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I have a question about a positive functional of the one-dimensional Brownian motion.

Let $(\{B_t\}_{t \ge 0},P_x)$ be the one-dimensional Brownian motion starting from $x >0$.

I would like to know whether the following equation holds \begin{align*} \text{(A)}\quad \lim_{x \to \infty}P_{x}\left(\int_{0}^{\sigma_1}\frac{1}{B(s)}\,ds \ge a\right)=1\quad \text{for any $a>0$.} \end{align*} Here, $\sigma_1=\inf\{t>0 \mid B_t=1\}$.

Namely, I would like to whether $\int_{0}^{\sigma_1}\frac{1}{B(s)}\,ds$ is very large.

Attempt

It holds that \begin{align*} &P_{x}\left(\int_{0}^{\sigma_1}\frac{1}{B(s)}\,ds \ge a\right) =P_{x}\left(\exp \left(-\int_{0}^{\sigma_1}\frac{1}{B(s)}\,ds \right) \le \exp(-a)\right). \end{align*}

Hence, to prove (A), it suffices to show \begin{align*} \lim_{x \to \infty} E_{x}\left(\exp \left(-\int_{0}^{\sigma_1}\frac{1}{B(s)}\,ds \right) \right)=0. \end{align*}

Let $f(x)=2x\log x-2x$. Then, $f'(x)=2\log x$ and $f''(x)=2/x$. Therefore, we have \begin{align*} f(B_t)=f(x)+2\int_{0}^{t}f'(B_s)\,dB_s+\int_{0}^{t}(1/B_s)\,ds \end{align*} and \begin{align*} &\exp\left(-\int_{0}^{\sigma_1}\frac{1}{B(s)}\,ds \right) \\ &=\exp(-f(B_t))+\exp(f(x))+\exp(2\int_{0}^{t}f'(B_s)\,dB_s)... \end{align*} However, I feel that this argument is not good.

Please let me know if you know a good proof method.

sharpe
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    If you want to have it quick, then you can take a look at the Handbook of Brownian motion, Table 1, formula 2.19, which gives an explicit expression for the Laplace transform which you are looking for. – saz Apr 13 '19 at 06:57
  • Thank you for your comment. I would like to look at the book. – sharpe Apr 13 '19 at 09:01
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    @saz I could find the formula. It seems that $\lim_{x \to \infty}P_x(\int_{0}^{\sigma_1}1/B_s^2,ds \ge a)=0$ holds too. – sharpe Apr 13 '19 at 09:32

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