Solve Itô integral with power

Question

$$\int_0^t e^{Ws} W_s^r dW_s$$ where $W_s$ is Wiener process and r> in $\mathbb{Z}$

My first approach would be to use Ito's lemma, however, coming up with the function $g(t,x)$ is difficult

The antiderivative of $$e^{Ws} W_s^r$$ is $$(-x)^{-n} x^n \Gamma (n+1,-x)$$ according to mathematica

I'm not sure what to do from here. Is there some kind of a trick?

Answer 1

Hints Fix $n \geq 0$.

1. Show by induction that $$\int^x e^y y^n \, dy = e^x \cdot \sum_{k=0}^n \frac{n!}{(n-k)!} \cdot (-1)^k x^{n-k} =: F(x). \tag{1}$$ (Hint: Use partial integration.)
2. By Itô's formula, $$\begin{align*} F(W_t)-F(W_0) &= \int_0^t F'(W_s) \, dW_s + \frac{1}{2} \int_0^t F''(W_s) \, ds. \end{align*}$$ By virtue of our choice, $$F'(x) = e^x \cdot x^n$$ and $$F''(x) = e^x (x^n + n \cdot x^{n-1}).$$ Morever, $(1)$ shows $$F(0) = F(W_0) = n!.$$ Hence, $$F(W_t)-n! = \int_0^t e^{W_s} W_s^n \, dW_s + \frac{1}{2} \int_0^t e^{W_s} (W_s^n + n W_s^{n-1}) \, ds,$$ i.e. $$\int_0^t e^{W_s} W_s^n \, dW_s = e^{W_t} \sum_{k=0}^n \frac{n!}{(n-k)!} \cdot (-1)^k W_t^{n-k}- \frac{1}{2} \int_0^t e^{W_s} (W_s^n + n W_s^{n-1}) \, ds.$$