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In this question, a comment from Lucian asserts that the convergence of the integral $$ I=\int_0^\infty\sin{(x^4)} dx $$ is due to the Riemann-Lebesgue lemma. However, I don't immediately see how to apply this lemma.

I know that $$\lim_{n \to \infty} \int_0^{\infty} f(t)\sin(nt)dt = 0$$ whenever $f \in L^1(\Bbb R)$. Should I write $\sin{(x^4)}$ as a product, like $f(t)\sin(nt)$ ?

[By the way, a simplier way to see the convergence of the integral $I$ is that « the humps for $x\mapsto \sin(x^4)$ go up and down. Each has an area smaller than that of the last. The areas converge to $0$ as you progress down the $x$-axis. By the alternating series test, this converges. » (adapted from this answer)].

Pang
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Watson
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    very good question... i don't think riemann lebesgue is a good way to go here – tired Jul 25 '16 at 10:31
  • @Watson I think he meant Dirichlet test for integral. Isn't it? – Olivier Oloa Jul 25 '16 at 10:34
  • @OlivierOloa: this is possible. However, in Wikipedia, the integrals of such "oscillatory" functions can be proved to be convergent thanks to the Riemann–Lebesgue lemma, which "states that the integral of a function like the above is small. The integral will approach zero as the number of oscillations increases." – Watson Jul 25 '16 at 10:36
  • Related: http://math.stackexchange.com/questions/1106606/, http://math.stackexchange.com/questions/413950/ – Watson Jul 25 '16 at 10:39
  • Let me have a look. – Olivier Oloa Jul 25 '16 at 10:39
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    i am not an expert in the area, but heuritically it seems that the integrals subject to rl-lemma might be bounded by something like $C/n$ which would indeed can be used to show convergence of the integral in question – tired Jul 25 '16 at 10:40
  • @Watson I don't see Riemann-Lebesgue there. – Olivier Oloa Jul 25 '16 at 10:41
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    I think a direct way to prove convergence of the given integral is $x=u^{1/4}$ (as below) followed by integration by parts. – Olivier Oloa Jul 25 '16 at 10:46
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    My point was similar to the one made by the user NC Math Sadist, whose answer you quoted, namely that the integrand basically tessellates itself into smithereens, much like in (but not identical to) the case of the Riemann-Lebesgue lemma. – Lucian Jul 25 '16 at 19:08
  • @Lucian: ok, thank you. So this is not a formal application of the Riemann-Lebesgue lemma, if I understand well. – Watson Jul 25 '16 at 19:10
  • Apparently not. – Lucian Jul 25 '16 at 19:14

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By the change of variable $x=u^{1/4}$, $dx=\dfrac14u^{-3/4}du$ you get $$ \int_0^\infty\sin{(x^4)}dx=\frac14\int_0^\infty\frac{\sin{u}}{u^{3/4}}du $$ then it is clearer how to apply the Dirichlet test for integral.

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Olivier Oloa
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