Is this stochastic integral convergent?

Question

Let $W(t)$ be a Wiener process. Is $$ X_{\delta}=\int_{\delta}^{1}\frac{W(u)}{u}du $$ convergent in $L^{2}$ while $\delta\rightarrow0^{+}?$

Answer 1

You can explicitly compute $E(X_s-X_t)^{2}$. For $s <t$ you get $E\int_s^{t}\int_s^{t} \frac {W(u)W(v)} {uv} dudv$. This can be written as $\int_s^{t}\int_s^{t} \frac {u \wedge v} {uv} dudv$. This becomes $(s-t) \ln t -(u\, ln u -u)|_s^{t}$. All the terms except tend to $0$. [$0 \geq s \, ln t \geq t \ln t \to 0$ etc]. Hence $X_\delta$ does converge in $L^{2}$ by completeness of $L^{2}$.