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Does $$\int_1^\infty \sin (x\log x) \ dx$$ converge or diverge?

I've tried some substitutions ($u = x\log x$, $u = \log x$). Also, I know that the integral $\int_1^\infty \sin x \ dx$ diverges, which leads me to think our integral behaves in the same manner.

egreg
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Elimination
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    http://math.stackexchange.com/questions/651951/does-the-integral-int-1-infty-sinx-log-x-mathrmdx-converge – StackTD Jan 21 '16 at 09:09
  • @Crostul, isn't that a risky theorem? since the integrand gets positive/negative values. – Elimination Jan 21 '16 at 09:15
  • @Crostul: That claim is false. $\int_0^\infty \sin(x^2)\ dx$ exists despite the integrand oscillating continuously back and forth between $0$ and $1$. See http://www.wolframalpha.com/input/?i=Integrate[Sin[x^2],{x,0,Infinity}] for an exact value! Edit: Crostul removed his comment; his claim was that the integral does not converge because the integrand does not converge to zero. I'll leave this here in case other people make the same mistake. – user21820 Jan 21 '16 at 09:19

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Hint: First prove that $f = ( x \mapsto x \ln(x) )$ is an increasing function on $[1,\infty)$. Next let $y$ be the increasing sequence such that $f(x) = n π$ for any $n \in \mathbb{N}^+$, and prove that $y_{n+1}-y_n \to 0$ as $n \to \infty$. Then since $f$ is increasing show that the sequence $A = ( \int_{y_n}^{y_{n+1}} \sin(f(x))\ dx )_{n\in\mathbb{N}^+}$ is alternating with absolute value going to zero. Thus the integral $\int_1^\infty \sin(f(x))\ dx$ converges, as it is bounded between adjacent partial sums of $A$.

user21820
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  • This $y_n$ sequence exists because $f$ is a continuous function, though I don't know how to explicitly isolate $y_n$ from $y_n\cdot \ln y_n = n\frac{\pi}{3}$. – Elimination Jan 21 '16 at 09:19
  • I guess it doesn't really matter if I can't explicitly express $y_n$ as long as we know there's such sequence. – Elimination Jan 21 '16 at 09:20
  • @Elimination: Right there is no need to explicitly express the sequence. All that is important is that $f$ is continuous as you said so that the intermediate value theorem works, and that it is monotonic so that the integral can be bounded. – user21820 Jan 21 '16 at 09:24
  • Why does the absolute value of the sequence not tend to zero? If I'm not mistaken, $y_{n+1}-y_n$ tends to $0$ and each integral is bounded by $1\times(y_{n+1}-y_n)$, hence tends to $0$ as well. – Wojowu Jan 21 '16 at 09:25
  • @Wojowu: Oh wait you are right; exactly the same reason as the example I gave in the comment. – user21820 Jan 21 '16 at 09:26
  • @Elimination: I've fixed my answer. Wojowu, thank you very much! – user21820 Jan 21 '16 at 09:32
  • How do you know that the integral's absolute value going to zero? is it because $(n+1)\pi - n\pi \to 0$? – Elimination Jan 21 '16 at 09:55
  • @Elimination: $| \int_{y_n}^{y_{n+1}} \sin(f(x))\ dx | \le \int_{y_n}^{y_{n+1}} |\sin(f(x))|\ dx$ – user21820 Jan 21 '16 at 09:56
  • Oops. I wrote a nonsense statement,but I still don't get it :/ – Elimination Jan 21 '16 at 10:06
  • @Elimination: What do you not get? The inequality I just wrote is a basic one, and also $|\sin(x)| \le 1$ for any $x \in \mathbb{R}$. – user21820 Jan 21 '16 at 10:08
  • Oh okay. So we have $\le \int_{y_n}^{y_{n+1}} 1 \ dx = y_{n+1} - y_n \to 0$ – Elimination Jan 21 '16 at 10:09
  • @Elimination: Yes and now check the sign of each integral. An alternating sequence with terms going to zero has convergent partial sums. Finally check that the in-between integrals are bounded between the adjacent partial sums. – user21820 Jan 21 '16 at 10:14
  • And the fact that $A_n$ is alternating sequence is a simple trigonometric fact. – Elimination Jan 21 '16 at 10:14
  • @Elimination: Right. – user21820 Jan 21 '16 at 10:15
  • Then, by Leibniz Theorem we conclude that $\sum A_n < \infty$. – Elimination Jan 21 '16 at 10:15
  • Many thanks @user21820! – Elimination Jan 21 '16 at 10:16