Assume we have a process given by $$ dY_t = - \frac{\beta^2}{2}dt + \beta dB_t $$ with $\beta \in \mathbb{R}$ and $(B_t)_{t\geq 0}$ being an $\mathcal{F}_t$-Brownian motion. We can rewrite the SDE in integral form as $$ Y_t =y_0 - \int_0^t \frac{\beta^2}{2} dt + \beta \int_0^t dB_t, \quad x_0 \in \mathbb{R} $$
If we use Itô's formula for the process $t\cdot Y_t$ we get that $$ \begin{aligned} t Y_t &= \int_0^t Y_s ds + \int_0^t s dY_s + [Y,t]_t \newline &= \int_0^t Y_s ds - \frac{1}{2}\beta^2 \int_0^t s ds + \beta \int_0^t s dB_s \end{aligned} $$ And if we rewrite this we get the expression $$ \int_0^t Y_s ds = t \cdot Y_t + \frac{1}{2}\beta^2 \int_0^t s ds - \beta\int_0^t s dB_s $$ How can I rewrite the last part to get $$ \int_t^T Y_s ds = Y_t (T-t) - \frac{1}{2}\beta^2 \int_t^T (T-s) ds + \beta \int_t^T (T-s) dB_s $$ I cannot understand how we would ever get a term with $(T-t)Y(t)$. Any help would be appreciated.
Hopefully the question is correct now. We can rewrite and get $$ \int_0^t Y_s ds = t Y_t + \frac{1}{2}\beta^2 \int_0^t (T-s) ds - \beta \int_0^t (T-s) dW_s $$ And for $T > t$ \begin{equation*} \begin{aligned} \int_0^T Y_s ds - \int_0^t Y_s ds &= \int_t^T Y_s ds \\ &= T Y_T - t Y_t + \frac{1}{2}\beta^2 \int_t^T (T-s) ds - \beta \int_t^T (T-s) dB_s \end{aligned} \end{equation*}
I do not know how you can rewrite the last part to get that last equation. From $$\tag1 \int_0^t Y_s\,ds = tY_t + \frac{1}{2}\beta^2 \int_0^t s\, ds - \beta \int_0^t s\, dB_s $$ I get $$\tag2 \int_t^T Y_s\,ds = TY_T-tY_t + \frac{1}{2}\beta^2 \int_t^T s\, ds - \beta \int_t^T s\, dB_s\,. $$ From $$\tag3 Y_t=Y_0-\frac12\beta^2t+\beta B_t $$ I get \begin{align}\tag4 TY_T&=TY_0-\frac12\beta^2T^2+T\beta B_T\,,\\ tY_t&=tY_0-\frac12\beta^2t^2+t\beta B_t\,.\tag5 \end{align} Therefore, \begin{align}\tag6 TY_T-tY_t&=Y_0(T-t)-\beta^2\int_t^Ts\,ds+\beta(TB_T-tB_t) \end{align} and so \begin{align}\tag7 \int_t^TY_s\,ds&=Y_0(T-t)-\frac12\beta^2\int_t^Ts\,ds+\beta(TB_T-tB_t)-\beta\int_t^Ts\,dB_s\\ &=Y_0(T-t)-\frac12\beta^2\int_t^Ts\,ds+\beta\int_0^T(T-s)\,dB_s-\beta\int_0^t(t-s)\,dB_s\,.\tag8 \end{align}