# About the convergence of $\int_{0}^{\pi/2}x^p(\sin x)^qdx$

For what values of $p,q\in\mathbb{R}$ the integral $\int_{0}^{\pi/2}x^p(\sin x)^qdx$ converges?

I think that we must use the limit comparison test. I tried to compare with the integral $\int_{0}^{\pi/2}x^pdx$. But this does not work because $\lim_{x\to 0}\frac{x^p(\sin x)^q}{x^p}=0$, and the test requires the limit to be positive.

Would anyone give me a hint?

Thank you.

• I'm not sure if this is helpful but if we consider the sequence $a_n = n^p (\sin n)^q$ then if the integral is finite we must have $\lim_{n \to \infty} a_n = 0$. – MathMajor May 14 '16 at 2:13
• the only problem is at $0$. and if $q = 1$ then $\lim_{\epsilon \to 0^+} \int_\epsilon^{\pi/2} x^p \sin (x) dx$ converges for $p > -2$, can you prove it ? – reuns May 14 '16 at 2:14
• Sorry, I don't know why that integral converges for $p>-2$. Could you give me a hint? And the case when $q\neq 1$ can be deduced from this particular case? @user1952009 – Talexius May 14 '16 at 2:38

The problem point is, as mentioned in the comments, $x = 0$. Try writing $$\int_{0}^{\pi/2}x^p (\sin x)^q\,d x = \int_{0}^{\pi/2}x^{p+q}\left(\frac{\sin x}{x}\right)^q$$ and then you do not need to worry about the sine term near $0$ (regardless of the sign of $q$).
• The second integral you wrote can be tested by limit comparison with $x^{p+q}$, is that what you mean? On the other hand, I don't understand why $x=\pi /2$ is a problem point. $x$ and $\sin (x)$ do not vanish at $x=\pi/2$. Thank you. – Talexius May 14 '16 at 18:57
• Yes, that was my thought for $x^{p+q}$. My mistake on $\pi/2$, corrected. – Lost in a Maze May 15 '16 at 1:05