On the application of Duhamel's formula The Duhamel's formula that I know of reads as follows:
$$ \frac{\mathrm d}{\mathrm dt} e^{A(t)} = \int_0^1 e^{sA} \dot A e^{(1-s)A} \mathrm ds \,,$$
where $A$ is a time-dependent operator.
I cannot for the life of me figure out how this translates to the following statement:
$$ e^{-\beta(H+V)}-e^{-\beta H} = -\int_0^\beta e^{-s(H+V)}V e^{(s-\beta) H} \mathrm ds \,,$$
where $0 \le \beta \lt \infty$ and $H$ and $V$ are operators as before.
Kindly help. This is a step in the proof of the structural stability of the Gibbs state with respect to local perturbations. 
 A: Lets assume $H,V$ are bounded operators. I think you don't want this in physics, since here the spectrum of the Hamiltonian is in general only bounded from below.
Calculate with the product rule:
$$\frac{d}{dt} e^{-t(H+V)}e^{t H} = - e^{-t(H+V)} Ve^{tH}\tag{1}$$
Integrate to get:
$$e^{-t(H+V)}e^{tH}-\Bbb1 = -\int_0^t ds\, e^{-s(H+V)}Ve^{sH}$$
which then gives
$$e^{-t(H+V)}-e^{-tH} = -\int_0^t ds\,e^{-s(H+V)}Ve^{(s-t)H}\tag{2}$$
In the case that $H$ is self-adjoint but not necessarily bounded $e^{tH}$ is another self-adjoint densly defined operator. If $H$ is bounded from below and we call the domain of $e^{tH}$ $D(t)$, then you have $D(t)\subset D(t')$ if $t≥t'$. Since $H$ is bounded from below (and lets assume $V$ is a bounded operator) then $e^{-t(H+V)}$ is a bounded operator, so it has domain the entire Hilbert space. This means that if you rewrite $(1)$ as
$$\frac{d}{dt}e^{-t(H+V)}e^{tH} \psi =- e^{-t(H+V)}Ve^{tH}\psi$$
where $\psi\in D(t)$ you can integrate etc (since $\psi\in D(s)$ for all $s$ in the integral) and you find that $(2)$ holds pointwise on a dense subspace. But all operators appearing in $(2)$ are continuous since $H$ is bounded from below, so it holds on the entire Hilbert space.
Edit: I think the integral in $(2)$ must only converge in the strong operator topology in the unbounded $H$ case, but I am not sure.
