Add 3.4 (f)

_psets/2.md

@@ -416,9 +416,16 @@ $$
 So our result above for $$p(\tau)$$ follows because $$E = k T \log(\tau / \tau_0)$$, so $$dE / d\tau
 = k T / \tau$$.
 
-
+Now we must show that $$S(f) \propto 1/f$$.
 
 $$
 \begin{align*}
+S(f) &= \int_0^\infty p(\tau) \frac{2 \tau}{1 + (2 \pi f \tau)^2} \mathrm{d} \tau \\
+&= \int_0^\infty \frac{2 p(E) k T}{1 + (2 \pi f \tau)^2} \mathrm{d} \tau \\
+&= \frac{1}{2 \pi f} \int_0^\infty \frac{2 p(E) k T}{1 + \tau^2} \mathrm{d} \tau \\
+&= \frac{p(E) k T}{\pi f} \int_0^\infty \frac{1}{1 + \tau^2} \mathrm{d} \tau \\
+&= \frac{p(E) k T}{\pi f} \int_0^{\pi/2} \frac{sec^2{\theta}}{1 + tan^2(\theta)} \mathrm{d} \theta \\
+&= \frac{p(E) k T}{\pi f} \int_0^{\pi/2} \mathrm{d} \theta \\
+&= \frac{p(E) k T}{2 f}
 \end{align*}
 $$