I have:
$dS_t=S_t\sigma_tdW_t$
$J_T=\frac{1}{T}\int_{0}^{T}\sigma_t^2dt$
And I need to prove that
$J_T=-\frac{2}{T}\ln(S_T/S_0)+\frac{2}{T}\int_0^T\sigma_tdW_t$
I know I need to use Itô's lemma but I don't really understand how. Here is what I did: $d\ln(S_t)=0dt+\frac{1}{S_t}dS_t-\frac{1}{2}\sigma_t^2dt$
$d\ln(S_t)=\sigma_tdW_t-\frac{1}{2}\sigma_t^2dt$
Integrate between $0$ and $T$:
$\int_0^T d\ln(S_t)=ln(S_T/S_0)=\int_0^T\sigma_tdW_t-\frac{1}{2}\int_0^T\sigma_t^2dt$1
and it is straightforward
You just integrate the SDE you got for $\log(S_t)$ from $0$ to $T$, giving $$\log(S_T/S_0) =\int_0^T\sigma_t dW_t - \int_0^T \frac{1}{2} \sigma^2_t dt $$ and then express the second term on the RHS in terms of $J_T$ and rearrange.