Operator integral Consider the following integral:
\begin{equation}
L =2 \int_{0}^{\infty} d t \exp\left\{-\hat{\rho}_{\lambda} t\right\} \partial_{\lambda} \hat{\rho}_{\lambda} \exp\left\{-\hat{\rho}_{\lambda} t\right\}
\end{equation}
where
\begin{equation}
\partial_\lambda \hat{\rho}_{\lambda}=\mathrm{i}\left[a^{\dagger}+a, \hat{\rho}_{\lambda}\right]
\end{equation}
with $a$ being the annihilator operator.
I cannot use the truncated Baker-Campbell-Haussdorf formula since the commutator is not a number. Is there anything I could do to evaluate this integral?
Any remark/tips are appreciated.
 A: It is frankly inconceivable the sign of the second exponential's exponent is not +, instead of -, so I'll switch it, by executive editorial action,
\begin{equation}
L =2 i\int_{0}^{\infty} d t \exp\left\{-\hat{\rho}_{\lambda} t\right\}     \left[a^{\dagger}+a, \hat{\rho}_{\lambda}\right]      \exp\left\{\hat{\rho}_{\lambda} t\right\}.
\end{equation}
To  bypass excessive notational clutter, the source of all mistakes, I'll also define
$ \hat{\rho}_{\lambda} =A$, and $  a^{\dagger}+a =B$, so that
$$
L =-2 i\int_{0}^{\infty}\!\! d t ~~\left ( e^{-t[A,}   [A,   \right )B    ~~~,
$$
where I am using the notation $[A, =\operatorname{ad}_A$ $\leadsto  \operatorname{ad}_A B= [A,B]$ which many physicists don't know, but is detailed in WP. Many call this the Hadamard formula.
The integrand already involves an antiderivative, so, formally (as a power series expansion), the integral is undone and the boundary terms are
$$
L =2 i \left ( e^{-\infty[A,}  -I   \right   )B    ~~~.
$$
Assuming every nested commutator $[A,...[A,B]]]...]$ is a positive operator, (your issue), the first term vanishes and you get $L=-2iB$. I doubt your problem is that pat, but, anyway, you might glean tricks from here.

Of course, regardless of the integral, your second relation
$$
\partial_\lambda A(\lambda) = i[B,A(\lambda)] ,
$$
has the customary ready solution for $\lambda$-independent B, (check it!),
$$
A(\lambda) = e^{i\lambda [B, } A(0)\equiv e^{i\lambda B} A(0) e^{-i\lambda B} .
$$
Are you positive you did not reverse symbols someplace?
