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I'm asked by the problem what real number $q$ makes the integral

$$\int _0 ^\infty \frac { \sin (x ^q ) } { x } \, dx$$

converges.

Using integral by parts I showed that the integral converges for all $q > 0$ (I'm not very sure about my proof). And I think the integral will not converge if $q \le 0 $. But I have trouble showing this. Any help is appreciated.

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    $\begingroup$ For $q = 0$, it's easy to see that the integral doesn't converge. For $q \neq 0$, you could substitute $u = x^q$. $\endgroup$ – Daniel Fischer Jan 3 '16 at 22:41
  • $\begingroup$ $\displaystyle\int_0^\infty\frac{\sin\big(x^q\big)}x~dx~=~\frac\pi{2|q|}$ $\endgroup$ – Lucian Jan 4 '16 at 6:26
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  • Assume $q>0$.

    Then, by the change of variable $u=x^q$, we have $\log u =q \log x$ thus $\dfrac{du}u=q\:\dfrac{dx}x$, we find that $$ \int_0^\infty \frac{\sin(x^q)}{x}\, dx=\frac1q\int_0^\infty \frac{\sin u}{u}\, du=\frac1q\int_0^\infty \frac{1-\cos u}{u^2}\, du $$ where we have integrated by parts in the last step, the latter integral converges.

  • Assume $q<0$.

    Then, by the change of variable $u=x^q$, we have $\log u =q \log x$ thus $\dfrac{du}u=q\:\dfrac{dx}x$, we find that $$ \int_0^\infty \frac{\sin(x^q)}{x}\, dx=-\frac1q\int_0^\infty \frac{\sin u}{u}\, du=-\frac1q\int_0^\infty \frac{1-\cos u}{u^2}\, du $$ where we have integrated by parts in the last step, the latter integral converges.

  • Assume $q=0$. Then the inital integral diverges.

Remark. The two first cases can be handled at the same time.

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