Show that if the Hamiltonian depends on time and $[H_{t1}, H_{t2}] = 0$, the time development operator is given by:
$$U(t) = exp [ \frac{-i}{\hbar} \int_{0}^{t} dt' H(t')]$$
I am having trouble to prove it:
Since $U(dt) = 1 - \frac{iHdt}{\hbar}$,
$U(t) = U(dt1)U(dt2)U(dt3)U(dt4)U(dt5)...$
$U(t) = 1 - \sum_{i = 0}^{N}\frac{iH(t_{i}) dt_{i}}{\hbar}$
The sum can be interpreted as an integral
$U(t) = 1 - \int_{i = 0}^{N}\frac{iH(t_{i}) dt_{i}}{\hbar}$
$U(t) = 1 + \frac{-i}{\hbar} \int_{i = 0}^{N}{H(t') dt'}$
But this is wrong, where is the error (i really think that the error is going from the sum to the integral, but i don't know how to fix it)?
The problem is when you wrote
$$U(t) = 1 - \sum_{i = 0}^{N}\frac{iH(t_{i}) dt_{i}}{\hbar}$$
Instead you should have ($\hbar=1$)
$$ U(t)=\prod_j(1-iH_jdt_j)$$
Which is useful if $H$ is time independent. Since here $H=H(t)$, we try a different approach. Since you know the infinitesimal form
$$ U(dt)=1-iHdt$$
You can find the equation obeyed by $U(t)$ as follows
$$ U(t+dt)=U(dt)U(t)=(1-iHdt)U(t)$$
$$ U(t+dt)-U(t)=-iHdtU(t)$$
$$ \frac{U(t+dt)-U(t)}{dt}=-iHU(t)$$
The left is simply a derivative when $dt \rightarrow0$
$$ \partial_t U(t)=-iH(t)U(t)$$
Since we are told that $[H(t_1),\ H(t_2)]=0$, everything in the above differential equation commutes, and we may integrate directly. Separating 'variables', we have
$$ U^{-1}\partial_tU=-iH$$
The left hand side is a logarithmic derivative, so integrating is simple
$$ \ln(U(t))=-i\int^t_0 dt' \ H(t')$$
Where we have used $U(0)=1$. Exponentiating both sides yields the answer
$$ U(t)=\exp\left( -i\int_0^t dt' \ H(t')\right) \tag*{$\blacksquare$}$$