= Logarithmic growth of an exponential Brownian clock with positive drift
{title2=$\mu>0\Longrightarrow\lim_{t\to\infty}\frac1t\log\int_0^te^{2B_s+2\mu s}ds=2\mu$}
The <strong law for Brownian motion> gives $B_s/s\to0$. For every $0<\varepsilon<\mu$, eventually $e^{(2\mu-2\varepsilon)s}\leq e^{2B_s+2\mu s}\leq e^{(2\mu+2\varepsilon)s}$. Integrating these bounds, taking logarithms and dividing by $t$ squeezes the growth rate between $2\mu-2\varepsilon$ and $2\mu+2\varepsilon$. Let $\varepsilon\downarrow0$. The corresponding inverse clock grows like $(\log t)/(2\mu)$.
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