Evaluating a Complex Integral.... I am trying to come up with a closed form for this integral:
$$\int_0^{\infty } \frac{\mu  x e^{-\mu  x}}{1-\frac{\lambda _R \left(1-e^{-\mu  x}\right)}{\mu }} \, dx$$
After playing around a bit, it turns out that:
$$\int_0^{\infty } \frac{\mu  x}{1-\frac{\lambda _R \left(1-e^{-\mu  x}\right)}{\mu }} \, dx = \frac{\text{Li}_2\left(1-\frac{\mu }{\lambda _R}\right)}{\mu -\lambda _R}$$
$$\int_0^{\infty } \frac{\mu ^2 x^2}{1-\frac{\lambda _R \left(1-e^{-\mu  x}\right)}{\mu }} \, dx = \frac{2 \text{Li}_3\left(1-\frac{\mu }{\lambda _R}\right)}{\mu -\lambda _R}$$
$$\int_0^{\infty } \frac{\mu ^3 x^3}{2 \left(1-\frac{\lambda _R \left(1-e^{-\mu  x}\right)}{\mu }\right)} \, dx = -\frac{3 \text{Li}_4\left(1-\frac{\mu }{\lambda _R}\right)}{\mu -\lambda _R}$$
and so on.
So, what I am trying to expand then numerator, $\mu  x e^{-\mu  x}$, in a Taylor series and then try to integrate the function term-by-term to get the result.
Using this approach, I think you can rewrite the integral 
$$\int_0^{\infty } \frac{\mu  x e^{-\mu  x}}{1-\frac{\lambda _R \left(1-e^{-\mu  x}\right)}{\mu }} \, dx$$
as
$$\int_0^{\infty } \frac{\mu  x}{1-\frac{\lambda _R \left(1-e^{-\mu  x}\right)}{\mu }} \, dx -\int_0^{\infty } \frac{\mu ^2 x^2}{1-\frac{\lambda _R \left(1-e^{-\mu  x}\right)}{\mu }} \, dx +\int_0^{\infty } \frac{\mu ^3 x^3}{2 \left(1-\frac{\lambda _R \left(1-e^{-\mu  x}\right)}{\mu }\right)} \, dx + \cdots$$
which, after you integrate term-by-term using the relation above and collecting terms, you get
$$\sum _{k=1}^{\infty } \frac{(-1)^{k+1} \left(k \text{Li}_{k+1}\left(1-\frac{\mu }{\lambda _R}\right)\right)}{\mu -\lambda _R}$$
But the problem is this diverges when $\mu = 1$ and $\lambda_R = 0.5$, which cannot be the case. 
Also, when I use a numerical mathematical package, with $\mu = 1$ and $\lambda_R = 0.5$ and evaluate the integral NOT the sum, I get 1.64493 as the answer with those parameters.
So, clearly, I am doing something wrong here. Any help would be great.
 A: $$\frac d{dx}1-e^{-\mu  x}=\mu e^{-\mu x}$$
Now apply integration by parts:
$$dv=\frac{\mu e^{-\mu  x}}{1-\frac{\lambda _R \left(1-e^{-\mu  x}\right)}{\mu }},\qquad u=x$$
$$\int_0^\infty\frac{\mu  x e^{-\mu  x}}{1-\frac{\lambda _R \left(1-e^{-\mu  x}\right)}{\mu }} \, dx=\lim_{a\to\infty}\left[-\mu x\ln|1-\lambda_R(1-e^{-\mu  x})|\bigg|_0^a+\int_0^a\mu \ln|1-\lambda_R(1-e^{-\mu  x})|\ dx\right]\\=\mu\int_0^\infty\ln|1-\lambda_R(1-e^{-\mu  x})|-\ln|1-\lambda_R(1-e^{-\mu  a})|\ dx$$
It should then be much easier to numerically evaluate $\int\ln|1-\lambda_R(1-e^{-\mu  x})|\ dx\sim-\frac\mu2x^2$.
If you wish, I suppose you could try expanding this, but note that you have to watch your radius of convergence's.  A series expansion and/or integration term by term has hidden requirements to perform, and if you get a diverging result, its likely because you didn't check if an interchange was valid.
A: $$I=\int^\infty_0\frac{\mu  x e^{-\mu  x}}{1-\frac{\lambda  \left(1-e^{-\mu  x}\right)}{\mu }} \, dx$$
Let $t = 1-e^{-\mu x}$ hence $-\mu x = \log(1-t)$
\begin{align}
\frac{1}{\mu}\int^1_0 \frac{-\log(1-t)(t-1)}{1-\frac{\lambda}{\mu}t}\frac{dt}{t-1} &=- \frac{1}{\mu}\int^1_0 \frac{\log(1-t)}{1-\frac{\lambda}{\mu}t}dt \\&= -\frac{1}{\mu}\sum_{n\geq 0}\left(\frac{\lambda}{\mu}\right)^n \int^1_0 t^n \log(1-t)\,dt \\&= \frac{1}{\lambda}\sum_{n\geq 0}\left(\frac{\lambda}{\mu}\right)^{n+1} \frac{H_{n+1}}{n+1}\\& = \frac{1}{\lambda}\sum_{n\geq 1}\left(\frac{\lambda}{\mu}\right)^{n} \frac{H_{n}}{n} \\&= \frac{\mathrm{Li}_2\left(\frac{\lambda}{\mu} \right)+\frac{1}{2}\log^2\left(1-\frac{\lambda}{\mu} \right)}{\lambda}
\end{align}
All evaluations are under the asummption $|\lambda/\mu|<1$.
