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I'm trying to show that $f:(0,1)\rightarrow \mathbb{R}$ given by $f(x)=\frac{1}{x}\sin(\frac{1}{x})$ is not Lebesgue integrable.

My initial thought is coming up with a smaller function $g(x)\leq |f(x)|$ and showing that $\int_{(0,1)}g(x)=\infty$.

I was thinking $g(x)=|\sin(\frac{1}{x})|$, then, $\int_{(0,1)}|\sin(\frac{1}{x})|=[x^2\cos(\frac{1}{x})]_0^1=1$ which doesn't help.

Can anyone help me?

Gregoire Rocheteau
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2 Answers2

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At an intuitive level, $\sin(\frac1x)$ fluctuates near $x\approx0$, such that roughly one third of the time, $\sin(\frac1x)>\frac12$. So the integral diverges at least as badly as $\frac{1}{6x}$. You can turn this into a precise argument by computing the region where $\sin(\frac1x)>\frac12$ and comparing the integral of $\frac1x$ there with the harmonic series.

Chris Culter
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Hint: For $n=1,2,\dots, $ let $a_n=1/(2n\pi + \pi/2),$ $b_n=1/(2n\pi + \pi/4).$ On the intervals $[a_n,b_n],$ which are disjoint, $\sin(1/x)\ge 1/\sqrt 2.$

zhw.
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  • Would you elaborate a little further? So to my understanding, $\int_{0}^{1}\frac{1}{x}|\sin(1/x)dx\geq\sum_{n=1}^{\infty}\int_{a_{n}}^{b_{n}}\frac{1}{x}|\sin(1/x)|dx$ – Gregoire Rocheteau May 07 '18 at 20:18
  • Right, now find a lower bound for the integrand on that interval and multiply by $b_n-a_n$ then add them up. – zhw. May 07 '18 at 20:54
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    Since both $a_n$ and $b_n$ are less than 1, the lower bound will be $sin(1/x)$, which is greater than or equal to $\frac{1}{\sqrt{2}}$, we have $\sum_{n=1}^{\infty}\int_{a_{n}}^{b_{n}}\frac{1}{x}|\sin(1/x)|dx\geq\sum_{n=1}^{\infty}|\sin(1/x)|\cdot(b_n-a_n)\geq\sum_{n=1}^{\infty}\frac{1}{\sqrt{2}}\cdot(b_n-a_n)=\infty$ am I missing something? – Gregoire Rocheteau May 07 '18 at 21:47
  • No, you left out an estimate for $1/x$ – zhw. May 07 '18 at 21:58
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    What would the estimate be? my guess is the lowerbound of the $1/x$ which is $1/b_n$ so would it be $1/b_n\cdot(b_n-a_n)$? – Gregoire Rocheteau May 07 '18 at 22:03
  • You need to decide if that works. – zhw. May 07 '18 at 22:08