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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.

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    $\begingroup$ Shouldn't be there $k!$ in the denominator ? $\endgroup$ – Zaid Alyafeai Feb 24 '17 at 0:49
  • $\begingroup$ I computed those out and they are in the denominator $\endgroup$ – PiE Feb 24 '17 at 0:59
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$$I=\int^\infty_0\frac{\mu x e^{-\mu x}}{1-\frac{\lambda \left(1-e^{-\mu x}\right)}{\mu }} \, dx$$

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 $|\lambda/\mu|<1$.

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  • $\begingroup$ Nice job!!! Thanks for your insight. $\endgroup$ – PiE Feb 24 '17 at 7:20
  • $\begingroup$ @PMF, thanks, happy to help. $\endgroup$ – Zaid Alyafeai Feb 24 '17 at 12:06
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$$\frac d{dx}1-e^{-\mu x}=\mu e^{-\mu x}$$

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.

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  • $\begingroup$ Maybe I'm not understanding, but your result seems to diverge as well...I plugged it into a numerical package $\endgroup$ – PiE Feb 24 '17 at 1:01
  • $\begingroup$ Could you give me a reference RE this comment: " 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" - I'm in the process of learning these things at this point - Thanks $\endgroup$ – PiE Feb 24 '17 at 1:04
  • $\begingroup$ @PMF My integral diverges, yes, but it is canceled by another diverging part, which yields a finite value. As per your second comment, try looking up "when can I interchange a sum and an integral". $\endgroup$ – Simply Beautiful Art Feb 24 '17 at 1:05
  • $\begingroup$ Ok - thanks - let me digest your answer a bit more... $\endgroup$ – PiE Feb 24 '17 at 1:06

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