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

Question

I am trying to compute this integral:

$$I = \int_0^{\infty} \frac{1}{e^x + x}dx$$

I don't see any obvious ways to integrate this using real methods.

So, now I'm trying to integrate with using complex analysis, I tried to transform the equation with $z = e^{-x}$ and get

$$I = \int_0^1 \frac{1}{1-z\log(z)}dz$$

I see that a pole of the above is $z = \frac{1}{W(1)}$, which is the Lambert W function

When I evaluate the residue, I get the value 0.

$$\text{Res}_{z \rightarrow \frac{1}{W(1)}}\left(\frac{1}{1-z \log (z)},f(z)\right) = 0$$

Assistance on next steps and a solution would be appreciated.

Answer 1

Of course, by Taylor series we have $$ I = \int_{0}^{+\infty}\frac{dx}{x+e^x}\,dz = \int_{0}^{1}\sum_{n\geq 0}(z\log z)^n\,dz = \sum_{n\geq 0}\frac{(-1)^n n!}{(n+1)^{n+1}} \tag{1}$$ and by integration by parts

$$ I = \int_{0}^{+\infty}\frac{1}{e^x+1}\cdot\frac{e^x+1}{e^x+x}\,dx = \int_{0}^{+\infty}\frac{e^x \log(e^x+x)}{(e^x+1)^2}\,dx \tag{2}$$ so $I$ has the following representation

$$ I = \log(2)+\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\int_{0}^{+\infty}\frac{x^{n+1}\,dx}{e^{(n-1)x}(e^x+1)^2}\tag{3} $$ where the involved integrals depend on values of the Riemann $\zeta$ function. However, $(3)$ is not a huge improvement over $(1)$. We may apply series acceleration techniques to $(1)$ in order to get $I\approx 0.8063956162$ really soon, but my opinion is that we cannot really hope in something "nicer" than the RHS of $(1)$.

Answer 2

$\int_0^\infty\dfrac{1}{e^x+x}~dx$

$=\int_0^\infty\dfrac{e^{-x}}{1+xe^{-x}}~dx$

$=\int_0^\infty\sum\limits_{n=0}^\infty(-1)^nx^ne^{-(n+1)x}~dx$

$=\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+1}n!x^ke^{-(n+1)x}}{(n+1)^{n-k+1}k!}\right]_0^\infty$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^nn!}{(n+1)^{n+1}}$