Prove that the improper integral $\int_0^{+\infty}\sin(\frac{\cos(x)}{\sqrt{x}})dx$ is convergent.

Question

The integral $$\int_0^{+\infty}\sin(\frac{\cos(x)}{\sqrt{x}})dx$$
is convergent near $ 0$ since the integrand is bounded.

But near $+\infty $, i cannot use comparison test, as the sign is changing.
Thanks in advance for any idea.

Answer 1

Using the inequality $|\sin t-t| \le |t|^3/6$ for $|t|\le 1$, we see that \begin{align*} \bigg| \int_1^X \sin\bigg( \frac{\cos x}{\sqrt{x}} \bigg) \,dx - \int_1^X \frac{\cos x}{\sqrt{x}} \,dx \bigg| &\le \int_1^X \bigg| \sin\bigg( \frac{\cos x}{\sqrt{x}} \bigg) - \frac{\cos x}{\sqrt{x}} \bigg| \,dx \\ &\le \frac16 \int_1^X \bigg| \frac{\cos x}{\sqrt x} \bigg|^3\,dx \le \int_1^X \frac1{6x^{3/2}}\,dx \end{align*} which converges. Therefore the integrals $\int_1^\infty \sin\big( \frac{\cos x}{\sqrt{x}} \big) \,dx$ and $\int_1^\infty \frac{\cos x}{\sqrt{x}} \,dx$ either both converge or both diverge.

There's a nice trick that sometimes works for showing that an integral of an oscillating integrand converges: integrate by parts, integrating the oscillating part of the integrand (which often doesn't increase its size significantly) while differentiating the monotone part (which often reduces its size significantly). Here, $$ \int_1^\infty \frac{\cos x}{\sqrt{x}} \,dx = \frac{\sin x}{\sqrt{x}} \bigg|_1^\infty - \int_1^\infty \frac{\sin x}{x^{3/2}} \,dx, $$ showing that this integral converges.