Premise
Let $W_t$ be the standard Wiener process, and let $X_t = \int_0^t W_s^2 \mathrm{d} s$. I am interested in determining the distribution of $X_t$.
What I did
My line of attack has been to determine moments $\mathbb{E}(X_t^r)$. To this end I set up a system of It$\bar{\text{o}}$ differential equations: $ \mathrm{d} X_t = Y_t^2 \mathrm{d} t, \qquad \mathrm{d} Y_t = \mathrm{d} W_t, \qquad X_0 = Y_0 = 0 $ Then $ \mathbb{E}(X_t^r) = \int_0^t \mathbb{E}( \mathcal{L}_0(X_s^r) ) \mathrm{d} s = \int_0^t \mathrm{d}s_1 \int_0^{s_1} \mathrm{d}s_2 \cdots \int_0^{s_{n-1}} \mathrm{d} s_n \mathbb{E}( \mathcal{L}_0^{ \circ n}(X_{s_n}^r) ) $ where $n$ is chosen so that $\mathcal{L}_0^{ \circ n}(X_{s_n}^r)$ is constant equal to $a_{n,r}$, then $\mathbb{E}(X_t^r) = \frac{t^r}{n!} a_{n,r}$, where $ \mathcal{L}_0(f(X_t, Y_t)) = \left(\frac{1}{2} \frac{\partial^2}{\partial Y_t^2} + Y_t^2 \frac{\partial}{\partial X_t} \right)(f(X_t, Y_t)) $ It is not hard to see that $ \mathbb{E}(X_t^r) = \frac{t^r}{(2r)!} \left( \frac{1}{2} \frac{\mathrm{d}^2}{\mathrm{d} y^2} + y^2 \frac{\mathrm{d}}{\mathrm{d} x} \right)^{2r} (x^r) $ Poking around OEIS, I stumbled across A126156, corresponding to $\frac{(2r)!}{r!} \mathbb{E}(X_1^r)$, from where I learnt the moment generating function for $X_t$: $ \phi(u)= \mathbb{E}(\exp(u X_t)) = \frac{1}{\sqrt{ \cos(\sqrt{2 u} t)}} = 1 + \frac{u t^2}{1!} \cdot \frac{1}{2} + \frac{u^2 t^4}{2!} \cdot \frac{7}{12} + \frac{u^3 t^6}{3!} \cdot \frac{139}{120} + \mathcal{o}(u^3) $ for $-\infty < u < \frac{\pi^2}{8 t^2}$, i.e. until the first zero of the cosine, which indicates that the main asymptotic term of the probability density is $f_{X_t}(x) \sim \exp\left(- \frac{\pi^2}{8 t^2} x \right)$. Indeed, using Weierstrass representation: $ \phi(u) = \prod_{n=0}^\infty \frac{1}{\sqrt{1-\frac{8 u t^2}{\pi^2} \frac{1}{(2n+1)^2}}} $ which suggests that $X_t = \sum_{n=0}^\infty Z_n$, where $Z_n$ are independent and $Z_n \sim \Gamma\left(\frac{1}{2}, 2 \left( \frac{2 t}{(2n+1) \pi} \right)^2 \right)$, which in turn means that $\kappa_r(X_t) = -(-8 t^2)^r (r-1)! (4^r-1) \frac{B_{2r}}{4 (2r)!} $
This unexpected success gives (slim) hope that the distribution function of $X_t$, or its probability density can be found in closed form.
Question
I am looking for references where $X_t$, or $X_1$, has been studied, or possible suggestions on how one would go about proving my guesswork.