Improper integral $\int^{1}_{0} \frac{x}{\sin{(x^{p})}} dx$

Question

I have an improper integral with $p > 0$,

$$\int^{1}_{0} \frac{x}{\sin{(x^{p})}} \ dx$$

and I want to find for which $p$ the integral exists. Now we should consider when $p = 1$ and when $p \not= 1$:

$p = 1$:

$$\lim_{t \to 0} \int^{1}_{t}\frac{x}{\sin{x}} dx$$

$p \not= 1$:

$$\lim_{t \to 0} \int^{1}_{t}\frac{x}{\sin{x^{p}}} dx$$

But I am not sure how to find either of these limits, can anyone help?

Answer 1

Hint: Use Comparison and the fact that for $0\lt t\le 1$ we have $\frac{t}{2}\lt \sin t\lt t$. Or use Limit Comparison and the fact that $\lim_{u\to 0}\frac{\sin u}{u}=1$.