Integral of $\int_0^\infty\frac{1}{e^{x}-x} dx$

Question

I am curious as to whether a closed form exists for the following integral:

$$\int_0^\infty\frac{1}{e^{x}-x} dx$$

I have tried a few elemetary methods on it, but I believe this integral (if it has a solution) can only be solved through complex analysis which I have no working knowledge of. Wolfram does not return any closed form. I'm not sure if it is much use, but this integral appears to be equivalent to $$\int_0^\infty\frac{x}{e^{x}-x} dx$$ Maybe this integral is related to the gamma function like $\int_0^\infty\frac{x^n}{e^{x}-1} dx$?

Answer 1

Here is an approach.

$$ \int_0^\infty\frac{1}{e^{x}-x} dx = \sum_{k=0}^{\infty} \int_{0}^{\infty}x^k e^{-kx-x}dx = \sum_{k=0}^{\infty}\frac{k!}{(k+1)^{k+1}}\sim 1.359098277. $$

See related techniques.

Answer 2

We have: $$\begin{eqnarray*}\int_{0}^{+\infty}\frac{dx}{e^x-x}&=&\int_{0}^{+\infty}\sum_{n=0}^{+\infty}x^{n}e^{-(n+1)x}dx=\sum_{n=0}^{+\infty}\frac{1}{(n+1)^{n+1}}\int_{0}^{+\infty}x^n e^{-x}dx\\&=&\sum_{n=1}^{+\infty}\frac{(n-1)!}{n^n}.\end{eqnarray*}$$ It is not a "closed" form, but it is pretty nice.