Finding a closed form expression for an integral with parametrization

Question

Is there a general expression if the integral below is evaluated? Note that $-\infty<q<1$ where $q$ can take either integer or non-integer values; and that $b<x$ where $b,x>0$. The integral is: $$\int_b^x \frac{1}{\left(1-(b/r)^{1-q}\right)^{1/2}}\,dr$$ The integral diverges for all values of $q<-1$, I think. I also tried feeding it up to Mathematica for every value of $q$, and the result is that for even integer values of $|q|$ except $q=-2$, the integral results to a hypergeometric function. For odd integer values of $|q|$ including $q=-2$, the integral diverges. Can someone help me figure this out. Thanks a lot.

Answer 1

By the substitution $g=(b/r)^{1-q}\implies dr=\frac{b}{q-1}g^{\frac1{q-1}-1}dg$ (and let $\frac1{q-1}-1=Q$), yielding $$ \int_b^x \frac{1}{\left(1-(b/r)^{1-q}\right)^{1/2}}\,dr=\frac{b}{q-1}\int_1^{x'}\frac{g^Q}{\sqrt{1-g}}dg=\frac{b}{q-1}\int_1^{x'}g^Q(1-g)^{-1/2}dg $$

where $x'=\left(\frac bx\right)^{1-q}$.

Let $$\overline B(x;a,b)=\int^1_x t^{a-1}(1-t)^{b-1}dt$$ be the upper incomplete beta function.

Then, $$ \frac{b}{q-1}\int_1^{x'}g^Q(1-g)^{-1/2}dg=\frac{b}{1-q}\cdot \overline B(x';Q+1,\frac12)=\frac{b}{1-q}\cdot \overline B\left(\left(\frac bx\right)^{1-q};\frac1{q-1},\frac12\right) $$

As a result, $$\color{red}{\int_b^x \frac{1}{\left(1-(b/r)^{1-q}\right)^{1/2}}\,dr=\frac{b}{1-q}\cdot \overline B\left(\left(\frac xb\right)^{q-1};\frac1{q-1},\frac12\right)}$$

which should be the best closed form you can obtain.

Luckily, the integral converges for any $b,x,q$ in your specified range. Although there is a pole near $1$, the integrand diverges as $(1-t)^{-1/2}$ near there and the integral does converge by p-test.