Let $f$ be a function $f:[0,1]\to\mathbb{R}$, $f(0)=1$ and $f(x)= \sqrt{x}\cos(\frac{1}{x}) $, if $x \neq 0$. I have to find out the Lebesgue measure of the set: $$A=\{x\in[0,1]~~\mid~~ f(x)<0\}$$ I have tried this one: we have that $0,1\notin A$. Then $A\subseteq(0,1)$.If $\sqrt{x}\cos(1/x)<0$, this implies that $\cos(1/x)<0$. We have that $x \in(2/3\pi,2/\pi)\cup (2/7\pi,2/5\pi)\cup \ldots$ Then the measure of the set $A$ is the measure of the union of this intervals, which are disjoint. Then the measure of $A$ will be the sum of the lenghts of this intervals. So:$$\lambda(A)=\frac{2}{\pi}\sum(1-1/3+1/5-1/7+\ldots)$$ And now I dont know if I calculate this sum correctly. Let $S=\frac{2}{3\cdot 1}+\frac{2}{3\cdot 5}+\ldots$. We have that $S=1$, then the sum in that paranthesis is $1/2$? Please help me! Thank you very much!
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A derivation of the sum $1-1/3+1/5-1/7+\ldots = \frac{\pi}{4}$ can be found in this answer – Winther Jan 31 '16 at 17:07
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Exactly, I wrote the sum differently.. – user308560 Jan 31 '16 at 17:32