I am dealing the ADF test process, and find the limit distribution of t stats. The context is, $(y_t)$ is given by \begin{equation}\label{eq1} y_t = \mu + y_t^0 \end{equation} and $(y_t^0)$ is generated as $y_t^0 = \alpha y_{t-1}^0 + u_t$. $(u_t)$ is given by $\alpha(L)u_t = \varepsilon_t$ where $\alpha(z)=1-\alpha_1 z - \cdots - \alpha_p z^p$ and $(\varepsilon_t)\sim\mbox{WN}(0,\sigma^2)$.
The testing for the presence of a unit root in $(y_t^0)$ is based on the augmented regression \begin{equation}\label{adf} \hat{y}_t^0 = \alpha \hat{y}_{t-1}^0 + \sum_{k=1}^p \alpha_k \triangle \hat{y}_{t-k} + \varepsilon_t \end{equation} where $(\hat{y}_t^0)$ are the fitted OLS residuals. Let $\hat{\alpha}$ be the OLS estimator of $\alpha$ and $T_n$ the $t$-test for testing $\alpha=1$. Show that
$$T_n \rightarrow_d \frac{\displaystyle\int_0^1 \tilde{W}d\tilde{W}} {\displaystyle\left(\int_0^1 \tilde{W}^2\right)^{1/2}} $$ where $\tilde{W} = W - \int_0^1 W$.
My current approach is that the form of $y_t$ is just a AR(1) with constant, and we need to test the unit root, so I end up with this
\begin{align*}
T_{n}
&\rightarrow_{d} \frac{\frac{1}{2}(W(1)^{2} - 1) - W(1) \int W(r)dr}{\{\int_{0}^{1} W^{2}(r)dr - [\int_{0}^{1} W(r)dr]^{2}\}^{1/2}}.
\end{align*}
My question is mainly in $\displaystyle\int_0^1 \tilde{W}d\tilde{W}$, as I could match the denominator. It is $d\tilde W$ and the integral is not normal one, I can only match it with my solution when I consider $d\tilde W$ = $dW$, but could it? Basically I may say $d\tilde W = (W-\int W)'dW = dW$, but the range of integral may change, that's where I'm confused. So is my solution wrong? or sth else, thank you !