I am currently studying Lebesgue integrability and encountered the following problem:
For which values of $a$ is the following integral Lebesgue integrable:
$$\int_{0}^{\infty}\frac{\sin(x)}{x^a}dx.$$
Of course this is equivalent to showing that the absolute value of the integrand converges.
Here's my approach and I would like to know if it is correct.
1.) $a\leq 0$:
For $a\leq 0$ one can estimate the integral in the following (standard) way $$\int_{k\pi}^{(k+1)\pi}\frac{|\sin(x)|}{x^a}dx>\frac{1}{((k+1)\pi)^a}\int_{k\pi}^{(k+1)\pi}|\sin(x)|dx=\frac{2}{((k+1)\pi)^a}.$$ Summing this over values of 0 to $\infty$ for $a\leq 0$ will diverge. Hence, for a\leq 0 not Lebesgue integrable.
2.) $a>0$ First I split the integral into two parts: $$\int_{0}^{\infty}\frac{|\sin(x)|}{x^a}dx=\int_{0}^{1}\frac{\sin(x)}{x^a}dx+\int_{1}^{\infty}\frac{|\sin(x)|}{x^a}dx=:I_1+I_\infty.$$
Consider first $I_1$.
This integral can be estimated in the following way: As $$\frac{Sin(x)}{x}\rightarrow 1$$ for $$x\rightarrow 0$$ the integrand effectively goes like $1/x^{a-1}$ and this is only convergent on [0,1] if $1<a<2$ (which can again be shown be standard means), otherwise it will blow up and the integral not Lebesgue integrable.
Thus it remains to show that $I_\infty$ is convergent for the range $1<a<2$ but this is obvious, as $$I_\infty\leq \int_{1}^{\infty}\frac{1}{x^a}dx<\infty$$
Finally: I have shown that
$$\int_{0}^{\infty}\frac{\sin(x)}{x^a}dx$$ is Lebesgue integrable for $1<a<2$.
Is my reasoning correct and mathematically bulletproof?