Taking the derivative of an integral of a discontinuous function When I took measure theory with Frank Jones' books years ago, I did every problem in the book because I loved its teaching style.
There was one problem that took me 4-5 years to solve. It was problem 23 of Chapter 16 (I've rewritten it to simplify some obscure notation):
Let $F(x)=x^2 \cos(\frac{1}{x})$ away from 0 and 0 at 0. Define the function $V_F(x)=\int_0^x |F'(t)|dt$. Show that $V_F'(0)=\frac{2}{\pi}$.
I finally solved it by assuming part of it went to 0, doing a u-substitution, getting the answer as a series, estimating the series using an integral test, and taking the derivative. I'm not sure I could recreate it from memory.
How are you supposed to solve this problem? Is there an easier way? All the other exercises in this book are pretty self contained or have a warning. Can this be solved with just techniques from the book?
 A: Since $F'(x)=2x\cos\frac{1}{x}+\sin\frac{1}{x}$ and since $V_F(x)=\int_0^x|F'(t)|dt$,
$$V_F'(0)=\lim_{x\to 0}\frac{1}{x}\int_0^x|2t\cos\frac{1}{t}+\sin\frac{1}{t}|dt.$$
Noting that
$$\frac{1}{x}\int_0^x|2t\cos\frac{1}{t}|dt\le \frac{2}{x}\int_0^xtdt=x,$$
we have 
$$V_F'(0)=\lim_{x\to 0}\frac{1}{x}\int_0^x|\sin\frac{1}{t}|dt\stackrel{s=\frac{1}{t}}{=}\lim_{x\to 0}\frac{1}{x}\int_{\frac{1}{x}}^\infty\frac{|\sin s|}{s^2}ds.$$
For every $k\in\mathbb{N}$, $\int_{k\pi}^{(k+1)\pi}|\sin s|ds=2$, so 
$$\frac{2}{\pi^2}(\frac{1}{k+1}-\frac{1}{k+2})<\frac{2}{(k+1)^2\pi^2}<\int_{k\pi}^{(k+1)\pi}\frac{|\sin s|}{s^2}ds<\frac{2}{k^2\pi^2}<\frac{2}{\pi^2}(\frac{1}{k-1}-\frac{1}{k}).$$
Therefore, when $n\pi\le\frac{1}{x}\le(n+1)\pi$,
\begin{eqnarray*}
\frac{2n}{(n+2)\pi}&=&\frac{2n}{\pi}\sum_{k=n+1}^\infty (\frac{1}{k+1}-\frac{1}{k+2})<n\pi\sum_{k=n+1}^\infty\int_{k\pi}^{(k+1)\pi}\frac{|\sin s|}{s^2}ds<\frac{1}{x}\int_{\frac{1}{x}}^\infty\frac{|\sin s|}{s^2}ds \\
&<&(n+1)\pi\sum_{k=n}^\infty\int_{k\pi}^{(k+1)\pi}\frac{|\sin s|}{s^2}ds<\frac{2(n+1)}{\pi}\sum_{k=n}^\infty(\frac{1}{k-1}-\frac{1}{k})=\frac{2(n+1)}{(n-1)\pi}.
\end{eqnarray*}
As $x\to 0$, $n\to\infty$, it follows that $V_F'(0)=\frac{2}{\pi}$.
