Is an integrated Wiener process recurrent or transient? Like the title says, if I take an integrated Wiener process / Brownian motion $\int ^t _0 W_s ds$, will it be recurrent or transient? Or, under what conditions will it be one or the other? 
I know that, for any $t$, the integral is a normal variable ~$N(0,\frac{t^3}{3})$. So it'll diverge to $\infty$ as $t\rightarrow \infty$. And it's not a martingale. But would it still be recurrent? And how would I show that, or show that it's not?
(If there are any theorems or discussions of this out there, even just a link to that would be great!)

It is recurrent.
Let $X_t := \int_0^t W_s \,ds$. I claim that $\varlimsup_{t\to\infty}X_t=\infty$ and $\varliminf_{t\to\infty}X_t=-\infty$, so every real number is visited infinitely often a.s.
Here is a sketch of a proof.
Write $\mathscr F_t := \sigma(W_s ; s \le t)$. For a finite stopping time $T$, write $X_t = X_{t \wedge T} + (t - T)^+ W_T + Y_{(t - T)^+}$ with $Y$ a copy of $X$ that is independent of $\mathscr F_T$. Use $T_n := \inf \{t \ge n ; W_t = 0\}$ to see that $[\sup_t X_t = \infty]$ is independent of $\mathscr F_n$ for each $n$, whence is trivial: $P[\sup_t X_t = \infty] \in \{0, 1\}$. If this event has probability $0$, then
use $T := \inf \{t \ge 1 ; W_t = -1\}$ to see that $P[\inf_t X_t = -\infty] = 1$, which contradicts symmetry. Therefore, $P[\sup_t X_t = \infty] = 1$.
A stronger result was proved by Khoshnevisan and Shi,
Chung's law for integrated Brownian motion,
Trans. Amer. Math. Soc. 350 (1998), no. 10, 4253–4264.
In two (and higher) dimension, integrated Brownian motion tends to infinity: Kolokoltsov, A Note on the Long Time Asymptotics of the Brownian Motion with Application to the Theory of Quantum Measurement, Potential Analysis 7 (1997), 759–764.
