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.

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    $\begingroup$ Are you sure this integral converges? It looks suspiciously divergent... $\endgroup$ – DonAntonio Mar 9 '17 at 22:48
  • $\begingroup$ Sure. It is stated that integral converges conditionally. $\endgroup$ – Kamil Mar 9 '17 at 22:49
  • $\begingroup$ 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? $\endgroup$ – DonAntonio Mar 9 '17 at 22:51
  • $\begingroup$ I mean that it is written in answers for this task. $\endgroup$ – Kamil Mar 9 '17 at 22:52
  • $\begingroup$ Have you considered integration by parts? $\endgroup$ – πr8 Mar 9 '17 at 23:04

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.

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

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