - \frac{\partial^2 \Psi^\star}{\partial x \partial t} \Psi = [\frac{i\hbar}{2m} \Psi^\star \frac{\partial^3 \Psi}{\partial x^3} + \frac{i\hbar}{2m} \Psi \frac{\partial^3 \Psi^\star}{\partial x^3}] - [\frac{i}{\hbar} V \Psi^\star \frac{\partial \Psi}{\partial x} + \frac{i}{\hbar} V \Psi \frac{\partial \Psi^\star}{\partial x}] - \frac{i}{\hbar} \frac{\partial V}{\partial x} |\Psi|^2 - \frac{i}{\hbar} \frac{\partial V}{\partial x} |\Psi|^2$$
We have to integrate this over the number line and so the bracketed expressions integrate to 0 using integration by parts and square integrability of $\Psi$. This simplifies the entire expression in (1) to:
$$2 \int_{-\infty}^{+\infty} |\Psi|^2 \frac{\partial V}{\partial x} dx$$
which is twice as much as we wanted to prove it was. Can somebody please tell me what I did wrong.
[Note: I have used Schwarz's theorem on equality of mixed derivatives and therefore have assumed the continuity of partial derivatives of $\Psi$]
Here's a faster way to prove Ehrenfest Theorem.
Set $\hbar = 2m = 1$. Let us multiple the Schrodinger equation by $\partial_x \Psi^\ast$ and integrate the expression with respect to $x$ yields \begin{align} \int^\infty_{-\infty}\frac{\partial \Psi^\ast(x, t)}{\partial x}\frac{\partial \Psi(x, t)}{\partial t}\ dx = i\int^\infty_{-\infty}\frac{\partial \Psi^\ast(x, t)}{\partial x}\frac{\partial^2 \Psi(x, t)}{\partial x^2}\ dx - i \int^\infty_{-\infty} \frac{\partial \Psi^\ast(x, t)}{\partial x}V(x, t) \Psi(x, t) \ dx \ \ \ (1) \end{align} with conjugate equation \begin{align} \int^\infty_{-\infty}\frac{\partial \Psi(x, t)}{\partial x}\frac{\partial \Psi^\ast(x, t)}{\partial t}\ dx =& -i\int^\infty_{-\infty}\frac{\partial \Psi(x, t)}{\partial x}\frac{\partial^2 \Psi^\ast(x, t)}{\partial x^2}\ dx\\ &+ i \int^\infty_{-\infty} \frac{\partial \Psi(x, t)}{\partial x}V(x, t) \Psi^\ast(x, t) \ dx. \ \ \ (2) \end{align} Using integration by parts, we see that the $(1)$ becomes \begin{align} \int^\infty_{-\infty} \Psi^\ast(x, t)\frac{\partial}{\partial t}\frac{\partial \Psi(x, t)}{\partial x}\ dx =&\ i\int^\infty_{-\infty}\frac{\partial^2 \Psi^\ast(x, t)}{\partial x^2}\frac{\partial \Psi(x, t)}{\partial x}\ dx \\ &- i \int^\infty_{-\infty} \Psi^\ast(x, t)\frac{\partial }{\partial x}[V(x, t) \Psi(x, t)] \ dx. \ \ \ (3) \end{align} Adding $(2)$ and $(3)$ yields \begin{align} \frac{d}{dt}\int^\infty_{-\infty} \Psi^\ast(x, t) \frac{\partial \Psi(x, t)}{\partial x}\ dx=&\ \int^\infty_{-\infty} \Psi^\ast(x, t)\frac{\partial}{\partial t}\frac{\partial \Psi(x, t)}{\partial x}+\frac{\partial \Psi(x, t)}{\partial x}\frac{\partial \Psi^\ast(x, t)}{\partial t} \ dx \\ =&\ -i \int^\infty_{-\infty} |\Psi(x, t)|^2\frac{\partial V(x, t)}{\partial x}\ dx. \ \ \ (4) \end{align} Now, multiply both side of $(4)$ by $i$ yields the desired identity.
Remark: You wrote \begin{align} \Psi^\star \frac{\partial^2 \Psi}{\partial x \partial t} - \frac{\partial^2 \Psi^\star}{\partial x \partial t} \Psi =&\ \left[\frac{i\hbar}{2m} \Psi^\star \frac{\partial^3 \Psi}{\partial x^3} + \frac{i\hbar}{2m} \Psi \frac{\partial^3 \Psi^\star}{\partial x^3}\right] \ \ \ (a) \\ &-\ \left[\frac{i}{\hbar} V \Psi^\star \frac{\partial \Psi}{\partial x} + \frac{i}{\hbar} V \Psi \frac{\partial \Psi^\star}{\partial x}\right]\ \ \ (b) \\ &-\ \frac{i}{\hbar} \frac{\partial V}{\partial x} |\Psi|^2 - \frac{i}{\hbar} \frac{\partial V}{\partial x} |\Psi|^2 \ \ \ (c) \end{align} but when you integrate the term $(b)$ you will see that $\partial_x$ will land on $V$ when performing an integration by parts. That term will cancel out your extra $|\Psi|^2 V$ term.