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Find all $\alpha \in \mathbb{R}$ such that $\int_0^\infty \sin(x^\alpha)dx$ converges.

There is an answer here that differs from mine (they claim for $-\infty<\alpha<-2$ and $1<\alpha<\infty$, but I'm not trusting it...they make an error in the third indented line of the solution...they claim that $$\int_0^\infty C u^\beta du$$ is finite for some $\beta$, which is false. Maybe it's just a slip of the pen, and they have the right answer, but here is my analysis:

My answer: I claim it is convergent iff $\alpha < -1$. (Update: iff $|\alpha|>1$.)

Case 1: $\alpha>0$:

As $x\to 0$, sine goes to zero, and we have that $\sin(x^\alpha)= x^\alpha + O(x^{3\alpha})$. Therefore we are fine at zero iff $\alpha<1$. Now to examine behavior at infinity, make the substitution $u=x^\alpha$, and $$\int_0^\infty \sin(x^\alpha)dx=\frac{1}{\alpha}\int_0^\infty\sin(u)u^{\frac{1-\alpha}{\alpha}}du.$$ As $x\to \infty$, also $u \to\infty$, so we consider the behavior at infinity of the above integral. By the Dirichlet test, it will converge iff $\frac{1-\alpha}{\alpha}<0$, which, along with $\alpha>0$, gives the condition $\alpha>1$. So it works for no $\alpha>0$, and since $\alpha =0$ is a non-starter, we are reduced to the case $\alpha<0$.

Case 2: $\alpha<0$: As $x\to \infty$, $x^{\alpha}\to 0$, so as $x\to\infty$, $\sin(x^\alpha)=x^{\alpha}+O(x^{3\alpha})$. So we have convergence as $x\to \infty$ iff $\alpha <-1$. Now to examine behavior at zero, again integrate by parts and get $$\int_0^\infty \sin(x^\alpha)dx=\frac{1}{\alpha}\int_0^\infty\sin(u)u^{\frac{1-\alpha}{\alpha}}du,$$ only this time as $x\to 0$, $u\to \infty$. So we find that it is finite as $x\to\infty$ iff $\frac{1-\alpha}{\alpha}$, which is always true when $\alpha<0$. Therefore it is finite iff $-\infty < \alpha < -1$.

Does this work? Can you see any errors in my analysis. I was confident before I read the other answer. Unfortunately Wolfram Alpha runs out of memory trying to check this..

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  • $\begingroup$ Yes, there are some mistakes on the page that you linked there. They are using the limit comparison, but they should be doing it for $1$ (or some positive number) to $\infty$, and the conclusion should be that $\alpha < -1$, not $\alpha <-2$. $\endgroup$ – Braindead Jan 4 '14 at 19:51
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    $\begingroup$ The case $\alpha=2$ is quite famous, it's called Fresnel integral, and it's related, through Euler's formula, to the Gaussian integral. $\endgroup$ – Lucian Jan 4 '14 at 23:10
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Therefore we are fine at zero iff $\alpha < 1$.

No, that is fine for all $\alpha > 0$ the sine is bounded for real arguments, so there never can be any problem at $0$. Only the behaviour at $\infty$ needs to be considered.

Your examination of that was correct, so the integral converges (as an improper Riemann integral) for $\alpha > 1$.

For $\alpha < 0$, we can on the one hand observe that again by boundedness of the sine, the behaviour at $0$ is harmless, and for large $x$, we have

$$\sin (x^\alpha) \sim x^\alpha,$$

whence the integral converges for $\alpha < -1$ and diverges for $-1\leqslant \alpha$. We can also make the substitution $t = x^{-1}$ to check that, the substitution leads to

$$\int_0^\infty \frac{\sin (t^{\lvert\alpha\rvert})}{t^2}\,dt,$$

and here the behaviour at $\infty$ is totally harmless due to the boundedness of the sine, and to have integrability at $0$, we must compensate the singularity of $t^{-2}$, and for that we need $\lvert\alpha\rvert > 1$, so this also leads to the

Conclusion:

$$\int_0^\infty \sin (x^\alpha)\,dx$$

converges if and only if $\lvert\alpha\rvert > 1$ (for $\alpha\in\mathbb{R}$, of course).

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  • $\begingroup$ Thanks! I don't know what I was thinking with the $\alpha<1$ thing. :) $\endgroup$ – Eric Auld Jan 4 '14 at 19:51

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