Surprisingly simple expected time for the "range" of a Brownian motion to extend beyond $a$ - is there a martingale method? I will call the range $R(t)$ of a standard Brownian motion the difference between its maximum $M(t) = \max_{0\leq s \leq t} B(t)$ and its minimum $m(t) = \min_{0\leq s \leq t} B(t)$. That is, I am calling $R(t) = M(t) - m(t)$.
Consider the first hitting time $T_a$ corresponding to when $R(t)$ hits $a$ for the first time. I am primarily interested in $E[T_a]$. I claim below that $E[T_a] = \frac{1}{2}a^2$, but I have to slog through a number of calculations to see it. In particular, it seems almost miraculous to get such a simple prefactor of $1/2$ after juggling many factors of $\pi$, etc.
This is a little funny, since one can often quickly calculate other first hitting times of similar flavors using martingales (such as the expected exit time out of the interval $(-a,a)$ being $a^2$). My question is: is there another, simpler way to calculate $E[T_a]$, such as with martingales?

Here are some thoughts. First, I believe that because of the $\sqrt{t}$ growth of the standard deviation of Brownian motion that $E[T_a] = a^2 E[T_1]$, and I indeed find this dependence on $a$.
According to Feller, we can find the probability distribution for the range $R(t)$, $f_{R,t}(r)$, via considering method of images on a relevant partial differential equation. He notes
$$f_{R,t}(r) =  \frac{8}{\sqrt{2\pi}} \sum_{k=1}^\infty (-1)^{k-1} k^2 e^{- \frac{k^2 r^2}{2t}} =  \sqrt{\frac{2}{\pi}} \frac{1}{r} \left[ \frac{d}{dz}\left(\frac{\sqrt{2 \pi}}{z}\sum_{k=1}^\infty e^{-\frac{(2k-1)^2 \pi^2}{8 z^2}} \right) \right]_{z=\frac{r}{2\sqrt{t}}}$$
where the judicious rewriting and likely use of Poisson summation formulae in the last equality (the part in the parentheses is the CDF of a probability distribution) allows for relatively straightforward computation of the 0th, 1st, and 2nd moments of $f_{R, t}(r)$.
Here is my calculation. Note that for non-negative random variables $G$ with probability densities $g(t)$ decaying fast enough at infinity, we have via integration by parts that $\int_0^\infty P(G>t) = \int_0^\infty t g(t) = E[G]$. Since the range is increasing in time, $P(T_a > t) = P(R(t) <a)$.
$$
\begin{split}
E[T_a] &= \int_0^\infty dt \int_0^{a}dr \sqrt{\frac{2}{\pi}} \frac{1}{r} \left[ \frac{d}{dz}\left(\frac{\sqrt{2 \pi}}{z}\sum_{k=1}^\infty e^{-\frac{(2k-1)^2 \pi^2}{8 z^2}} \right) \right]_{z=\frac{r}{2\sqrt{t}}}
\\ &= \int_0^a dr \frac{2}{r} \int_0^\infty dt  \sum_{k=1}^\infty \left(-\frac{4t}{r} + \frac{2(2k-1)^2 \pi^2 4^2 t^2}{8r^4} \right)  e^{-\frac{(2k-1)^2 \pi^2 t}{2 r^2}}
\\ &= \frac{1}{\pi^4}\sum_{k=1}^\infty \frac{1}{(2k-1)^4} \int_0^a dr \frac{2}{r}(-16r^2+64 r^2)
\\ &=\frac{1}{\pi^4} \frac{15 \zeta(4)}{16} 48 a^2
\\ &= \frac{1}{2} a^2
\end{split}
$$
 A: Let $\sigma_k:=\min\bigl\{t\ge 0: B(t) \in \{-1,k\}\bigr\}$. By using optional stopping for the martingale $B_{t}^2-t$, it is easy to see (and standard) that
$$E_0[\sigma_k]=E_0[B(\sigma_k)^2]=\frac{k}{k+1}(-1)^2+\frac{1}{k+1}k^2 =k\,. \tag{1}$$
Let $$\tau_k:=\min\{t \ge 0:  \lfloor{M(t)}\rfloor-\lceil{m(t)}\rceil=k \}\,$$
denote the first time that the image $B[0,t]$ contains $k+1$ integers.
By $(1)$ we have $E_0 \tau_1=1$. Next, fix $k \ge 1$ and observe that     $B(\tau_{k })$ is an integer, and it must equal either $M(\tau_{k })$ or $m( \tau_{k })$. On either one of these events, we find that
$$E[\tau_{k+1}-\tau_{k} |{\mathcal F}_{\tau_{k}}]=E[\sigma_{k+1}]=k+1 \,. \tag{2}$$
For example, if $B(\tau_{k})=m( \tau_{k})=h$, then
$M(\tau_{k}) \in [h+k,h+k+1)$, so given
${\mathcal F}_{\tau_{k}}$, the increment $\tau_k-\tau_{k-1}$ is the  time  it takes a Brownian motion (started at $h$) to reach one of the levels $\{h-1,h+k\}$. Taking expectation in (2) and summing over $k<n$, we find that $$E[\tau_n]=\frac{n(n+1)}{2} \,. \tag{3}$$
For every $a>0$ and $m \ge 1$, Brownian scaling gives $E[T_{ma}]=m^2E[T_a]$. Moreover,
$$ \tau_{\lfloor ma \rfloor-1} \le T_{ma} \le \tau_{\lceil ma \rceil} \,.$$
Thus
$$\frac {E\tau_{\lfloor ma \rfloor-1}}{m^2} \le E[T_a]=\frac{E[T_{ma}]}{m^2} \le \frac{E[\tau_{\lceil ma \rceil}]}{m^2} \tag{4} \,.$$
Taking a limit in $(4)$ as $m \to \infty$, using  $(3)$,we obtain
$$E[T_a]=a^2/2 \,.$$
