The direct way to prove the convergence of an integral

Question

For the integral $$\int_1^{\infty} \frac{\cos(2x)}{2x}$$

It is possible to prove the convergence by the integration by parts but I am looking for another direct way which could be simple and more clever.

For example, to benefit from the fact $|\cos x| \leq 1$ then to do something like $$ \int_1^{\infty} \frac{\cos(2x)}{2x} = \int_1^{\infty} \frac{\cos(y)}{y} \leq \int_1^{\infty} \frac{1}{y} $$ but I can't conclude something from this!

Answer 1

Direct application of Dirichlet's test works: antiderivative of $\cos 2x$ is bounded and $\frac{1}{2x}$ monotonically tends to zero. That's why $$ \int_1^{\infty} \frac{\cos(2x)}{2x} $$ converges.

Answer 2

Consider the contributions of successive half periods,

$$\int_{k\pi}^{k\pi+\pi/2}\frac{\cos 2x}{2x}dx$$ and

$$\int_{k\pi+\pi/2}^{k\pi+\pi}\frac{\cos 2x}{2x}dx.$$

As the signs alternate, the sum equals

$$\int_{k\pi}^{k\pi+\pi/2}\cos 2x\left(\frac1{2x}-\frac1{2x+\pi}\right)dx=\int_{k\pi}^{k\pi+\pi/2}\frac{\pi\cos 2x}{2x(2x+\pi)}dx.$$

Now you easily see that these terms are bounded by $\dfrac1{4\pi k^2}$ and the global sum converges.