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..

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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


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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.