Convergence/divergence of $\int_0^1 \frac{\sin(\frac{1}{x})}{\sqrt{x}}dx$?

Question

Does the following converge or diverge:

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

I was thinking simply

$$\int_0^1 \bigg |\frac{\sin(\frac{1}{x})}{\sqrt{x}}dx \bigg | \le \int_0^1 \frac{1}{\sqrt{x}}$$ and the latter converges.

It can't be this simple though?

Answer 1

Substitute $\;u=\frac1x\;$ to get the improper integral

$$\int_\infty^1-\frac{du}{u^2}\,\sqrt u\,\sin u=\int_1^\infty\frac{\sin u}{u^{3/2}}du$$

and now use the comparison theorem to show absolute convergence:

$$\left|\frac{\sin u}{u^{3/2}}\right|\le\frac1{u^{3/2}}\implies\;\;\text{since}\;\;\int_1^\infty\frac{du}{u^{3/2}}=\left.-\frac2{\sqrt u}\right|_1^\infty=2\;\;\;\text{converges}$$

then so does our integral absolutely