Computing $\int_{0}^{+\infty}\frac{\text{d}x}{1+xe^x}$ I've shown that the following integral exists
$$
I=\int_{0}^{+\infty}\frac{\text{d}x}{1+xe^x}
$$
WolframAlpha tells me that $I \approx 0,767$ but I can't find a way to compute the exact form. Any tips ?
 A: Maybe another way to find some nicer form is the following:
$$
\int_{0}^{\infty}\frac{1}{1+xe^x}dx = \int_{0}^{W(1)}\frac{1}{1+xe^x}dx + \int_{W(1)}^{\infty}\frac{1}{1+xe^x}dx
$$
where ${W(1)}$ is the number satisfying ${W(1)e^{W(1)}=1}$. Why? Well - we can now exploit some geometric series,
$$
\frac{1}{1+xe^x} = \frac{1}{1-(-xe^x)} = \sum_{n=0}^{\infty}(-1)^nx^ne^{nx}
$$
this will converge on ${[0,W(1))}$. So
$$
\int_{0}^{W(1)}\frac{1}{1+xe^x}dx = \int_{0}^{W(1)}\sum_{n=0}^{\infty}(-1)^nx^ne^{nx}dx
$$
now you can make some arguments about interchanging sum and integral to get
$$
=\sum_{n=0}^{\infty}(-1)^n\int_{0}^{W(1)}x^ne^{nx}dx = W(1) - \sum_{n=1}^{\infty}(-1)^{n+1}\int_{0}^{W(1)}x^ne^{nx}dx
$$
with substitution ${u=nx}$, this becomes
$$
=W(1) - \sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n^{n+1}}\int_{0}^{nW(1)}u^ne^{u}dx
$$
now I do sub ${t=-u}$ (I want to make use of the LOWER incomplete gamma function):
$$
=W(1) - \sum_{n=1}^{\infty}\frac{1}{n^{n+1}}\int_{0}^{-nW(1)}t^ne^{-t}dt
$$
the definition of ${\int_{0}^{-nW(1)}t^ne^{-t}dt}$ is ${\gamma(n+1,-nW(1))}$ (https://en.wikipedia.org/wiki/Incomplete_gamma_function). So this sum is now
$$
=W(1) - \sum_{n=1}^{\infty}\frac{\gamma(n+1,-nW(1))}{n^{n+1}}
$$
this is only half the battle. Now we must deal with ${\int_{W(1)}^{\infty}\frac{1}{1+xe^x}dx}$. Well - this is
$$
=\int_{W(1)}^{\infty}\frac{1}{xe^x}\left(\frac{1}{1+(xe^x)^{-1}}\right)dx
$$
you guessed it - geometric series:
$$
=\int_{W(1)}^{\infty}\frac{1}{xe^x}\sum_{n=0}^{\infty}\frac{(-1)^n}{x^ne^{nx}}dx=\int_{W(1)}^{\infty}\sum_{n=0}^{\infty}\frac{(-1)^n}{(xe^x)^{n+1}}dx
$$
making some interchange argument once again,
$$
=\sum_{n=0}^{\infty}(-1)^{n}\int_{W(1)}^{\infty}\frac{1}{(xe^x)^{n+1}}dx
$$
this can be written as
$$
=\sum_{n=0}^{\infty}(-1)^{n}\int_{W(1)}^{\infty}x^{-n-1}e^{-(n+1)x}dx
$$
the aim is to now use the UPPER incomplete gamma function. Do substitution ${u=(n+1)x}$:
$$
=\sum_{n=0}^{\infty}(-1)^{n}\frac{1}{n+1}\int_{(n+1)W(1)}^{\infty}\left(\frac{u}{n+1}\right)^{-n-1}e^{-u}du=\sum_{n=0}^{\infty}(-1)^n(n+1)^{n}\int_{(n+1)W(1)}^{\infty}u^{-n-1}e^{-u}du
$$
This is now
$$
=\sum_{n=0}^{\infty}(-1)^{n}(n+1)^{n}\Gamma(-n,(n+1)W(1))
$$
giving us our answer overall as
$$
\int_{0}^{\infty}\frac{1}{1+xe^x}dx = W(1) - \sum_{n=1}^{\infty}\frac{\gamma(n+1,-nW(1))}{n^{n+1}} + \sum_{n=0}^{\infty}(-1)^{n}(n+1)^{n}\Gamma(-n,(n+1)W(1))
$$
Not sure if this can be simplified any further, but it also seems this answer agrees numerically with ${0.767\dots}$.
EDIT I guess we could bring those infinite sums together,
$$
=W(1) + \sum_{n=1}^{\infty}(-1)^{n+1}n^{n-1}\Gamma(-n+1,nW(1)) - \sum_{n=1}^{\infty}\frac{\gamma(n+1,-nW(1))}{n^{n+1}}
$$
becomes
$$
=W(1) + \sum_{n=1}^{\infty}\frac{(-1)^{n+1}n^{2n}\Gamma(-n+1,nW(1))-\gamma(n+1,-nW(1))}{n^{n+1}}
$$
so finally
$$
\int_{0}^{\infty}\frac{1}{1+xe^x}dx = W(1) + \sum_{n=1}^{\infty}\frac{(-1)^{n+1}n^{2n}\Gamma(-n+1,nW(1))-\gamma(n+1,-nW(1))}{n^{n+1}}
$$
where ${W(1)}$ is the product log function at $1$, ${\Gamma(s,x)}$ is the upper incomplete gamma function and ${\gamma(s,x)}$ is the lower incomplete gamma function. You can see this final answer does have a sort of nice form in the sum, "something times upper gamma minus lower gamma over ${n^{n+1}}$", with a constant term of ${W(1)}$ at the front. I doubt this simplifies any further, but I could be wrong.
A: I should not expect a closed form.
The inverse symbolic calculator proposes, as an approximation
$$\left(\frac{\Gamma \left(\frac{11}{24}\right)}{\Gamma
   \left(\frac{1}{4}\right)}\right)^{K_0(1)}= 0.7672292583\cdots$$ while the "exact" value is
$$I=\int_{0}^{+\infty}\frac{dx}{1+x\,e^x}=0.7672292594\cdots$$
