# Finding values for which $\int_{1}^{\infty}\frac{\sin (x^\alpha)}{x} \mathrm{dx}$ converges

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 $$x=y^{1/\alpha}.$$

• I don't like it because we can use comparison only with non-negative or non-positive functions – Eugene Sirkiza Jul 1 '19 at 14:36
• @EugeneSirkiza In case 1, the integrand is positive. – zhw. Jul 1 '19 at 14:46
• I see. Thank you. – 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 $$\alpha < 0$$.

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