I am trying to solve this integral $\int_{1}^{\infty} \frac{\sin^3(x)}{x}$ to see if it converges or not with the $p$-test.
This is what I have so far :
$L = \lim_{x\to\infty}$ = $\frac{\frac{\sin^3(x)}{x}}{\frac{1}{x^p}}$ solving this I get $\lim_{x\to\infty}$ = $\frac{\sin^3(x)}{x^{1-p}}$
Then I make a distinction between $1-p > 0$ and $1 - p < 0$ for these I get that when $p > 1$ this limit tends to infinity and when $p < 1$ the limit tends to 0.
Now, my problem is that I am not sure how I am supposed to make a conclusion about the convergence. I know that if the limit is $0$ that $\frac{1}{x^p}$ is greater than $\frac{\sin^3(x)}{x}$ and the contrary is true when the limit is $\infty$. From there I am not sure what I am supposed to do. Can someone please help out?
To prove the convergence of this integral you can use Dirichlet's test. \begin{equation}f(x)=\frac{1}{x}\ decreasing\ on\ [1,+\infty)\ and\ tends\ to\ 0\end{equation} \begin{equation}g(x)=sin^3(x)\ has\ limited\ integral\ on\ [1,+\infty)\end{equation} \begin{equation}(\int sin^3(x)dx=\frac{cos^3(x)}{3}-cos(x)+C)\end{equation} These two imply that: \begin{equation} \int_1^{\infty}f(x)g(x)dx\ converges\end{equation} The Dirichlet test is very useful when it comes to these types of integrals.