Stochastic integral in closed form

Let $(W_t)_{t\geq 0}$ be a Brownian motion and $\alpha>0$ be a constant. Consider the following quantity: $$\mathbb{E}\Big(\int_0^tsdW_s{\bf 1}_{\{t^{-\frac{1}{2}}W_t>\alpha\}}\Big).$$ Can a closed form expression be found? Without the indicator function, the quantity inside the expectation is Gaussian, so can this e.g. be written as the positive part of a Gaussian random variable?

Let $X_t=\displaystyle\int_0^ts\mathrm dW_s$ and $Y_t=\displaystyle\int_0^tW_s\mathrm ds$, then $X_t+Y_t=tW_t$ hence $E(X_t\mid W_t)=tW_t-E(Y_t\mid W_t)$. Furthermore, $E(W_s\mid W_t)=sW_t/t$ for every $s$ in $(0,t)$ hence $$E(Y_t\mid W_t)=\displaystyle\int_0^tE(W_s\mid W_t)\mathrm ds=(W_t/t)\displaystyle\int_0^ts\mathrm ds=tW_t/2.$$ Thus, $E(X_t\mid W_t)=tW_t/2$ hence, for every $x$, $$E(X_t;W_t\gt x)=E(E(X_t\mid W_t);W_t\gt x)=(t/2)E(W_t;W_t\gt x).$$ Since $W_t$ is distributed like $\sqrt{t}W_1$, when $x=\alpha\sqrt{t}$, one gets $$E(X_t;W_t\gt\alpha\sqrt{t})=(t\sqrt{t}/2)E(W_1;W_1\gt\alpha),$$ and a standard computation yields the expectation in the RHS, leading to $$E(X_t;W_t\gt\alpha\sqrt{t})=\mathrm e^{-\alpha^2/2}t\sqrt{t}/(2\sqrt{2\pi}),$$