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I need to prove that $$ \lim\limits_{x\to 0^+}\int\limits_{x/2}^{x}\frac{\cos \frac{1}{t}}{t} \,\mathrm{d}t=0 \;, $$ but I have no idea how to prove it. I have tried substitution by letting $y=\frac{1}{t}$, but it doesn't seem to work. Also, I have tried to make a bound of $\cos \frac{1}{t}$, but as $t\to0^+$, $\cos \frac{1}{t}$ oscillates, so that it is of no use. How can I prove it?

Thank you.

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  • $\begingroup$ L'Hospital should help here $\endgroup$
    – tired
    Commented Nov 25, 2016 at 11:14

1 Answer 1

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Hint. The substitution $s=1/t$ and $y=1/x$ is fine. We have that $$\lim\limits_{x\to 0^+}\int\limits_{x/2}^{x}\frac{\cos \frac{1}{t}}{t} \,\mathrm{d}t=\lim\limits_{y\to +\infty}\int\limits_{y}^{2y}\frac{\cos s}{s} \,\mathrm{d}s\\ =\lim\limits_{y\to +\infty}\int\limits_{1}^{2y}\frac{\cos s}{s} \,\mathrm{d}s-\lim\limits_{y\to +\infty}\int\limits_{1}^{y}\frac{\cos s}{s} \,\mathrm{d}s=L-L=0$$ where $$L=\int\limits_{1}^{+\infty}\frac{\cos s}{s} ds=\left[\frac{\sin s}s\right]_{1}^{+\infty}+\int_{1}^{+\infty}\frac{\sin s}{s^2}ds<+\infty$$ because $$\int_{1}^{+\infty}\frac{|\sin s|}{s^2}ds\leq \int_{1}^{+\infty}\frac{1}{s^2}ds=1$$ and absolute convergence implies convergence.

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  • $\begingroup$ Thank you for your answer, but as far as I have tried, I can't get how $L$ converges. All I can get is $-1-\sin1 < L < 1-\sin1$, which could possibly indicates that L oscillates between those two endpoints. $\endgroup$ Commented Nov 25, 2016 at 13:16
  • $\begingroup$ Note that if an integral is absoluteley convergent then it is convergent $\endgroup$
    – Robert Z
    Commented Nov 25, 2016 at 13:18
  • $\begingroup$ see en.wikipedia.org/wiki/… $\endgroup$
    – Robert Z
    Commented Nov 25, 2016 at 13:20

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