Improper integral with parameter and square root on denominator I can't find a way to show for which values of $\alpha$ the following improper integral converges:
$$\large{\int_{0}^2{x^\alpha\over{\sqrt{4-x^2}}}dx}$$
Thank you for your help.
 A: Your integral reads
$$\int_0^2\frac{x^{\alpha}}{\sqrt{4-x^2}}dx=\frac{1}{2}\int_0^2\frac{x^{\alpha}}{\sqrt{1-\frac{x^2}{4}}}dx$$
If you peform the change $\frac{x^2}{4}\rightarrow y$ (then $dx=\frac{1}{\sqrt{y}}dy$):
$$\int_0^2\frac{x^{\alpha}}{\sqrt{4-x^2}}dx=\frac{1}{2}\int_0^2\frac{x^{\alpha}}{\sqrt{1-\frac{x^2}{4}}}dx=2^{\alpha-1}\int_0^1y^{\frac{\alpha-1}{2}}(1-y)^{-1/2}dy$$
wich is a Euler Beta function $\beta\left(\frac{\alpha+1}{2},\frac{1}{2}\right)$ and then, it converges if and only if both real part parameters are positive, that is, if $\Re(\alpha)>-1$. In this case, integral converges to
$$\boxed{\beta\left(\frac{\alpha+1}{2},\frac{1}{2}\right)=\frac{2^{-1+\alpha} \sqrt{\pi}\,\, \Gamma\left(\frac{1+\alpha}{2}\right)}{\Gamma\left(1+\frac{\alpha}{2}\right)}}$$
A: The function $\frac{x^\alpha}{\sqrt{4-x^2}}$ is comparable to $\frac{1}{\sqrt{4-x^2}}$ around the point $2$, and to $x^{\alpha}$ around the point $0$. So the only point we should be worried about is $0$. For $\alpha \neq -1$ we have $$\int_0^1 x^{\alpha} dx = \frac{1}{1+\alpha}x^{1+\alpha}|_{x=0}^{x=1} = \frac{1}{1+\alpha}-\frac{1}{1+\alpha}\lim_{x\to 0^+} x^{1+\alpha}  $$ converges for $\alpha>-1$. $$\int_0^1\frac{1}{x}dx = -\lim_{x\to 0}\ln x = \infty. $$
Hence the integral converges if and only if $\alpha>-1$ by comparison test.
