# Show $\int _0^t \frac{\left|B_u \right|}{u}du < \infty \ a.e.$

How to show that for all $t\geq 0$ $$\int _0^t \frac{\left|B_u \right|}{u}du < \infty \ a.e.,$$ where $\left( B_t \right)_{t\geq 0}$ is the real standard brownian motion starting from zero ?

Using Fubini-Tonelli theorem and Cauchy-Schwarz's inequality, we have $$E\left[\int_0^t \frac{|B_u|}{u}\,du\right] = \int_0^t\frac{E(|B_u|)}{u}\,du \leq \int_0^t \frac{E(B_u^2)^{1/2}}{u}\,du = \int_0^t \frac{du}{\sqrt{u}} =2\sqrt{t} <\infty.$$

Edit: As noted by did, the expectation can be computed exactly:

$$E\left[\int_0^t \frac{|B_u|}{u}\,du\right] = \sqrt{\frac{8t}{\pi}}.$$

• By scaling, $E(|B_u|)=\sqrt{u}E(|B_1|)$ hence Cauchy-Schwarz is not required.
– Did
Jan 16 '13 at 13:08
• You're right of course, but it gives the bound $E(|B_1|) \leq 1$. Jan 16 '13 at 13:24
• Which is perfect, no? Jan 16 '13 at 13:25
• @StefanHansen: It's not perfect since $E(|B_1|) = \sqrt{2/\pi} < 1$. Jan 16 '13 at 13:34
• Which changes nothing to show the wanted result. Enven if it's refined estimation. Thank you, everybody!
– Paul
Jan 16 '13 at 13:38

Hint: Show that $$E\left[\int_0^t \frac{|B_u|}{u}\,\mathrm du\right]<\infty$$ by the use of Tonelli's theorem.