Sojourn time for Brownian motion with drift 
Assume that $X(t) = \mu t + \sigma W(t)$, where $W$ is a standard Brownian motion and $\mu > 0$, and define 
  $$
T = \int_0^\infty I(X(t) \in (0,\delta))\ dt  
$$ 
  where $\delta >0$. What is $E(T)$?

There seems to be a number of questions (and answers) on first passage times, but nothing out there on sojourn times.
This is taken from Joseph T. Chang's course on stochastic processes (see Chang's webpage).
 A: Note that, by definition of $T$ and by the scaling properties of Brownian motion,
$$
E(T)=\int_0^\infty P(0\lt X(t)\lt\delta)\,\mathrm dt=\int_0^\infty P(x_\delta(t)\lt \xi\lt x_0(t))\,\mathrm dt,
$$
where $\xi$ denotes a standard normal random variable and, for every nonnegative $\delta$,
$$
x_\delta(t)=\mu\sqrt{t}/\sigma-\delta/(\sigma\sqrt{t}).
$$
Let $t_\delta$ denote the inverse function of $x_\delta$, that is, for every nonnegative $\delta$ and every $x$, $t_\delta(x)$ denotes the unique nonnegative solution $t$ of the equation
$$
\mu\sqrt{t}/\sigma-\delta/(\sigma\sqrt{t})=x.
$$
For example, $t_0(x)=x^2\sigma^2/\mu^2$, $t_\delta(x)\geqslant t_0(x)$ for every nonnegative $\delta$ and every $x$, and
$$
[x_\delta(t)\lt \xi\lt x_0(t)]=[t_0(\xi)\lt t\lt t_\delta(\xi)],
$$
hence
$$
E[T]=E[t_\delta(\xi)]-E[t_0(\xi)].
$$
Note that $t_\delta(x)$ solves
$\mu t-x\sigma\sqrt{t}-\delta=0$ and that $\mu\gt0$,
thus,
$$
t_\delta(x)=\left(\frac{x\sigma+\sqrt{x^2\sigma^2+4\delta\mu}}{2\mu}\right)^2=\frac{x^2\sigma^2+2\delta\mu+x\sigma\sqrt{x^2\sigma^2+4\delta\mu}}{2\mu^2}.
$$
The distribution of $\xi$ is even hence
$$
E\left(\xi\sqrt{\xi^2\sigma^2+4\delta\mu}\right)=0,
$$
which implies that
$$
E[t_\delta(W_1)]=\frac{\sigma^2+2\delta\mu}{2\mu^2},
$$
and one is left with
$$
E(T)=\frac\delta\mu.
$$
More generally, for every Borel subset $B$ of $\mathbb R_+$, the time $T_B$ spent by $X$ in $B$ has expectation
$$
E(T_B)=\frac{\mathrm{Leb}(B)}\mu.
$$
