$\int_0^\infty \frac{x^k}{k!} \frac{1}{\mu }e^{-(1+ \frac1\mu)x}.\,dx= \frac{\mu^k}{(1+\mu)^{k+1}}$
where $\mu > 0$ and $k$ is a constant
I haven't been able to prove this. My way so far is to find that there is a pattern to this $\int {x^k} e^{cx} \,dx= e^{cx} \sum_{i=0}^k (-1)^i \frac{k!}{(k-i)!} \frac{x^{k-i}}{c^{i+1}}$
I know that $e^{cx}$ approaches zero for negative c. However I can't reduce the summation to a term that eliminates $x$ in the numerator where $\infty$ is not acceptable nor eliminate $x$ from the denominator where $0$ dividing by zero is not acceptable.
It's a little odd since there's a factorial term.
Is there an easier way? Is this a known series?
Let $\Gamma(k) = \int_0^\infty e^{- a x}x^k.dx$ where $a > 0$
Then $\Gamma(k) = \frac{k}{a} \Gamma(k-1)$
We find that
$\Gamma(1) = \left[ - \frac {ax+1}{a^2} e^{-ax} \right]_0^\infty=\frac 1 {a^2}$
then
$\Gamma(k) = \frac {k!} {a^{k+1}} $
using that to solve the original question (with $a= 1 + {\frac 1 \mu}$):
$\int_0^\infty \frac {x^k} {k!} \frac 1 {\mu} e^{- (1+\frac 1 \mu)} .dx = \frac 1 {k!}. \frac 1 {\mu} \frac {k!} {(1+ \frac 1 \mu )^{k+1}} = \frac {{\mu}^k}{{\mu + 1}^{k+1}} $