In dimension 3 we have that for $T=\int_{[0,\infty)}1_{B_{t}\in B(0,1)}dt$ has the Laplace transform $$E[e^{-\lambda T}]=sech(\sqrt{2\lambda}).$$

And in dimension 1 we have the same for $\tau=\min\{t: |B(t)|=1\}$:

$$E[e^{-\lambda \tau}]=sech(\sqrt{2\lambda}).$$

Inverting this in Mathematica didn't give a clean answer, but I found in "An Atlas of Functions: with Equator, the Atlas Function Calculator" that

$$\int_{a-i\infty}^{a+i\infty}\frac{sech(\nu \sqrt{\lambda})}{\sqrt{\lambda}}e^{\lambda s}\frac{ds}{2\pi i}=\frac{1}{\nu}\hat{\theta}_{2}(\frac{1}{2},\frac{t}{\nu^{2}}),$$

where $$\hat{\theta}_{2}(x,t):=\frac{1}{\sqrt{\pi t}}[\sum_{0}^{\infty}(-1)^{n}\exp(\frac{-(x+n)^{2}}{t})-\sum_{-1}^{-\infty}(-1)^{n}\exp(\frac{-(x+n)^{2}}{t})].$$

Q1:Has there been any further work on relating this to behaviour for Brownian motion?

Q2:Is there a closed formula for the inverse laplace transform of $sech( \sqrt{2\lambda})$?