Expectation of Ito Integral to the nth power

Question

I have this Ito integral (with respect to Brownian motion):

$$ X(t) = \int^t_0 g(s)dW(s) $$

where $g$ is a deterministic function. I know that $X(t)$ must be a Gaussian process due to $W$. However, I need to deal with $X(t)^n$, and specifically to find the expectation of $X(t)^n$:

$$ \mathbb{E}[(X(t))^n] = \mathbb{E}\Bigl[\Bigl(\int^t_0 g(s)dW(s)\Bigr)^n\Bigr] $$

When $n=2$ we can apply Ito isommetry to deal with this, but when $n \geq 2$, how do we deal with this?

Thanks

Answer 1

$\int_0^{t}g(s)dW(s)$ is normal with mean $0$ and variance $\int_0^{t}g(s)^{2}ds$. So $E[X(t)^{n}]$ is nothing but $[\int_0^{t}g(s)^{2}ds]^{n/2} EY^{n}$ where $Y\sim N(0,1)$.