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


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

*If $\alpha >0,$ make the change of variables $x=y^{1/\alpha}.$
A: 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 $\alpha=0$.
A: 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 $g(x) = \frac{1}{x^{\alpha}}$


*

*Function F(x) = $\int_{1}^{x} f(t)  \mathrm{dt}$ bounded for every $x \in [1, \infty)$

*g in monotonic

*$\lim_{x\to\infty} g(x) = 0$ 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 $\alpha < 0$. 
