# Does the improper integral converges?

Proof the convergenсe of the integral: $\int\limits_{1}^{\infty} \sin (x \ln x) d x.$ Tried to change variables, no result. May you suggest something? And may you also suggest some books with a focus on examples? Feeling trouble with this type of integrals.

• Are you sure this integral converges? It looks suspiciously divergent... – DonAntonio Mar 9 '17 at 22:48
• Sure. It is stated that integral converges conditionally. – Kamil Mar 9 '17 at 22:49
• I doubt it since $\;x\log x\xrightarrow[x\to\infty]{}\infty\implies \sin x\log x\;$ wobbles between $\;-1\;$ and $\;1\;$ ... But perhaps it converges. And who is "it" that stated the integral converges conditionally? – DonAntonio Mar 9 '17 at 22:51
• I mean that it is written in answers for this task. – Kamil Mar 9 '17 at 22:52
• Have you considered integration by parts? – πr8 Mar 9 '17 at 23:04

## 1 Answer

Hint: Let $f(x) = x\ln x.$ Then $f'(x) \ge 1,$ so $f$ is a nice 1-1 map from $[1,\infty)$ t0 $[0,\infty).$ Make the change of variables $x= f^{-1}(y).$ You get

$$\int_0^\infty (\sin y)(f^{-1})'(y)\,dy.$$

You are set up for Dirichlet's test.

• But for Dirichlet's test you need $(f^{-1})'(y)$ to be monotonous and $(f^{-1})'(y) \to 0$. Why is that true? – Kamil Mar 9 '17 at 23:05
• There's a well known formula for the derivative of the inverse. Time to dust it off. – zhw. Mar 9 '17 at 23:09
• Get it! Thanks! – Kamil Mar 9 '17 at 23:10
• Any reason for the down vote? – zhw. Mar 10 '17 at 16:37
• I voted for your answer. – Kamil Mar 10 '17 at 18:24