With ${\cal T^{-1}}$ being reverse time ordering and $a(t)$ being any operator (or matrix):
Is ${\cal T^{-1}}\exp\left[\int_0^ta^{\dagger}(t')dt'\right]{\cal T}\exp\left[-\int_0^ta(t')dt'\right]$ equal to
${\cal T}\exp\left[\int_0^ta^{\dagger}(t')-a(t')dt'\right]$? (Time ordering disappear or not? It is wrong if $[a^{\dagger}(t),a(t)] = 0$ is not satisfied?)
How can I simplify it further?
I know that \begin{align} {\cal T}\exp\left[-\int_0^ta(t')dt'\right] &= I - \int_{0}^{t} dt' a(t') + \left(-\right)^2 \frac{1}{2} \mathcal{T}\left(\int_{0}^{t} dt' a(t')\right)^2 \\& \quad + \left(-\right)^3 \mathcal{T}\left(\frac{1}{3!}\left(\int_{0}^{t} dt' a(t')\right)^3\right) \cdots \\ & = I - \int_{0}^{t}dt' a(t') + \left(-\right)^2 \int_{0}^{t}dt' \int_{0}^{t'}dt'' a(t') a(t'') \\ & \quad + \left(-\right)^3 \int_{0}^{t}dt' \int_{0}^{t'}dt'' \int_{0}^{t''}dt''' a(t') a(t'') a(t''') +\cdots \end{align}
It involves many terms and I do not see them cancel out when calculating the product.