Integral or Closed form expression for $\int^\infty_r \left( 1-\frac{1}{1 + Av^{-\alpha}} \right)v \, dv$

Question

How can I find the integral of (or closed form expression for) the following

$$\int^\infty_r \left( 1-\frac{1}{1 + Av^{-\alpha}} \right)v \, dv$$

where $A$ and $\alpha$ are some constants.

Answer 1

Let $\alpha>2$, $A>0$, and $r>0$, then \begin{align} \int^\infty_r \left( 1-\frac{1}{1 + Av^{-\alpha}} \right)v \, dv&=\int^\infty_r \frac{v}{(\frac v{A^{1/\alpha}})^{\alpha} + 1} \, dv\\ &=A^{2/\alpha}\int^\infty_{r/A^{1/\alpha}} \frac{t}{t^{\alpha} + 1} \, dt\tag{$t=\frac v{A^{1/\alpha}}$}\\ &=A^{2/\alpha}F(\frac r{A^{1/\alpha}}) \end{align} where $$F:x\mapsto\int^\infty_x \frac{t}{t^{\alpha} + 1} \, dt$$ which cannot be simplified except some special $\alpha$.

Answer 2

Continuing on from where @Aforest left off, the integral $$F(x) = \int_x^\infty \frac{t}{t^\alpha + 1} \, dt,$$ can be expressed in terms of the incomplete Beta function $\text{B}(x;a,b)$, namely $$\text{B}(x;a,b) = \int_0^x t^{a - 1} (1 - t)^{b - 1} \, dt,$$ as follows. Enforcing a substitution of $t \mapsto t^{1/\alpha}$ gives $$F(x) = \frac{1}{\alpha} \int_{x^\alpha}^\infty \frac{t^{\frac{2}{\alpha} - 1}}{t + 1} \, dt.$$ If we next enforce a substitution of $1 + t \mapsto \dfrac{1}{t}$ one has $$F(x) = \frac{1}{\alpha} \int_0^{\frac{1}{x^\alpha + 1}} t^{-2/\alpha} (1 - t)^{2/\alpha - 1} \, dt = \frac{1}{\alpha} \text{B} \left (\frac{1}{x^\alpha + 1}; 1 - \frac{2}{\alpha}, \frac{2}{\alpha} \right ),$$ an expression for $F(x)$ in terms of the incomplete Beta function.