Scaled integrated Brownian motion has limit Let $B$ be a standard Brownian motion and put $$X(t)=\frac{1}{\sqrt{t}}\int_{0}^{t}f(B(s))ds,$$ where $f \in L_1(\mathbb{R}^{1})$ and $\int f(x)dx=1$. Show that $$ \lim_{t \rightarrow \infty} EX(t) =\sqrt{2/\pi}$$ and $$ \lim_{t \rightarrow \infty} E[X(t)^2]=1.$$ 
I was thinking to try writing them as Riemann sums, but it seems that it does not work out well. Any hints or help would be appreciated.
 A: Let 
$$X_t := \frac{1}{\sqrt{t}} \int_0^t f(B_s) \, ds.$$
Using that $B_s \sim N(0,s)$, $s \geq 0$, and applying Fubini's theorem yields
$$\begin{align*} \mathbb{E}(X_t) &= \frac{1}{\sqrt{t}} \int_0^t \mathbb{E}(f(B_s)) \, ds \\ &= \frac{1}{\sqrt{t}} \int_{\mathbb{R}} \int_0^t f(x) \frac{1}{\sqrt{2\pi s}} \exp \left(- \frac{x^2}{2s} \right) \, ds \, dx. \end{align*}$$
Now we substitute $r := \frac{\sqrt{s}}{\sqrt{t}}$ and obtain
$$\mathbb{E}(X_t) = \frac{2}{\sqrt{2\pi}} \int_{\mathbb{R}} \left[ \int_0^1 \exp \left( - \frac{x^2}{2r^2 t} \right) \, dr \right] f(x) \, dx.$$
From the dominated convergence theorem, it follows that
$$\lim_{t \to \infty} \mathbb{E}(X_t) = \sqrt{\frac{2}{\pi}} \int f(x) \, dx = \sqrt{\frac{2}{\pi}}.$$
For the second equation, write
$$\left| \int_0^t f(B_r) \, dr \right|^2 = \int_0^t \int_0^t f(B_r) f(B_s) \, dr \, ds,$$
and use that $(B_r,B_s)$ is Gaussian with mean vector $(0,0)$ and covariance matrix 
$$\begin{pmatrix} r & r \\ r & s \end{pmatrix}$$
for $r \leq s$. By choosing a suitable substitution, we obtain - as in the first part - an integral which can be calculated explicitely.
