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

Question

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 $\infty$?

Answer 1

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 $|f(x)|\leqslant 1/x^{3/2}$.