Question about obtaining Ito differential of stochastic integral How do you obtain the Ito differential of $X_t = \int_0^t W_s ds$, $Y_t = t\exp{(\int_{0}^{t} W_s ds)}$, and $Z_t = t\exp{(\int_{0}^{t} W_s dW_s)}$? I'm not sure how to even go about applying Ito's lemma to stochastic integrals - any help would be appreciated!
I might be getting confused on notation, but is it the case that if $X_t = \int_{0}^{t} W_s ds$, then $dX_t = W_t dt$ and if $Y_t = \int_{0}^{t} W_s dW_s$, then $dY_t = W_t dW_t$? I have also seen that $\int_0^t W_s ds = tW_t -\int_0^t s dW_s = \int_0^t (t-s) dW_s$ and $\int_0^t W_s d W_s = \frac{1}{2} (W_t^2 - t)$. However, I'm not sure if/how to incorporate this.
 A: I think you have to recall that an equality of Itô differentials actually means after it's Itô integration.
And you remind that $\int_0^t dA_s = A_t - A_0$ for any Itô process $A_t$.
So, by the definition of the Itô differential and the definition of $X_t := \int_0^tW_sds$, $dX_t = W_tdt$.
First one can easily verify that $(dX_t)^2 = 0, dtd(\exp(X_t)) = 0$.
Hence $d\exp(X_t) = \exp(X_t)dX_t$, and $d(t\exp(X_t)) = \exp(X_t)dt + td\exp(X_t)$.
Since $Y_t = t\exp(X_t)$, we obtain
$$
\begin{align*}
dY_t &= d(t\exp(X_t)) \\
&= \exp(X_t)dt + td\exp(X_t) \\
&= \exp(X_t)dt + t\exp(X_t)dX_t \\
&= \exp(\int_0^tW_sds)(t+W_t)dt. 
\end{align*}
$$
Similarly, we can compute $dZ_t$ as follows:
As you computed, $A_t : = \int_0^t W_sdW_s = \frac{1}{2}(W_t^2 - t)$.
Then $dA_t = W_t dW_t, (dA_t)^2 = W_t^2 dt$.
Hence, by Itô's formula,
$$
d(\exp(A_t)) = \exp(A_t)dA_t + \frac{1}{2}\exp(A_t)(dA_t)^2 
= \exp(A_t)W_tdW_t + \frac{1}{2}\exp(A_t)W_t^2dt.
$$
Thus
$$
\begin{align*}
dZ_t &= d(t\exp(A_t)) \\
&= \exp(A_t)dt + td(\exp(A_t)) \\
&= \exp(A_t)dt + t\exp(A_t)W_tdW_t + \frac{1}{2}\exp(A_t)W_t^2dt\\
&= t\exp(A_t)W_tdW_t + \exp(A_t)(\frac{1}{2}W_t^2+1)dt \\
&= t\exp(\frac{1}{2}(W_t^2 - t))W_tdW_t + \exp(\frac{1}{2}(W_t^2 - t))(\frac{1}{2}W_t^2+1)dt
\end{align*}
$$
I think no further calculations are necessary.
