Can $\int_{b}^{\infty}\frac{v^a}{(v+1)^k}\mathrm{d}v$ be expressed in a closed-form? $b>0,k\ge1,a<0$ 
Can
  $$\int_{b}^{\infty}\frac{v^a}{(v+1)^k}\mathrm{d}v$$
  where $b>0$, $k\geq1$, $a<0$, be expressed in a closed-form, and if so, what is it?

I appreciate any comments or hints.
 A: In closed-form your integral can be written in terms of the incomplete beta function $\text{B}(x;a,b)$ which is defined as
$$\text{B}(x; a,b) = \int^x_0 t^{a - 1} (1 - t)^{b - 1} \, dt.$$
To show this, as $a < 0$, for convenience, we will set $a = - \alpha$ where $\alpha > 0$. Now let $1 + v = 1/y, dv = -dy/y^2$ and for the limits of integration we have when $v = b, y = 1/(1+b)$ and when $v \to \infty, y \to 0^+$. Thus
\begin{align*}
\int^\infty_b \frac{v^a}{(1 + v)^k} \, dv &= \int^\infty_b \frac{dv}{v^\alpha (1 + v)^k}\\
&= \int^{\frac{1}{1 + b}}_0 y^{k + \alpha - 2} (1 - y)^{-\alpha} \, dy\\
&= \int^{\frac{1}{1 + b}}_0 y^{(k + \alpha - 1) - 1} (1 - y)^{(1 - \alpha) - 1} \, dy\\
&= \text{B} \left (\frac{1}{1 + b}; k + \alpha - 1, 1 - \alpha \right ).
\end{align*}
or
$$\int^\infty_b \frac{v^a}{(1 + v)^k} \, dv = \text{B} \left (\frac{1}{1 + b}; k -a - 1, 1 + a \right ),$$
in terms of your $a$. 
Notice that if $b \to 0^+$, the incomplete beta function reduces to the (complete) beta function and the integral can be written in terms of gamma functions. In this case we will have
$$\lim_{b \to 0^+}\int^\infty_b \frac{v^a}{(1 + v)^k} \, dv = \text{B} (k - a - 1, 1 + a) = \frac{\Gamma (k - a - 1) \Gamma (1 + a)}{\Gamma(k)}.$$
