Let us consider the following partial differential equation for $f = f(x,v,t),$ with $x \in \mathbb{T}^1, v \in \mathbb{R}$ and $t \in [0, \tau]$:$$ \partial_t f(x,v,t) + v \cdot \partial_x f (x,v,t) = \rho [f] (x,t) \cdot M[T(x,t)] - f(x,v,t) $$ and positive initial condition$ f(x,v,0) = f_0(x,v) \in L^1(\mathbb{T}^1 \times \mathbb{R}).$$\mathbb{T}^1$ is the one-dimensional flat torus, represented by $[0,1]$ and periodic boundary conditions. $T(x,t) > 0$ is a given continuous function and $M[T(x,t)] > 0$ denotes the Maxwellian, defined by $ M[T] = \frac{1}{\sqrt{\pi T}} \exp \left(- \frac{v^2}{T} \right)$ fulfilling $ \int_{\mathbb{R}} M [T] (v) \ dv = 1$ and $\rho[f] (x,t):= \int_{\mathbb{R}} f(x,v,t) \ dv.$
The mild solution is given by\begin{equation} f_{}(x,v,t) = \mathrm{e}^{-t}f_0(x-vt,v) + \int_{0}^{t} \mathrm{e}^{s-t} \rho_{}(x-v(t-s),s) M_{}[T_{}(x-v(t-s),s)] (v) \, \mathrm{d}s.\end{equation}
It is known that there exists a (unique) mild solution $f \in L^{\infty}((0, \tau); L^1(\mathbb{T}^1 \times \mathbb{R}))$ which is absolutely continuous with respect to $t,$ hence differentiable almost everywhere with respect to $t$ and no further regularity.
I would like to show that$$ \frac{d}{dt} \int_{\mathbb{T}^1} \int_{\mathbb{R}} f(x,v,t) \ dvdx = 0 $$ by just using the mild formulation.
My idea: Interchanging derivative and integral and first calculating the derivative would give using Leibniz-Rule\begin{align*}\partial_t f(x,v,t) = -\mathrm{e}^{-t}f_0(x-vt,v) - \mathrm{e}^{-t} v \cdot \partial_x f_0(x-vt,v) + \rho [f] (x,t) M[T(x,t)] (v) + \int_{0}^{t} \partial_t \left(\mathrm{e}^{s-t} \rho_{}(x-v(t-s),s) M_{}[T_{}(x-v(t-s),s)] (v) \right) \ \mathrm{d}s \\= -\mathrm{e}^{-t}f_0(x-vt,v) - \mathrm{e}^{-t} v \cdot \partial_x f_0(x-vt,v) + \rho [f] (x,t) M[T(x,t)] (v) - \int_{0}^{t} \left(\mathrm{e}^{s-t} \rho_{}(x-v(t-s),s) M_{}[T_{}(x-v(t-s),s)] (v) \right) \ \mathrm{d}s.\end{align*}Then integration out $x$ and $v$ gives:\begin{align*}\int_{\mathbb{T}^1} \int_{\mathbb{R}}\partial_t f(x,v,t) \ dvdx= -\mathrm{e}^{-t} \int_{\mathbb{T}^1} \int_{\mathbb{R}}f_0(x-vt,v) \ dv dx + \int_{\mathbb{T}^1} \rho [f] (x,t) \ dx \\ - \int_{0}^{t} \int_{\mathbb{T}^1} \int_{\mathbb{R}} \left(\mathrm{e}^{s-t} \rho_{}(x-v(t-s),s) M_{}[T_{}(x-v(t-s),s)] (v) \right) \ dv dx ds.\end{align*}After interchanging the order of the integrals and changing the variables in the last expression of the r.h.s, I end up with\begin{align*}\int_{\mathbb{T}^1} \int_{\mathbb{R}}\partial_t f(x,v,t) \ dvdx= -\mathrm{e}^{-t} \int_{\mathbb{T}^1} \int_{\mathbb{R}}f_0(x,v) \ dv dx + \int_{\mathbb{T}^1} \rho [f] (x,t) \ dx \\ - \int_{0}^{t} \int_{\mathbb{T}^1} \left(\mathrm{e}^{s-t} \rho_{}(x,s) \right) \ dx ds.\end{align*}Unfortunately I can't get any further here. Where is my mistake? Or is there another more elegant way? Thanks for help in advance!