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I am looking for a solution or a method of approximation for : $$\int \frac{1}{1-we^w}dw$$ that came up while working on an ODE problem.

Got any suggestions?

Note: $w$ is also a one variable function

Thanks to anyone who can lend a hand

Update:

The original ODE is: $$xdw=(e^{-w}-w)dx$$

Carlos Seda
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    There is no solution in elementary functions. The methods of approximations will depend on the restrictions on the variable (for example if $|w|<1$ a series solution seems best) – Yuriy S Jan 29 '17 at 20:25
  • What's the original ODE? Perhaps you made an error? – Teh Rod Jan 29 '17 at 20:38
  • Updated the question with original ODE, thanks to both for taking a moment of your time to help me out – Carlos Seda Jan 29 '17 at 21:01
  • Kind of important, there is a constant solution, there is a real number $w_0$ with $e^{-w_0} = w_0.$ Solutions for first order ODE cannot cross, so there are solutions with $w > w_0$ and others with $w < w_0.$ You should expect some horizontal asymptotes at $w \rightarrow w_0,$ along with evident vertical asymptotes at $x \rightarrow 0$ – Will Jagy Jan 29 '17 at 21:28

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How to find $\mathrm{\sum_{n=1}^\infty E_{n}(n)= \lim_{k\to\infty}\int_1^\infty \frac{e^{kt+t}t^k-e^t}{e^{kt+2t}t^{k+1}-e^{kt+t}t^k}dt=.26929…}$? shows a nice result from your integral which will be generalized here using geometric series and gamma type functions. The result is simple, but has the restriction of x<Ω=W(1) which is the Omega constant:

$$\int \frac{dx}{1-xe^x}=-\int \sum_{n=0}^\infty \left(xe^x\right)^ndx=-\sum_{n=0}^\infty \int x^n e^{xn}dx$$

This is a close definition of the gamma function and exponential integral function mentioned in the bolded link:

$$\sum_{n=0}^\infty \int x^n e^{xn}dx=C-\sum_{n=0}^\infty \frac{ (e^{-n x} (e^x x)^n (-n x)^{-n} Γ(n + 1, -n x))}{n}\mathop=^{x\in\Bbb R} C-\sum_{n=0}^\infty x^{n+1} \text E_{-n}(-n x)=C+\sum_{n=0}^\infty \frac{(-1)^nΓ(1 + n, -n x)}{ n^{n+1}} $$

If there was some function, like a hypergeometric function to give a closed form, please post it. Please correct me and give me feedback!

Тyma Gaidash
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