Let $x_t = \int_{s=0}^t dW_s$ where $W_t$ is the standard Weiner process. Now I define $$ f(x_t) = \frac{1}{(2-x_t)^4} $$ Using Ito's Lemma, $$ df(x_t) = \frac{4}{(2-x_t)^5}dW_t - \frac{10}{(2-x_t)^6}dt. $$ and $$ f(x_t) = \int_{0}^{T}\frac{4}{(2-x_t)^5}dW_t - \int_{0}^{T}\frac{10}{(2-x_t)^6}dt. $$ My main question is if the Ito integral $\int_{t=0}^T \frac{4}{(2-x_t)^5}dW_t$ well-defined or even bounded for $T=1$?
I have tried to show $L^2$ integrability in expectation by using Fubini's Theorem, i.e., proving $\mathbb{E}[g^2(t)] < \infty$ for all $0 <t <1$ and therefore, $\mathbb{E}[\int_{t=0}^T g^2(t)dt] < \infty$. $$ \mathbb{E}[g^2(t)] = \mathbb{E}\left[\frac{16}{(2-x_t)^{10}}\right] = 16\int_{0}^{\infty} \Pr\left(\frac{1}{(2-x_t)^{10}} \geq r\right)dr \leq 16\int_{0}^{\infty}\Pr\left(1 + r^{-1/10}\geq x_t \geq 2 - r^{-1/10}\right)dr $$ Let $u = r^{-1/10}$, then $$ \int_{0}^{\infty}\Pr\left(2 - r^{-1/10}\leq x_t \leq 2 + r^{-1/10}\right)dr = 10\int_{0}^{\infty}u^{-11}\Pr\left(2 - u\leq x_t \leq 2 + u\right)du $$
which I don't think is bounded. Any suggestions about how to prove this integral is well-defined or bounded would be appreciated. Thanks!
Let $\tau:=\inf\{t: x_t=2\}$, so that $\lim_{t\uparrow \tau} f(x_t)=+\infty$. Your stochastic integral also runs into trouble at time $\tau$. The quadratic variation of $\int_0^t (2-x_s)^{-5} dW_s$ is $$ Q_t:=\int_0^t (2-x_s)^{-10} ds,\qquad 0\le t<\tau, $$ so that $\lim_{t\uparrow\tau}Q_t=+\infty$. In short, the Ito integral $I_t:=\int_0^t (2-x_s)^{-5} dW_s$ is only well-defined on the random interval $[0,\tau)$, and $$ \limsup_{t\uparrow\tau}I_t=+\infty,\qquad \liminf_{t\uparrow\tau}I_t=-\infty. $$