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I came across an integral $\int_0^\infty x^\alpha \cos x\, dx.$ Does this integral exists for any real value of $\alpha$?

Gregory Grant
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OKPALA MMADUABUCHI
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    Isn't it obvious that, for example, it does not exist for $;\alpha=0;$ ? What have you done so far in this? – Timbuc May 12 '15 at 22:32
  • In which sense exist? Riemann or Lebesgue? – Lukas Betz May 12 '15 at 22:38
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    If you are curious about Lebesgue integration, see my related question and answer: http://math.stackexchange.com/questions/764841/for-what-values-of-alpha-beta-is-x-alpha-sinx-beta-in-l10-1. – Alex Schiff May 12 '15 at 22:39

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Hint: you can only have problems near $0$ or near $\infty$. Near $0, \cos x \approx 1$, so you can ignore it. What values of $\alpha$ are trouble? Near $\infty$, the oscillation of $\cos x$ is a problem, so you need $x^\alpha$ to control that. What values of $\alpha$ will work for this purpose? What values of $\alpha$ work at both ends?

Ross Millikan
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HINT 1:

Abel's test can be used to show that $\int_1^{\infty}\frac{\cos x}{x^a}dx$ converges for $0<a$.

HINT 2:

$\int_0^{1}\frac{\cos x}{x^a}dx$ converges for $a<1$

Mark Viola
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