It can be shown that $[dW(t)]^\alpha=0$ with $\alpha\in\mathbb{R}$ and $\alpha\ge 3$ generalizing the integer case.

Let us consider the stochastic differential equation $dX(t)=[dW(t)]^\alpha$ with $\alpha>0$. We can write the solution in the form $X(t)=X(t_0)+\int_{t_0}^t[dW(t)]^\alpha$ with the integral in the Ito sense. Then, we have to evaluate this integral with the sum
\begin{equation}
S_n=\sum_{k=1}^n[W(t_k)-W(t_{k-1})]^\alpha.
\end{equation}
The power of the Brownian process can be evaluated in the following way
\begin{equation}
[W(t_k)-W(t_{k-1})]^\alpha = [(1+W(t_k)+W(t_{k-1}))-1]^\alpha=
\end{equation}
\begin{equation}
(-1)^\alpha\sum_{l_1=0}^\infty\left(\begin{array}{c}
\alpha \\
l_1
\end{array}\right)(-1)^{l_1}(1+W(t_k)+W(t_{k-1}))^{l_1}=
\end{equation}
\begin{equation}
(-1)^\alpha\sum_{l_1=0}^\infty\sum_{l_2=0}^\infty\left(\begin{array}{c}
\alpha \\
l_1
\end{array}\right)\left(\begin{array}{c}
l_1 \\
l_2
\end{array}\right)(-1)^{l_1}
[W(t_k)-W(t_{k-1})]^{l_2}
\end{equation}
provided $|W(t_k)-W(t_{k-1})|<1$. Now, we can use stochastic calculus to remove powers higher than 2 and it is easy to see that
\begin{equation}
S_n=(-1)^\alpha\sum_{k=1}^n\sum_{l_1=0}^\infty\left(\begin{array}{c}
\alpha \\
l_1
\end{array}\right)(-1)^{l_1}-
(-1)^\alpha\sum_{l_1=0}^\infty\left(\begin{array}{c}
\alpha \\
l_1
\end{array}\right)l_1(-1)^{l_1}\sum_{k=1}^n[W(t_k)-W(t_{k-1})]+
\end{equation}
\begin{equation}
(-1)^\alpha\sum_{l_1=0}^\infty\left(\begin{array}{c}
\alpha \\
l_1
\end{array}\right)\frac{l_1(l_1-1)}{2}(-1)^{l_1}
\sum_{k=1}^n[W(t_k)-W(t_{k-1})]^2.
\end{equation}
So, we have the required expansion with coefficients
\begin{eqnarray}
\mu_0&=&\sum_{l_1=0}^\infty\left(\begin{array}{c}
\alpha \\
l_1
\end{array}\right)(-1)^{l_1} \nonumber \\
\mu_1&=&\sum_{l_1=0}^\infty\left(\begin{array}{c}
\alpha \\
l_1
\end{array}\right)l_1(-1)^{l_1} \nonumber \\
\mu_2&=&\sum_{l_1=0}^\infty\left(\begin{array}{c}
\alpha \\
l_1
\end{array}\right)\frac{l_1(l_1-1)}{2}(-1)^{l_1}
\end{eqnarray}
Now we see immediately that $\mu_0=\left.(1-x)^\alpha\right|_{x=1}=0$. Besides, we get immediately the result that, for any real $\alpha\ge 3$, we have again $[dW(t)]^\alpha=0$ as in this case the coefficients are all zero when $\mu_1$ and $\mu_2$ are evaluated thorugh Abel summation. Finally, when $0<\alpha<1$ both the coefficients $\mu_1$ and $\mu_2$ are divergent and maybe no meaning can be attached to them (I have in mind summable divergent series here, any suggestion is greatly appreciated).

any"textbooks result" "given for integer $\alpha$" (even for $\alpha=1$, the integral as written above does not make sense). – Did Jan 26 '12 at 20:42the discussion with George. // Unsurprisingly, no answer to my request for (at least) one example of the textbooks mentioned in your question. – Did Jan 26 '12 at 20:49