I am trying to prove that the normalization constant is independent of time. If we have fixed it for a particular time then it will remain constant for all time.
Suppose $\psi(x,t)$ is a wavefunction.
Let $A(t)$ be the normalization constant of $\psi(x,t)$
Then $\displaystyle A^*A\int_{-\infty}^{\infty}|\psi(x,t)|^2dx=1\tag{1}$
$\displaystyle \implies \frac{d}{dt}A^*A\int_{-\infty}^{\infty}|\psi(x,t)|^2dx=0\tag{2}$
$\displaystyle \implies\int_{-\infty}^{\infty}|\psi(x,t)|^2dx\frac{d}{dt}A^*A+A^*A\frac{d}{dt}\int_{-\infty}^{\infty}|\psi(x,t)|^2dx=0\tag{3}$
Now first analyze, $\displaystyle \frac{d}{dt}\int_{-\infty}^{\infty}|\psi(x,t)|^2dx$
$\displaystyle \frac{d}{dt}\int_{-\infty}^{\infty}|\psi(x,t)|^2dx=\int_{-\infty}^{\infty}\frac{\partial}{\partial t}|\psi(x,t)|^2dx\tag{4}$
$\displaystyle \frac{\partial}{\partial t}|\psi(x,t)|^2dx=\frac{\partial}{\partial t}(\psi^*\psi)dx=\psi^*\frac{\partial\psi}{\partial t}=\psi\frac{\partial\psi^*}{\partial t}\tag{5}$
By time dependent Schrodinger equation
$\displaystyle \frac{\partial\psi}{\partial t}=\frac{i\bar{h}}{2m}\frac{\partial^2\psi}{\partial x^2}-\frac{i}{\bar{h}}V\psi\tag{6}$
Also, $\displaystyle \frac{\partial\psi^*}{\partial t}=-\frac{i\bar{h}}{2m}\frac{\partial^2\psi^*}{\partial x^2}+\frac{i}{\bar{h}}V\psi^*\tag{7}$
So, $\displaystyle \frac{\partial|\psi|^2}{\partial t}=\psi^*\Big(\frac{i\bar{h}}{2m}\frac{\partial^2\psi}{\partial x^2}-\frac{i}{\bar{h}}V\psi\Big)+\psi\Big(-\frac{i\bar{h}}{2m}\frac{\partial^2\psi^*}{\partial x^2}+\frac{i}{\bar{h}}V\psi^*\Big)\tag{8}$
$\displaystyle \implies\frac{\partial|\psi|^2}{\partial t}=\frac{i\bar{h}}{2m}\Big(\frac{\partial^2\psi}{\partial x^2}-\frac{\partial^2\psi^*}{\partial x^2}\Big)\tag{9}$
$\displaystyle \implies\frac{\partial|\psi|^2}{\partial t}=\frac{\partial}{\partial x}\Big[\frac{i\bar{h}}{2m}\Big(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\Big)\Big]\tag{10}$
$\displaystyle \frac{d}{dt}\int_{-\infty}^{\infty}|\psi(x,t)|^2dx=\int_{-\infty}^{\infty}\frac{\partial}{\partial x}\Big[\frac{i\bar{h}}{2m}\Big(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\Big)\Big]dx\tag{11}$
$\displaystyle \implies\frac{d}{dt}\int_{-\infty}^{\infty}|\psi(x,t)|^2dx=\frac{i\bar{h}}{2m}\Big(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\Big)\Big|_{-\infty}^\infty\tag{12}$
As $\displaystyle \psi(x,t)\to0$ as $x\to\pm\infty$.
So, $\displaystyle \frac{d}{dt}\int_{-\infty}^{\infty}|\psi(x,t)|^2dx=0\tag{13}$
So, $(3)$ becomes
$\displaystyle \implies\int_{-\infty}^{\infty}|\psi(x,t)|^2dx\frac{d}{dt}A^*A=0\tag{14}$
As $\psi$ is square integrable, so $\int_{-\infty}^{\infty}|\psi(x,t)|^2dx=c$ where $c\in\mathbb R$ and $c\neq 0$
So, $\frac{d}{dt}A^*A=0\tag{15}$
$\implies |A|^2=constant\tag{16}$
I have the following doubts from the proof
(i) From $(11)$ to $(12)$, in the RHS, we have used fundamental theorem of calculus.
Ingtegrating of the derivative is the antiderivative. $\int_a^bf'(x)dx=f(x)$. But here the condition is that f has to continuous and differentiable on $[a,b]$ with $f'$ integrable on $[a,b]$.
So, in $\int_{-\infty}^{\infty}\frac{\partial}{\partial x}\Big[\frac{i\bar{h}}{2m}\Big(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\Big)\Big]dx$,
we take $\Big(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\Big)$ to be continuous. As $\psi$ and $\psi^*$ is continuous, this means that $\frac{\partial\psi}{\partial x}$ is also continuous.
But we know in general that first derivative of $\psi$ can be discontinuous also. So how we have used fundamental theorem of calculus here?
(ii) From $(12)$ to $(13)$, we have taken $\frac{i\bar{h}}{2m}\Big(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\Big)\Big|_{-\infty}^\infty=0$ using the fact that $\psi\to 0$ as $x\to\pm\infty$. But this also means that $\frac{\partial\psi}{\partial x}\to 0$ as $x\to\pm\infty$. But how we can be sure that $\frac{\partial\psi}{\partial x}$ is bounded?
(iii) In $(16)$, we get $|A|^2=constant$. But from this how we get $A(t)=constant$.
$|A|^2$ constant means that magnitude of the vector in complex plane is constant but it might happen that the angle of $A$ changes. This angle changes as a function of $t$. So how we conclude that $A$ is independent of time?