# Determine the $\alpha$'s for which the integral converges $\int _0^4\:\frac{\sqrt{x}}{\left(4-x\right)^{\alpha }}dx$

I have a problem that I don't know how to solve. I have to determin all $\alpha$'s for which this integral converges: $$\int _0^4\:\:\frac{\sqrt{x}}{\left(4-x\right)^{\alpha }}dx$$ What I've tried: $$\int _0^4\:\:\frac{\sqrt{x}}{\left(4-x\right)^{\alpha }}dx = \lim _{x\to b}\left(\int _0^b\:\frac{\sqrt{x}}{\left(4-x\right)^{\alpha }}dx\right)$$ But from here I don't know how to solve this integral.

• Consider $|\frac{\sqrt{x}}{(4-x)^\alpha)}| \leq \frac{2}{(4-x)^\alpha}$. If the right hand side is integrable then the left hand side is integrable. It is equivalent to consider the condition for the integrability of $\frac{1}{x^\alpha}$. – Xianjin Yang Jul 6 '16 at 11:54

Look at the neighborhood of 0 and 4. In the neighborhood of 0 this integral equivalent to $\sqrt{x}/4^{\alpha}$. Now let us look at the neighborhood of 4. Integral equivalent to $2/(4-x)^{\alpha}$. This integral equals to $2/(-\alpha+1)(4-x)^{\alpha-1}$. So, $\alpha-1<0$