Tail estimation for $\int{\cos\sqrt{x}}\cdot \frac{1}{x}dx$

Question

How to prove or disprove $$\int_x^{+\infty}{\frac{\cos\sqrt x}{x}}dx \leq\frac{K}{\sqrt{x}}$$ as $x$ goes to $+\infty$.

Answer 1

For starters, you should always use different variables for clarity like so:

$$\int_x^{+\infty}\frac{\cos(\sqrt t)}t~\mathrm dt\le\frac K{\sqrt x}$$

Secondly I presume you mean to ask more specifically whether the integral is $\mathcal O(1/\sqrt x)$. To see this, first substitute $t\mapsto t^2$ and integrate by parts to get

$$\int_x^{+\infty}\frac{\cos(\sqrt t)}t~\mathrm dt=\int_{\sqrt x}^{+\infty}\frac{2\cos(t)}t~\mathrm dt=-\frac{2\sin(\sqrt x)}{\sqrt x}+\int_{\sqrt x}^{+\infty}\frac{2\cos(t)}{t^2}~\mathrm dt$$

By directly integrating you can show $\displaystyle\int_{\sqrt x}^\infty\frac1{t^2}~\mathrm dt$ is $\mathcal O(1/\sqrt x)$, and hence the original integral is $\mathcal O(1/\sqrt x)$.