Prove $\int_1^\infty \frac{\cos(\sqrt{x})}{x} dx$ converges

Question

I want to use the Dirichlet criterion, i.e. if $f(x)$ integrable on $(a,A)$ for all $A>a$, $\int_{(a,A)} f(x) dx$ is finite and $g(x)$ is monotone and $\to 0$, then $\int f(x)g(x)$ converges. It should work with $f(x)=\cos(\sqrt{x})$ and $g(x)=1/x$. But how can I prove $\int_{(a,A)} \cos(\sqrt{x}) dx$ is bbd? I mean intuitively I can obviously see it, but how to write it down formally?

Answer 1

I suggest you do the following trick first:

$t=\sqrt x\implies dt=\frac{1}{2\sqrt x} dx\implies 2tdt=dx\implies\int_{1}^{\infty}\frac{\cos(\sqrt x)}{x}dx=\int_{1}^{\infty}2\frac{\cos(t)}{t}dt$

You could use Dirichlet test for this integral rather than the original

$$\int_{1}^{\infty}\frac{\cos(t)}{t}dt$$

Boundness of $\int_{1}^{r} \cos(t)dt$ is trivial as this integral is $\sin(r)-\sin(1)$ which is between $-2$ and $2$. You can now use Dirichlet's test.

Answer 2

Let's use by parts criterion with

$$u(x)=\frac{2}{\sqrt{x}}$$

$$v'(x)=\frac{\cos(\sqrt{x})}{2\sqrt{x}}$$

$$\lim_{x\to+\infty}u(x)v(x)=0$$

thus

$$\int^\infty u(x)v' (x)dx\text{ and } \int^\infty u'(x)v(x)dx$$ have same nature.

the second is absolutely convergent since $$|u'(x)v(x)|=|\frac{\sin(\sqrt{x})}{x^\frac32}|\le \frac{1}{x^\frac32}$$