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 $sech( \sqrt{2\lambda})$?



Inverse Laplace transform of $sech(\sqrt{2\lambda})$ and Brownian motion occupation time
**Q2** $\qquad$ *"Is there a closed formula for the inverse Laplace transform of ${\rm sech}(\sqrt{2\lambda})$"*  

might still benefit from an explicit answer in terms of a special function.


The inverse Laplace transform of ${\rm sech}(\sqrt{2\lambda})=1/\cosh(\sqrt{2\lambda})$ follows from an entry in Table 2 in [Theta functions; transform tables and examples for electrochemists:](https://www.sciencedirect.com/science/article/pii/0022072888870013)


$$\int\_0^\infty dt\, e^{-\lambda t}\,\frac{\partial}{\partial x}\theta\_1\left(\frac{x}{2c}\biggl|\frac{i\pi t}{c^2}\right)=\frac{c\cosh(x\sqrt{\lambda})}{\cosh(c\sqrt{\lambda})},\;\;|x|<c, $$
where $\theta\_1$ is a [Jacobi theta function](https://en.wikipedia.org/wiki/Theta_function#Jacobi_theta_function).  

The inverse Laplace transform of $1/\cosh(\sqrt{2\lambda})$ is thus given by
$${\cal L}^{-1}\left(\frac{1}{\cosh(\sqrt{2\lambda})}\right)=\frac{1}{\sqrt 2}\lim\_{x\rightarrow 0}\frac{\partial}{\partial x}\theta\_1\left(\frac{x}{2\sqrt{2}}\biggl|\frac{i\pi t}{2}\right).$$


