Prove that $\int\limits_1^{\infty} \frac{\cos(x)}{x} \, \mathrm{d}x$ converges

Question

Prove the convergence of

$$\int\limits_1^{\infty} \frac{\cos(x)}{x} \, \mathrm{d}x$$

First I thought the integral does not converge because

$$\int\limits_1^{\infty} -\frac{1}{x} \,\mathrm{d}x \le \int\limits_1^{\infty} \frac{\cos(x)}{x} \, \mathrm{d}x$$

But in this case

$$\int\limits_1^{\infty} \frac{\cos(x)}{x} \, \mathrm{d}x \le \int\limits_1^{\infty} \frac{1}{x^2} \, \mathrm{d}x$$

it converges concerning the majorant criterion. What's the right way?

Answer 1

You might want to use integration by parts, obtaining for $M\ge1$, $$ \int_{1}^M \frac{\cos x}{x}\: dx=\left[\frac{\sin x}{ x}\right]_1^M+ \int_1^M \frac{\sin x}{x^2}\: dx $$ letting $M \to \infty$ gives $$ \int_{1}^\infty \frac{\cos x}{x} \:dx=\lim_{M \to \infty}\int_1^M \frac{\cos x}{x} \:dx= -\sin 1+\int_1^\infty \frac{\sin x}{ x^2}\: dx $$then one may conclude by the absolute convergence of the latter integral: $$ \left|\int_1^\infty \frac{\sin x}{ x^2}\: dx\right|<\int_1^\infty \frac{|\sin x|}{ x^2}\: dx<\int_1^\infty \frac{1}{x^2}\: dx<\infty. $$