# Does $\int^1_0\frac{(1-s)^\alpha}{s^\beta} \operatorname d \!s$ converge?

I want to ask about the finitess of the following integral

$$\int^1_0\frac{(1-s)^\alpha}{s^\beta}ds$$

when $\alpha>\beta>1$. This integral is very similar to the Beta function, aside from the negative power $-\beta$ here. Can anyone confirm if there really exists some big $\alpha$ such that the integral is finite? Thank you.

• For $\beta \geqslant 1$, you have a non-integrable singularity at $0$. $(1-s)^\alpha$ is essentially $1$ there, so you are left with $$\int_0^\varepsilon \frac{ds}{s^\beta}.$$ – Daniel Fischer Feb 14 '14 at 21:25
• Check $\beta$ function. – Mhenni Benghorbal Feb 14 '14 at 21:26
• @canis89: You are welcome. – Mhenni Benghorbal Feb 14 '14 at 21:28

The integral is finite if and only if $$\beta<1\quad\text{and}\quad -\alpha<1\iff \beta<1\quad\text{and}\quad \alpha>-1$$