What is the expected inverse stopping time for an Brownian Motion? Let $B_t$ be standard Brownian motion and $\tau_a=\inf\{t\geq 0 : B_t \geq a\}$ be the stopping time where $B_t$ exceeds some value $a$.
Is there an analytic form for $\mathbb{E}\left[\frac{1}{\tau_a}\right]$ ?

Context:
Suppose $B_t$ represents the price of a financial instrument. If an investor buys this instrument today at a price of $p$ and plans to sell at a profit of $a$, then the expected (continually-compounded) return $r$ of the investor is given by
$$r = \log\left(1+\frac{a}{p}\right)\mathbb{E}\left[\frac{1}{\tau_a}\right]$$
I wish to find an analytic form for $r$ so that I may then maximise $r$ over all $a$. That is, I wish to find the optimal price at which the investor should sell the instrument in order to maximise her return.

First, note that for all $t\ge0$, we have the equality of events :
$$\{\tau_a\le t\} = \left\{\sup_{0\le s \le t} B_s \ge a\right\}$$
(In words, this says that the Brownian Motion $B$ reaches level $a$ before time $t$ if and only if its supremum on $[0,t]$is at least $a$).
By the reflection principle and the above equality, we thus have
$$\mathbb P\left(\tau_a\le t\right)=\mathbb P\left(\sup_{0\le s \le t} B_s \ge a\right)=2\mathbb P\left(B_t\ge a\right) $$
The above is true for any $t>0$ and in particular it is true for $t'\equiv 1/t$, so we have
$$\begin{align}
F_a(t)\equiv\mathbb P\left(\frac {1}{\tau_a}\le t\right) &= \mathbb P\left(\tau_a\ge \frac 1 t\right)\\
&= 1 - \mathbb P\left(\tau_a\le \frac 1 t\right)\\
&= 1 - 2\mathbb P\left(B_{1/t}\ge a\right) \\
&= 1 - 2\mathbb P\left(tB_{1/t}\ge ta\right)\\
&= 1 - 2\mathbb P\left(B_t\ge ta\right)\tag1\\
&= 1 - 2\mathbb P \left(Z\ge a\sqrt t\right)\tag2\\
\end{align}$$
Where I used the time inversion property in $(1)$ and the fact that $Z\equiv B_t/\sqrt t\sim \mathcal N(0,1)$ in $(2)$.
At this point, it is quite straightforward to express the CDF of $\frac {1}{\tau_a}$ $F_{a} $ in terms of the standard normal CDF. In particular, if you want to compute $\mathbb E\left[\frac {1}{\tau_a}\right] $, you have at least two options :

*

*By a well-known formula, you can integrate $1-F_a$ directly :
$$\mathbb E\left[\frac {1}{\tau_a}\right] = \int_0^\infty (1-F_a(t))\ dt = \int_0^\infty \left[1-\text{erf}\left(\frac{a\sqrt t}{\sqrt 2}\right)\right]\ dt $$

*You can compute the PDF $f_a(t) :=dF_a(t)/dt$ and integrate
$$\mathbb E\left[\frac {1}{\tau_a}\right] = \int_0^\infty tf_a(t)\ dt $$
Both of these integrals involve very standard functions and by digging up the literature you should be able to find a way to express them in terms of $\text{erf}, \text{erfc}$ or other well-known quantities, which should allow you to approximate $\mathbb E\left[\frac {1}{\tau_a}\right] $ well enough for your purposes.
