Integrate equation (15.3.14) to obtain [ boldsymbol{r}(t)=boldsymbol{v}(0) tauleft(1-e^{-t / tau} ight)+tau int_{0}^{t}left{1-e^{(u-t) / tau} ight} boldsymbol{A}(u) d

Integrate equation (15.3.14) to obtain

\[
\boldsymbol{r}(t)=\boldsymbol{v}(0) \tau\left(1-e^{-t / \tau}\right)+\tau \int_{0}^{t}\left\{1-e^{(u-t) / \tau}\right\} \boldsymbol{A}(u) d u
\]

so that \(\boldsymbol{r}(0)=0\). Taking the square of this expression and making use of the autocorrelation function \(K_{A}(s)\), derive formula (15.3.31) for \(\left\langle r^{2}(t)\rightangle\).