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Prove $\frac{1}{\sqrt{t}}\log\left(\int_0^t \exp(B_s)\mathrm{d}s\right)$ converges to $\sup\limits_{t\in [0,1]}B_t$ in distribution as $t\to\infty$. I have a sense to use scaling invariance, but no idea how to derive this whole thing.

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Note that \begin{align*} \int_0^t e^{B_s}ds &= t\int_0^1 e^{B_{tu}}du\\ &=t\int_0^1 e^{\sqrt{t}\frac{1}{\sqrt{t}}B_{tu}}du\\ &=t\int_0^1 e^{\sqrt{t}W_u}du, \end{align*} where $\{W_u=\frac{1}{\sqrt{t}}B_{tu}, \, u\ge 0\}$ is a standard Brownian motion. Then \begin{align*} \frac{1}{\sqrt{t}} \ln \int_0^t e^{B_s}ds &= \frac{\ln t}{\sqrt{t}} + \frac{1}{\sqrt{t}}\ln \int_0^1 e^{\sqrt{t}W_u}du\\ &= \frac{\ln t}{\sqrt{t}} + \ln \bigg(\int_0^1 \left(e^{W_u}\right)^{\sqrt{t}}du \bigg)^{\frac{1}{\sqrt{t}}}\\ &= \frac{\ln t}{\sqrt{t}} + \ln\, \big\lVert{e^{W_u}}\big\rVert_{\sqrt{t}}\\ &\rightarrow\ln\, \big\lVert{e^{W_u}}\big\rVert_{\infty}\\ &= \ln\Big( \max_{0\le u \le 1} e^{W_u}\Big)\\ &=\max_{0\le u \le 1} W_u. \end{align*} That is, $\frac{1}{\sqrt{t}} \ln \int_0^t e^{B_s}ds$ converges to $\max_{0\le t \le 1} B_t$ in distribution, as $t\rightarrow \infty$.

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