brownian motion - Partial derivative of an integral

Suppose I have a model for the short rate $r$ as ($W(t)$ is standard Brownian motion)

$r(t) = c+ \int_0^t \sigma (s) ^2 (t-s) ds+ \int_0^t \sigma (s) dW(s)$

I then want to find the dynamics of $r$, but how do I do that when the process itself contains integrals w.r.t Brownian motion? I get stuck when using Ito's formula and trying to calculate the integral $\frac{\partial}{\partial t} \int_0^t \sigma (s) dW(s)$

Answer

Let

$$X_t = \int_0^t \sigma(s) dW_s$$ denote a stochastic integral in the Itô sense. In that case one can write $r_t = f(t,X_t)$ where $$f:(t,x) \to c + \int_0^t \sigma^2(s)(t-s) ds + x \tag{1}$$ and use Itô's lemma to compute the differential $$dr_t = \partial_t f(t,X_t) dt + \partial_x f(t,X_t) dX_t + \partial_{xx} f(t,X_t) d\langle X \rangle_t$$ where from $(1)$ $$\begin{align} \partial_t f(t,X_t) &= \partial_t \int_0^t \sigma^2(s)(t-s) ds \\ &= \int_0^t \sigma^2(s) ds + 1 (\sigma^2(t)(t-t)) - 0 (\sigma^2(0)(0-s)) \\ &= \int_0^t \sigma^2(s) ds \end{align}$$ from Leibniz integral rule and $$\partial_x f(t,X_t) = 1,\quad \partial_{xx} f(t,X_t) = 0$$ along with, by definition of the Itô integral: $$dX_t = \sigma(t) dW_t,\quad d\langle X \rangle_t = \sigma^2(t) dt$$ such that finally: $$dr_t = \left(\int_0^t \sigma^2(s) ds\right) dt + \sigma(t) dW_t$$

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And indeed by integrating this last equation from $0$ to $t$ one gets:

$$r_t - r_0 = \int_0^t \left( \int_0^u \sigma^2(s) ds\right) du + \int_0^t \sigma(u) dW_u$$ and noting that $$\begin{align} \int_0^t \int_0^u \sigma^2(s) ds du &= \int_0^t \int_s^t \sigma^2(s) du ds\\ &= \int_0^t \sigma^2(s) (t-s) ds \end{align}$$ by Fubini theorem, one gets $$r_t = r_0 + \int_0^t \sigma^2(s) (t-s) ds + \int_0^t \sigma(s) dW_s$$