$$\int_0^\infty \frac{\sin (x^m)}{x^n}dx $$
Putting $x^m = t$
$$ \begin{align} \frac{1}{m}\int_0^\infty \frac{\sin t}{t^{(\frac{m+n-1}{m})}}dt \end{align} $$
By applying Dirichlet Test I've been able to prove that the integrand is convergent when $1-m < n$.
But the answer given is $1-m < n < m+1$. Can you give me any hint on how to get the second condition?