Itô integral with brownian motion powers

Question

I am looking to compute the following integral for $n \geq 1$ where $W_t$ is a standard brownian motion: $$\int_0^T W_t^n dW_t$$ Applying Itô's lemma with $f(x)=x^{n+1}$ reduces the power to $n-1$ in the integral term. Is there a more straightforward way to do this instead of applying Itô $n$ times?

Answer 1

Integration by parts looks like $$\int_a^b f(t)g'(W_t)dW_t = f(t)g(W_t)|_a^b - \int_a^b f'(t)g(W_t)dt - \frac{1}{2}\int_a^b f(t)g''(W_t)dt,$$

where $f(t)$ is a deterministic function and $g(W_t)$ is a function of a Wiener process. For the integral you want we can let $f(t) = 1$ and we get

$$\int_a^b g'(W_t)dW_t = g(W_t)|_a^b - \frac{1}{2}\int_a^b g''(W_t)dt.$$

So for the integral $\int_a^b W_tdW_t$ we can let $g(x) = \frac{x^{n+1}}{n+1}$ and this gives

$$\int_a^b W^n_tdW_t = \frac{W_t^{n+1}}{n+1}|_a^b - \frac{n}{2}\int_a^bW^{n-1}_t dt.$$

Answer 2

$\int_0^T B_tdB_t = \frac{1}{2} B_T^2 - \frac{1}{2}T$ by direct computation and also $\int_0^T B_t^2dB_t = \frac{1}{3} B_T^3 - \int_0^T B_td_t$