This is a question from Exercise 5.10 of Arbitrage Theory in Continuous Time (2009) by Tomas Bjork (rest in peace). The problem states that,
Consider the following boundary value problem in the domain $[0,T]×R$: $$\frac{\partial F}{\partial t} + \mu \frac{\partial F}{\partial x} + \frac{1}{2}\sigma^2(t,x)\frac{\partial^2 F}{\partial x^2} + k(t,x) = 0$$ $$F(T,x) = \Phi(x)$$
Here $\mu, \sigma, k, \Phi$ are assumed to be known functions.
Prove that this problem has the following stochastic representation formula $$F(t,x) = E^{t,x}[\Phi(X(T))] + \int_t^T E^{t,x}[k(s, X(s))]ds$$
where as usual $X$ has the dynamics
$$dX(s) = \mu(s, X(s))ds + \sigma(s, X(s))dW(s)$$ $$X(t) = x$$
The problem also gives a hint: Define $X$ as above, assuem that F actually solves the PDE and consider the process $Z(s) = F(s, X(s))$.
Now, using the hint, one could easily derive the stochastic representation formula from the PDE. However, I recently read the proof of Feynman-Kac Theorem. The proof uses the Markov property $E^{t,X(t)}[\Phi(X(T))] = E[\Phi(X(T)) | \mathcal{F}(t)]$ and thus $\Phi(X(T))$ is a Doob Martingale, then applies Ito's formula on it and let coefficient of $dt$ be zero to derive $g_t+\beta g_x + \frac{1}{2}\gamma^2 g_{xx} = 0$, where $g(t,x) = E^{t,x}[\Phi(X(T))]$. So I wanted to use the same method here and solve the problem the other way around: deriving the SDE from the stochastic representation formula.
So my sketch of proof as follows: Assume $F(t, x)$ follows the stochastic representation formula, then we know that $F(t,x) - \int_t^T E^{t,x}[k(s, X(s))]ds$ is a Doob Martingle. We thus apply Ito's formula and let coefficient of $dt$ be zero.
However, this would involve $\frac{d}{dx}\int_t^T E^{t,x}[k(s, X(s))]ds$ and $\frac{d^2}{dx^2}\int_t^T E^{t,x}[k(s, X(s))]ds$ and I need to prove that they are both zero. I have no idea how to calculate these derivatives, or to prove that they are zero. Can I just say that this is an integral of $ds$, so its derivative w r.t. $dx$ is 0?