To determine whether the integral $\int_0^{\infty} \frac{\sin{(ax+b)}}{x^p} \,\mathrm dx$ converges for $p>0$ 
If $p >0$, determine if the following integral converges:
  $$\int_1^{\infty} \frac{\sin{(ax+b)}}{x^p} \,\mathrm dx$$

What i did so far is:


*

*If $p>1$, then $$f(x)= \frac{|\sin{(ax+b)}|}{x^p}<\frac{1}{x^p}$$ hence the integral converges.

*If $0<p\le1$, The above is false, And then what can i do?

 A: Since you know that $$\frac{|\sin{(ax+b)}|}{x^p}<\frac{1}{x^p}$$ you can just consider $\frac{1}{x^p}$. If $\frac{1}{x^p}$ converges, so will $\frac{|\sin{(ax+b)}|}{x^p}$. So, lets integrate $\frac{1}{x^p}$:
$$\int_1^{\infty} \frac{1}{x^p} \,\mathrm dx=\lim_{a \to +\infty}(\int_1^{\infty} \frac{1}{x^p}dx)=\lim_{a \to +\infty}\Big(\frac{1}{1-p}x^{1-p}|_1^a\Big)=\lim_{a \to +\infty}\Big(\frac{1}{1-p}(a^{1-p}-1^{1-p})\Big)=$$
So, limit of one to any power is one. The only tricky thing here is that there is a negative sign in front of one, but we can get rid of it. As a result the last expression above reduces to:
$$=\lim_{a \to +\infty}\Big(\frac{1}{1-p}(a^{1-p})\Big)+\frac{1}{p-1}$$
Forgot to mention that $\frac{|\sin{(ax+b)}|}{x^p}<\frac{1}{x^p}$ holds regardless of whether p is greater or smaller than 1 because $|\sin{(ax+b)}|<1$ always. So, now you have to work with this limit only. If $p>1$, then the limit equals zero and we are left with $\frac{1}{p-1}$ which means the integral converges (to $\frac{1}{p-1})$. If $p<1$ then the limit goes to infinity (diverges) and the integral diverges. 
To see why the integral in question converges for $p<1$ read Baby Dragon's comment.
A: Let us consider the integral, $\int_0^\infty\frac{\sin x}{x^p}dx$ (I changed the lower limit of integration. Note that the function $\frac{\sin x}{x^p}$ has a limit at zero if $0<p\leq1$). We may rewrite this integral as $$\Sigma_{n=0}^{\infty}\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x^p}dx.$$ Note that this is an alternating series so the 
alternating series test  applies. We must show that

1 $|\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x^p}dx|>|\int_{(n+1)\pi}^{(n+2)\pi}\frac{\sin x}{x^p}dx|$
2 $\lim_{n\to\infty}\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x^p}dx=0$

Let us start with (1).
Let us assume that $\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x^p}dx>0$ (the other case is similar). Note that in this case that  $\frac{\sin x}{x^p}\geq 0$. This also means that $\int_{(n+1)\pi}^{(n+2)\pi}\frac{\sin x}{x^p}dx<0$, so
$$|\int_{(n+1)\pi}^{(n+2)\pi}\frac{\sin x}{x^p}dx|=-\int_{(n+1)\pi}^{(n+2)\pi}\frac{\sin x}{x^p}dx$$ We will now make a change of variables, $u=x-\pi$, which makes
$$-\int_{(n+1)\pi}^{(n+2)\pi}\frac{\sin x}{x^p}dx=-\int_{(n)\pi}^{(n+1)\pi}\frac{\sin (u+\pi))}{(u+\pi)^p}du=\int_{(n)\pi}^{(n+1)\pi}\frac{\sin x}{(x+\pi)^p}dx$$ (note that $\int_a^bf(x)dx=\int_a^bf(u)du$). Again note that $\frac{\sin (x+\pi))}{(x+\pi)^p}>0$ in the interval $[n\pi,(n+1)\pi]$. We must now compare the integrals, $$\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x^p}dx\mbox{ and }\int_{n\pi}^{(n+1)\pi}\frac{\sin (x+\pi))}{(x+\pi)^p}dx. $$
Since $x^p<(x+\pi)^p$, we get that
$$\frac{\sin x}{x^p}>\frac{\sin (x)}{(x+\pi)^p}. $$ Therefore
$$\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x^p}dx>\int_{n\pi}^{(n+1)\pi}\frac{\sin (x+\pi))}{(x+\pi)^p}. $$, which shows (1).
Now to show (2) simply note that $|\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x^p}dx|\leq\int_{n\pi}^{(n+1)\pi}\frac{1}{x^p}dx$ and that $0<\int_{n\pi}^{(n+1)\pi}\frac{1}{x^p}dx$. An elementary calculation will $\lim_{n\to\infty}\int_{n\pi}^{(n+1)\pi}\frac{1}{x^p}dx=0$
A: For simplicity, we can just set $a=1$ and $b=0$.
If $0<p\le1$, we can use Dirichlet's test.
Because $x^{-p}$ is monotonically decreasing and tends to $0$, and $|\int_s^t \sin x\ dx|\le 2$ for arbitrary $t>s\ge1$, then $\int_1^{\infty}x^{-p}\sin x\ dx$ converges.
Moreover, we can prove that this integral is conditionally convergent. 
$\int_1^{\infty}x^{-p}|\sin x|\ dx \ge \int_1^{\infty}x^{-p} (\sin x)^2\ dx =\frac{1}{2}\int_1^{\infty}x^{-p}[1-\cos (2x)]\ dx$ 
Using Dirichlet's test again, we know that $\int_1^{\infty}x^{-p}\cos (2x)\ dx$ is convergent.
However, $\int_1^{\infty}x^{-p}\ dx$ is divergent, making the whole integral $\int_1^{\infty}x^{-p}|\sin x|\ dx$ is divergent.
Therefore, we can conclude that:


*

*If $p>1$, the integral is absolutely convergent.

*If $0<p\le 1$, the integral is conditionally convergent. 

