Question
Let $X_t = \int_0^t W_\tau , d\tau$, where $W_t$ is a standard Wiener process.
What is the distribution of $X_t$?
Solution

Using integration by parts, $$ X_t = \int_0^t W_s\, ds = t W_t - \int_0^t s\, dW_s = \int_0^t (t-s)\, dW_s. $$

For deterministic square-integrable $f(s)$, $$ \int_0^t f(s), dW_s \sim \mathcal{N}\!\left( 0,\; \int_0^t f(s)^2\, ds \right). $$

Here $f(s) = t - s$, so $$ \int_0^t (t-s)^2\, ds = \frac{t^3}{3}. $$

Thus, $$ X_t \sim \mathcal{N}\!\left( 0,\; \frac{t^3}{3} \right). $$