Is there a closed form for the integral $\int_u^{\infty}(x+a)^ve^{-bx}dx$ I'm looking for the close form of the integral $\int_u^{\infty}(x+a)^ve^{-bx}dx$, where $u,a,b$ are positive real numbers (b could be integer in special case), $v$ is a complex number. Are there anyone aware of it? Thanks in advance!
 A: $$
\int_u^\infty(x+a)^{\nu}\mathrm{e}^{-bx}dx
$$
lets make the sub $t = x+ a$ we find
$$
\mathrm{e}^{ab}\int_{u+a}^\infty t^\nu\mathrm{e}^{-b(t-a)}dt = \mathrm{e}^{ab}\int_{u+a}^\infty t^\nu\mathrm{e}^{-bt}dt =\frac{\mathrm{e}^{ab}}{b^{\nu+1}}\int_{b(u+a)}^\infty s^{\nu}\mathrm{e}^{-s}ds
$$
If we map $\nu \to \alpha - 1$ we find
$$
\frac{\mathrm{e}^{ab}}{b^{\nu+1}}\int_{b(u+a)}^\infty s^{\nu}\mathrm{e}^{-s}ds = \frac{\mathrm{e}^{ab}}{b^{\nu+1}}\int_{b(u+a)}^\infty s^{\alpha-1}\mathrm{e}^{-s}ds
$$
The latter is the form of the incomplete gamma function
$$
\Gamma(s,x) = \int_x^\infty t^{s-1}\mathrm{e}^{-t}dt
$$
so we have
$$
\frac{\mathrm{e}^{ab}}{b^{\nu+1}}\int_{b(u+a)}^\infty s^{\alpha-1}\mathrm{e}^{-s}ds = \frac{\mathrm{e}^{ab}}{b^{\nu+1}}\Gamma(\alpha,b(u+a)) = \frac{\mathrm{e}^{ab}}{b^{\nu+1}}\Gamma(\nu+1,b(u+a))
$$
We have suitable conditions on $\alpha$ and thus $\nu + 1$.
A: Wolfram alpha shows that a antiderivative is given by
$$\int (x+a)^\nu e^{-b x} dx= -b^{-(\nu+1)} e^{a b}  \Gamma(\nu+1,b(a+x)). $$
Evaluating the antiderivative at the boundary values, we obtain
$$\int_u^\infty (x+a)^\nu e^{-b x} dx = b^{-(\nu+1)} e^{a b}  \Gamma(\nu+1,b(a+u))$$
with $\Gamma$ the incomplete Gamma function.
