For a standard Wiener Process/Brownian Motion, $W$, for the usual integrals $\int_0^t\sigma(u)dW(u)$ and $\int_0^tW(u)du$, I know how to manipulate them using Ito's Lemma/normal calculus rues like Leibniz to calculate the derivatives/differential forms.
One of my confusions is regarding what taking a differential means when applied to a time integral of an Ito process. I know that $dX_t$ for some Ito Process $X_t$ is a heuristic form, but when using normal calculus rules, I can't seem to figure out what an expression like $d\int_0^tW(u)du$ is 'doing'. $\frac{d}{dt}\int_0^tW(u)du$ is something I can calculate using Leibniz, and this leads to $W(t)$ and so in that sense if you multiply both sides by $dt$, $d\int_0^tW(u)du = W(t)dt$. But is it possible to get to that point without involving the time variable at all?
And what about the case where the inner stochastic process depends on $t$? For example, the following integral: $\int_0^tL(u,t)du$, where $L$ is an Ito process: $dL = \mu(u,t)dt + \sigma(u,t)dW_t$, I was trying to calculate $d\int_0^tL(u,t)du$, but could not figure out how to proceed after applying Leibniz's Integral rule in the sense of calculating $\frac{d}{dt}\int_0^tL(u,t)du$. I have the first term as $L(t,t)$, ($L(t,t)dt$ if only the differential is used), but the second term looks to be $\int_0^t\frac{d}{dt}L(u,t)du$, which I am not sure how to proceed with since time derivatives of stochastic processes are something I am unfamiliar with.
The answer (Not including any terms involving the brownian motion, $W$) according to the book I am reading is supposed to be $\frac{d}{dt}\int_0^tL(u,t)du = L(t,t) + \int_0^t\mu(s,t)ds$ where $\mu$ is $L$'s drift term. But I have no idea how they reached that point.
$\def\si{\sigma}$ $\def\th{\theta}$
Let $M(u,t)=\int_0^t\si(u,r)\,dW_r$ and assume the sample paths of $\si$ have continuous partial derivatives. In this answer, I will repeatedly use the fact that if $\th$ is a bounded variation process, then $$ \int_0^t\th_r\,dW_r=\th_tW_t-\int_0^tW_r\,d\th_r.\tag{IBP} $$
Proof. Fix $s$. Let $Y_t=\int_0^t\int_0^s\si(u,r)\,du\,dW_r$. By (IBP), we have $$ Y_t = W_t\int_0^s\si(u,t)\,du - \int_0^tW_r\int_0^s\si_2(u,r)\,du\,dr, $$ where $\si_2$ is the partial derivative of $\si$ with respect to the second argument. For each sample path, the above integrals are ordinary integrals. Thus, by Fubini's theorem, \begin{align} Y_t &= W_t\int_0^s\si(u,t)\,du - \int_0^s\int_0^tW_r\si_2(u,r)\,dr\,du\tag{1}\\ &= \int_0^s\left({W_t\si(u,t) - \int_0^tW_r\si_2(u,r)\,dr}\right)\,du. \end{align} By (IBP), $$ \int_0^t W_r\si_2(u,r)\,dr = W_t\si(u,t) - \int_0^t \si(u,r)\,dW_r.\tag{2} $$ Thus, $Y_t=\int_0^s\int_0^t \si(u,r)\,dW_r\,du$. $\square$
Let $X_t=\int_0^t M(u,t)\,du$. By (1) with $s=t$, $$ X_t = W_t\int_0^t\si(u,t)\,du - \int_0^t\int_0^tW_r\si_2(u,r)\,dr\,du. $$ Thus, again using (IBP), \begin{align} dX_t &= \int_0^t\si(u,t)\,du\,dW_t + W_t\left({ \si(t,t) + \int_0^t\si_2(u,t)\,du }\right)\,dt\\ &\qquad - \left({ \int_0^tW_r\si_2(t,r)\,dr + \int_0^tW_t\si_2(u,t)\,du }\right)\,dt\\ &= \int_0^t\si(u,t)\,du\,dW_t + \left({ W_t\si(t,t) - \int_0^tW_r\si_2(t,r)\,dr }\right)\,dt. \end{align} By (2) with $u=t$, this gives \begin{align} dX_t &= \int_0^t\si(u,t)\,du\,dW_t + \int_0^t \si(t,r)\,dW_r\,dt\\ &= \int_0^t\si(u,t)\,du\,dW_t + M(t,t)\,dt. \end{align} Finally, let \begin{align} L(u,t) &= L(u,0) + \int_0^t\mu(u,r)\,dr + \int_0^t\si(u,r)\,dW_r\\ &= L(u,0) + \int_0^t\mu(u,r)\,dr + M(u,t). \end{align} Then \begin{align} \int_0^t L(u,t)\,du &= \int_0^t L(u,0)\,du + \int_0^t\int_0^t\mu(u,r)\,dr\,du + \int_0^t M(u,t)\,du\\ &= \int_0^t L(u,0)\,du + \int_0^t\int_0^t\mu(u,r)\,dr\,du + X_t. \end{align} Therefore, \begin{align} d\left({\int_0^t L(u,t)\,du}\right) &= L(t,0)\,dt + \int_0^t\mu(t,r)\,dr\,dt + \int_0^t\mu(u,t)\,du\,dt\\ &\qquad + \int_0^t\si(u,t)\,du\,dW_t + M(t,t)\,dt\\ &= L(t,t)\,dt + \int_0^t\mu(u,t)\,du\,dt + \int_0^t\si(u,t)\,du\,dW_t. \end{align} This formula has an extra $dW$ term which your formula does not. In hindsight, this seems reasonable. By (1), $\int_0^s M(u,t)\,du$ is a process such that if we fix $t$, then it is bounded variation in $s$, but if we fix $s$, then it is quadratic variation in $t$. Your formula implies that this process is bounded variation on the diagonal $s=t$, which seems fairly counterintuitive. (Edit: I just noticed the phrase, "not including any terms involving the brownian motion", so I guess the formula you posted at the end was intentionally incomplete.)
Edit 2: If we interpret expressions such as $\int_0^t f(u,t)\,dZ_t\,du$ as meaning $\left({\int_0^t f(u,t)\,du}\right)\,dZ_t$, then we have just proved that under suitable assumptions on $\mu$ and $\si$, $$ d\left({\int_0^t L(u,t)\,du}\right) = L(t,t)\,dt + \int_0^t dL(u,t)\,du. $$