I have trouble to understand how to understand the equality between the 2 terms equation (1)
\begin{equation} f: \mathbb{R^+} \times \mathbb{R} \rightarrow \mathbb{R} \end{equation} \begin{equation} f \in \mathbb{C}^{1,1} \end{equation}
(1)
\begin{equation} d(\int_{0}^{t} \frac{\partial}{\partial s} f(s,x) ds) = \frac{\partial}{\partial t} f(t,x) dt \end{equation}
This result comes from the two differents way to present Ito's Lemma:
\begin{equation} f(t,X_t) = f(0,X_0) + \int_{0}^{t} \frac{\partial f}{\partial s}(s,X_s) ds + \int_{0}^{t} \frac{\partial f}{\partial x}(s,X_s) dX_s + \frac{1}{2} \int_{0}^{t} \frac{\partial^2 f}{\partial x^2}(s,X_s) d[X]_s \end{equation}
Which can be differentiate in:
\begin{equation} df(t,X_t) = \frac{\partial f}{\partial t}(t,X_t) dt + \frac{\partial f}{\partial x}(t,X_t) dX_t + \frac{1}{2} \frac{\partial^2 f}{\partial x^2}(t,X_t) d[X]_t \end{equation}
I find a different result for equation (1) because I have always learnt that:
\begin{equation} df(x,y) = \frac{\partial f}{\partial x}(x,y) dx + \frac{\partial f}{\partial y}(x,y) dy \end{equation}
Applying this to (1) I find:
\begin{equation} d(\int_{0}^{t} \frac{\partial}{\partial s} f(s,x) ds) = \frac{\partial}{\partial t} f(t,x) dt \end{equation} $\iff$ \begin{equation} d(\int_{0}^{t} \frac{\partial}{\partial s} f(s,x) ds) = d( f(t,x) - f(0,x)) \end{equation}
Admitting that $f(0,x) = 0$ we obtain:
\begin{equation} df(t,x) = \frac{\partial f}{\partial t}(t,x) dt + \frac{\partial f}{\partial x}(t,x) dx \end{equation}
Which is different to equation (1). I don't understand why we seem to be able to remove \begin{equation} \frac{\partial f}{\partial x}(t,x) dx \end{equation}
I feel like this is something basic but I'm really struggling to understand, can anyone help?
Thank you a lot.
The ant-derivative, with respect to $s$, of $\frac{\partial}{\partial s}f(x,s)$ is $f(s,x)$. Thus the integral is $f(t,x)-f(0,x)$ . $$\frac{\partial}{\partial t}(f(t,x)-f(0,x))=\frac{\partial}{\partial t}(f(t,x))$$ $$\frac{\partial}{\partial x}(f(t,x)-f(0,x))=\frac{\partial}{\partial x}(f(t,x))-\frac{\partial}{\partial t}(f(0,t))$$