Calculate $ \int_{0}^{1}\frac{r^{p+1}}{(1-r^{2})^{\frac{p}{2}-\frac{1}{2}}}dr $

Question

$$ \int_{0}^{1}\frac{r^{p+1}}{(1-r^{2})^{\frac{p}{2}-\frac{1}{2}}}dr=\frac{2\Gamma(\frac{3}{2}-\frac{p}{2})\Gamma(\frac{p}{2}+1)}{3\sqrt{\pi}}\text{ for }-2<p<3, $$

and if $p\geq3$, the above integration diverges.(This is the result from wolframalpha and I just want to know the above is convergent only when $-2<p<3$ and divergent for $p\geq 3$)

I want to calculate this. I know the definition of Gamma function which is $$\Gamma(z)=\int_{0}^{\infty}x^{z-1}e^{-x}dx.$$ Any suggention? Thanks in advance.

Answer 1

$$I=\int_{0}^{1}\frac{r^{p+1}}{(1-r^{2})^{\frac{p}{2}-\frac{1}{2}}}dr $$ Let $\quad r^2=x \quad\to\quad dr=\frac{1}{2x^{1/2}}dx\quad\to\quad I=\int_{0}^{1}\frac{x^{(p+1)/2}}{(1-x)^{(p-1)/2}}\frac{1}{2x^{1/2}}dx$ $$I=\frac{1}{2}\int_{0}^{1}\frac{x^{p/2}}{(1-x)^{(p-1)/2}}dx= \frac{1}{2}\int_{0}^{1}\frac{(1-x)^{(1-p)/2} }{x^{-p/2} }dx$$ Obviously, if$\quad p\leq -2 \quad\to\quad -p/2\geq 1\quad$the integral is not convergent for $x\to 0$.

Let $\quad 1-x=t$ $$I=\frac{1}{2}\int_{0}^{1}\frac{(1-t)^{p/2}}{t^{(p-1)/2}}dt$$ Obviously, if$\quad p\geq 3 \quad\to\quad (p-1)/2\geq 1\quad$the integral is not convergent for $t\to 0\quad (x\to 1)$.

Thus the integral is convergent only if $-2<p<3$.

$$I=\frac{1}{2}\int_{0}^{1}t^{(1-p)/2}(1-t)^{p/2}dt$$ Beta function : https://en.wikipedia.org/wiki/Beta_function

$$\quad B(a,b)=\int_0^1 t^{a-1}(1-t)^{b-1}=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$

With $\quad a=\frac{1-p}{2}+1\quad$ and $\quad b=\frac{p}{2}+1$ $$I=\frac{1}{2}B\left(\frac{3-p}{2}\:,\:\frac{p+2}{2} \right)= \frac{1}{2} \frac{\Gamma(\frac{3-p}{2})\Gamma(\frac{p+2}{2} )}{\Gamma(5/2)}$$

$\Gamma(5/2)=\frac{3\sqrt{\pi}}{4}$

$$I=\frac{2\Gamma(\frac{3}{2}-\frac{p}{2})\Gamma(\frac{p}{2}+1)}{3\sqrt{\pi}}$$

Answer 2

I am not sure to properly answer your question; so, please forgive me if I am off topic.

Considering $$I_p=\int_{0}^{1}\frac{r^{p+1}}{(1-r^{2})^{\frac{p}{2}-\frac{1}{2}}}\,dr$$ let $r=\sin(t)$ making $$I_p=\int_{0}^{\frac \pi 2}\sin ^{p+1}(t) \cos ^{2-p}(t)\,dt$$

Looking in the seventh edition of "Table of Integrals, Series, and Products" by I.S. Gradshteyn and I.M. Ryzhik, you will find this integral in section $3.621$ (the fifth formula).

Applied to your case, $\mu=p+2$, $\nu=3-p$, this then write

$$I_p=\frac{1}{2} B\left(\frac{p+2}{2},\frac{3-p}{2}\right)$$ and using the relation between beta and gamma functions $$I_p=\frac{2}{3 \sqrt{\pi }} \Gamma \left(\frac{3}{2}-\frac{p}{2}\right) \Gamma \left(1+\frac{p}{2}\right)$$ as given by Wolfram Alpha.