(a) From the defining equation of the variable (Y(t)), we get [ begin{equation*} leftlangle Y^{2}(t) ightangle=int_{u}^{u+t} int_{u}^{u+t}leftlangle...

(a) From the defining equation of the variable \(Y(t)\), we get

\[
\begin{equation*}
\left\langle Y^{2}(t)\rightangle=\int_{u}^{u+t} \int_{u}^{u+t}\left\langle y\left(u_{1}\right) y\left(u_{2}\right)\rightangle d u_{1} d u_{2} \tag{1}
\end{equation*}
\]

Since \(y(u)\) is statistically stationary, we may write

\[
\begin{equation*}
\left\langle y\left(u_{1}\right) y\left(u_{2}\right)\rightangle=\int_{0}^{\infty} w(f) \cos (2 \pi f s) d f \quad\left(s=u_{2}-u_{1}\right) \tag{2}
\end{equation*}
\]

see eqn. (15.5.15). Substituting (2) into (1), we get

\[
\begin{align*}
\left\langle Y^{2}(t)\rightangle & =\int_{0}^{\infty} w(f) I(f, t) d f, \text { where }  \tag{3}\\
I(f, t) & =\int_{u}^{u+t} \int_{u}^{u+t}\left\{\cos \left(2 \pi f u_{2}\right) \cos \left(2 \pi f u_{1}\right)+\sin \left(2 \pi f u_{2}\right) \sin \left(2 \pi f u_{1}\right)\right\} d u_{1} d u 2 \\
& =\left[\int_{u}^{u+t} \cos (2 \pi f u) d u\right]^{2}+\left[\int_{u}^{u+t} \sin (2 \pi f u) d u\right]^{2} \\
& =\frac{1}{4 \pi^{2} f^{2}}\left[[\sin \{2 \pi f(u+t)\}-\sin (2 \pi f u)]^{2}+\right. \\
& \left.=\frac{1}{2 \pi^{2} f^{2}}[1-\cos (2 \pi f u)-\cos \{2 \pi f(u+t)\}]^{2}\right]
\end{align*}
\]

Substituting (4) into (3), we obtain the desired result for \(\left\langle Y^{2}(t)\rightangle\).

Next, it follows that

\[
\begin{align*}
\frac{\partial}{\partial t}\left\langle Y^{2}(t)\rightangle & =\frac{1}{\pi} \int_{0}^{\infty} \frac{w(f)}{f} \sin (2 \pi f t) d f, \text { and }  \tag{5}\\
\frac{\partial^{2}}{\partial t^{2}}\left\langle Y^{2}(t)\rightangle & =2 \int_{0}^{\infty} w(f) \cos (2 \pi f t) d f . \tag{6}
\end{align*}
\]

Taking the sine transform of (5) and the cosine transform of (6), we obtain the other quoted results. Finally, a comparison of eqns. (2) and (6) shows that

\[
\begin{equation*}
K_{y}(s)=\frac{1}{2} \frac{\partial^{2}}{d s^{2}}\left\langle Y^{2}(s)\rightangle \tag{7}
\end{equation*}
\]

(b) If the variable \(y(u)\) is the \(x\)-component of the velocity of a Brownian particle, with power spectrum (15.5.21), then eqns. (3) and (4) give

\[
\begin{align*}
\left\langle x^{2}(t)\rightangle & =\frac{2 k T \tau}{\pi^{2} M} \int_{0}^{\infty} \frac{1-\cos (2 \pi f t)}{f^{2}\left\{1+(2 \pi f \tau)^{2}\right\}} d f \\
& =\frac{4 k T \tau^{2}}{\pi M} \int_{0}^{\infty} \frac{1-\cos (x t / \tau)}{x^{2}\left(1+x^{2}\right)} d x \\
& =\frac{2 k T \tau^{2}}{M}\left[\frac{t}{\tau}-\left(1-e^{-t / \tau}\right)\right] \tag{8}
\end{align*}
\]

in complete agreement with eqn. (15.3.7) for the quantity \(\left\langle r^{2}(t)\rightangle\). We also note that

\[
\begin{equation*}
\frac{1}{2} \frac{\partial^{2}}{\partial s^{2}}\left\langle x^{2}(s)\rightangle=\frac{k T}{M} e^{-s / \tau} \quad(s>0), \tag{9}
\end{equation*}
\]

which indeed is equal to the autocorrelation function \(K(s)\) of the variable \(\mathrm{v}_{x}\); see eqn. (15.6.20).