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


See p. 342 of Feller, vol. 2: Here, B starts at x and is absorbed at 0 and a.

half page from Feller

What you wrote about T is not correct: They are not equal in dimensions 1 and 3. Rather, this total occupation time in dimension 3 equals in distribution the exit time in dimension 1 (the Ciesielski-Taylor identity).


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:

$$\int_0^\infty dt\, e^{-\lambda t}\,\frac{\partial}{\partial x}\theta_1\left(\frac{x}{2c}\biggl|\frac{i\pi t}{c^2}\right)=\frac{c\cosh(x\sqrt{\lambda})}{\cosh(c\sqrt{\lambda})},\;\;|x|<c, $$ where $\theta_1$ is a Jacobi theta function.
The inverse Laplace transform of $1/\cosh(\sqrt{2\lambda})$ is thus given by $${\cal L}^{-1}\left(\frac{1}{\cosh(\sqrt{2\lambda})}\right)=\frac{1}{\sqrt 2}\lim_{x\rightarrow 0}\frac{\partial}{\partial x}\theta_1\left(\frac{x}{2\sqrt{2}}\biggl|\frac{i\pi t}{2}\right).$$


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