# Proof of $\lim\limits_{x\to 0^+}\int\limits_{x/2}^{x}\frac{\cos \frac{1}{t}}{t} \,\mathrm{d}t = 0$

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.

• L'Hospital should help here Commented Nov 25, 2016 at 11:14

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.
• 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. Commented Nov 25, 2016 at 13:16