$\mathbb{E} \int_a^b W^3(t)\,dW(t)=?$

Question

Is it true that

$\mathbb{E} \int_a^b W^3(t)\,dW(t)=0$, for $a < b \in \mathbb{R}$

I know that for an adapted process $\Delta(t), t\geq 0$, the integral

$\int_0^t \Delta(u)dW(u)$ is a martingale in $t$, and therefore had expectation $0$, if

$\mathbb{E} \int_0^T \Delta^2(t) dt < \infty$

How can I show the square-integrability condition, and is there an other way to show the expectation?

Answer 1

Interchanging integral and mean (tonelli) gives us $\int_0^T E[W(t)^6]dt=\int_0^T 15 t^3 dt =15/4 \cdot T^4< \infty$ where we have used this scheme to find the 6'th moment.