I'm trying to figure out for which values of $\alpha$ the improper integral converges:

$$ \int_{1}^{\infty}\frac{\sin (x^\alpha)}{x} \mathrm{dx}. $$

I figured I could use Dirichlet, by showing that $\int_{1}^{b}\sin (x^\alpha) \mathrm{dx}$ is bounded, but I'm having trouble showing that this is indeed the case.

My intuition tells me that while $\sin (x^\alpha)$ is not technically periodic, its "period" is growing in a predictable way, and that I can somehow divide up the integral into a sum of integrals which I can find an upper bound for.

I would appreciate only answers that use fairly elementary analysis theorems.


There's a simpler method:

If $\alpha>0$, then you can make substitution $y=x^{\alpha}$ to get $$ \int_1^\infty \frac{\sin(x^\alpha)}{x} dx = \int_1^\infty \frac{\sin y}{\alpha y} dy$$ which is convergent.

If $\alpha<0$, the same substitution gives $$ \int_1^\infty \frac{\sin(x^\alpha)}{x} dx = \int_0^1 \frac{\sin y}{|\alpha| y} dy$$ which is also convergent.

For $\alpha=0$ we have $$ \int_1^\infty \frac{\sin(x^\alpha)}{x} dx = \int_1^\infty \frac{\sin(1)}{x} dx $$ which is not convergent.


Hints: 1. If $\alpha <0$ then the integrand is $\sim 1/x^{1+|\alpha|}$ at $\infty.$

  1. If $\alpha =0,$ the integrand equals $\sin (1)/x.$

  2. If $\alpha >0,$ make the change of variables $x=y^{1/\alpha}.$

  • $\begingroup$ I don't like it because we can use comparison only with non-negative or non-positive functions $\endgroup$ – Eugene Sirkiza Jul 1 '19 at 14:36
  • $\begingroup$ @EugeneSirkiza In case 1, the integrand is positive. $\endgroup$ – zhw. Jul 1 '19 at 14:46
  • $\begingroup$ I see. Thank you. $\endgroup$ – Eugene Sirkiza Jul 1 '19 at 14:58

If $\alpha=0$, $$\int_{1}^{\infty}\frac{\sin (x^\alpha)}{x} \mathrm{dx} = \int_{1}^{\infty}\frac{\sin (1)}{x} \mathrm{dx}$$ diverges. If $\alpha>0$, Under $u=x^\alpha$ and integration by part, one has $$ \int_{1}^{\infty}\frac{\sin (x^\alpha)}{x} \mathrm{dx}=\frac1\alpha\int_1^\infty\frac{\sin u}{u}du=-\frac1\alpha\frac{\cos u}{u}\bigg|_1^\infty+\frac1\alpha\int_1^\infty\frac{\cos u}{u^2}du $$ which implies $$\int_{1}^{\infty}\frac{\sin (x^\alpha)}{x} \mathrm{dx}$$ Converges. If $\alpha<0$, $0<\sin (x^\alpha)\le x^\alpha$ in $[1,\infty)$ and hence $$0<\int_{1}^{\infty}\frac{\sin (x^\alpha)}{x} \mathrm{dx}\le \int_{1}^{\infty}\frac{x^\alpha}{x} \mathrm{dx}\le\int_{1}^{\infty}\frac{1}{x^{1-\alpha}} \mathrm{dx}<\infty. $$ So $0<\int_{1}^{\infty}\frac{\sin (x^\alpha)}{x} \mathrm{dx}$ converges if $\alpha\neq0$ and diverges if $\alpha=0$.


It's obvious, that when $\alpha=0$ integral doesn't converge. There is simple hint, how to solve this kind of problem.

$\int_{1}^{\infty}\frac{\sin (x^\alpha)}{x} \mathrm{dx} = \int_{1}^{\infty}\frac{x^{\alpha-1} \sin (x^\alpha)}{x^{\alpha}} \mathrm{dx}$

Now we need to apply Dirichlet theorem to functions:

$f(x) = x^{\alpha-1} \sin (x^\alpha)$ and $g(x) = \frac{1}{x^{\alpha}}$

  1. Function F(x) = $\int_{1}^{x} f(t) \mathrm{dt}$ bounded for every $x \in [1, \infty)$
  2. g in monotonic
  3. $\lim_{x\to\infty} g(x) = 0$ iff $\alpha > 0$

So integral converges if $\alpha > 0$

UPD: If $\alpha < 0$, we have $\frac{sin(x^\alpha)}{x} \sim {x^{\alpha-1}}$, which converges iff $\alpha < 0$.

  • $\begingroup$ The integral DOES converge for $\alpha<0$. $\endgroup$ – Mark Viola Jul 1 '19 at 14:32
  • $\begingroup$ @MarkViola, okey, i see. If have updated my post. Usually in this kind of tasks, when Dirichlet not work, there is no converge. $\endgroup$ – Eugene Sirkiza Jul 1 '19 at 14:56

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