$\int _{2}^{+\infty }\ \dfrac{\sin x}{x\ln x }dx$ is convergent but not absolutely convergent

Question

I have known $\frac{\sin{x}}{x}\>\text{is convergent but not absolutely convergent(Conditional convergence)}$,but I don't know the connection of these two integrals. I was wondering how to prove the promblem: $$\int _{2}^{+\infty }\ \dfrac{\sin x}{x\ln x }dx$$is convergent but not absolutely convergent.

I know that it can be shown to be convergent using integral by parts and trigonometric inequalities, but I don't know how to show that it is not absolutely convergent.

Answer 1

This is a very classic question. To prove the conditional convergence, note the following: \begin{align*} \int_2^{+\infty} \left\vert \frac {\sin (x)}{x \log (x)} \right\vert \mathrm dx &= \int_2^{+\infty} \frac {|\sin (x)|}{x \log (x)} \mathrm dx \\ &\geqslant \int_2^{+\infty} \frac {\sin (x)^2}{x \log (x)} \mathrm dx \\ &= \int_2^{+\infty} \frac {1}{x \log (x)} \mathrm dx - \frac 12 \int_2^{+\infty} \frac {\cos(2x)}{x \log (x)} \mathrm dx, \end{align*} and $\int_2^\infty \mathrm dx/(x \log(x))$ diverges by direct computation, while $\int_1^\infty \cos(2x)\,\mathrm dx/x \log(x)$ converges by either integration by parts or the famous Dirichlet's test, where $\cos(2x)$ has a bounded integral over closed intervals and $1/x \log(x)$ decreases to $0$ monotonically. Conclusively the original $\int_2^\infty |\sin(x)|\mathrm dx/x \log(x)$ is greater than a sum of a divergent integral and a convergent integral, and the original one diverges.