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How to show $\int_0^{\infty} \frac{1}{\sqrt{x}}\sin({\frac{1}{x}})dx$ converges?

I have that $$\frac{-1}{\sqrt{x}}\le \frac{\sin({\frac{1}{x}})}{\sqrt{x}} \le \frac{1}{\sqrt{x}}$$ but when you integrate you get

$$-\infty \le \int_0^{\infty} \frac{1}{\sqrt{x}}\sin({\frac{1}{x}})dx \le \infty$$

Is there another function you could bound it by to show convergence at $\infty$?

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  • $\begingroup$ It's not enough that $\sin(1/x)$ is bounded for large $x$; you need it to go to zero (and fairly quickly) for the integral to converge. $\endgroup$ – vadim123 Feb 1 '15 at 18:38
  • $\begingroup$ Yes there is and it's fairly similar to your approach. Just note that $\sin(x)\leq x$ for all $x\geq 0$. $\endgroup$ – AndreGSalazar Feb 1 '15 at 23:08
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Define $f(x):= \frac{ \sin(1/x) } { \sqrt x} $.

  • If $0\lt x \lt 1$, then $|f(x)|\leqslant 1/\sqrt x$ and $\int_0^1 1/\sqrt x\mathrm dx$ is convergent.
  • If $x\gt 1$, due to the inequality $|\sin t|\leqslant |t|$ for any $t$, we have $|f(x)|\leqslant 1/x^{3/2}$.
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  • $\begingroup$ Thank you very much. I have one doubt and please help me. $|f(x)|\leqslant 1/\sqrt x$ taken only for $0\lt x \lt 1$. Does it hold for $x>1$? Also does $|\sin t|\leqslant |t|$ hold for $0\lt x \lt 1$? $\endgroup$ – rama_ran Apr 9 '18 at 16:43
  • $\begingroup$ The inequality you mention are true for all positive $t$, $x$, but not always useful. $\endgroup$ – Davide Giraudo Apr 9 '18 at 17:33

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