Hi there! I state that I'm an electronic engineer (undergraduate), then the my knowledges about the world of economics are almost null. A colleague asked to me an help about one point of the proof of the theorem 1 in this paper. The critical point is how to obtain [latex]\text{ d}Y_t [/latex] that at first glance it seemed to me enough trivial. Unfortunately I get a wrong result, so I ask if is possible, without using advanced tools like stochastic calculus, get the solution, and in that case, how to get it. I post my (absolutely not rigorous) calculations.
[latex]
\begin{aligned}
\text{d}Y_t &= \lim_{\tau \to 0} \,\, [ Y_{t+\tau}-Y_{t} ] \\ &= \lim_{\tau \to 0} \left[- \int_{t+\tau}^s f(t+\tau,u) \text{ d}u -\left( - \int_t^s f(t,u) \text{ d}u\right) \right] \\ &=\lim_{\tau \to 0} \left[- \left( \int_t^s f(t+\tau,u) \text{ d}u - \int_t^{t+\tau} f(t+\tau,u) \text{ d}u \right) -\left( - \int_t^s f(t,u) \text{ d}u\right) \right] \\
&=\lim_{\tau \to 0} \left[ \int_t^{t+\tau} f(t+\tau,u) \text{ d}u -\left( \int_t^s f(t+\tau,u)-f(t,u) \text{ d}u\right) \right] \\
&=\lim_{\tau \to 0} \int_t^{t+\tau} f(t+\tau,u) \text{ d}u - \int_t^s \lim_{\tau \to 0} \, [f(t+\tau,u)-f(t,u)] \text{ d}u \\ &=f(t,s)\text{ dt}- \int_t^s \frac{\partial f}{\partial t} (t,u) \text{ d}t \text{ du} \end{aligned}
[/latex]
The question is: why it appear the total differential of [latex] f(t,u) [/latex] under the sign of integral? Thank you in advance and sorry for my English.