Computing $\int_{0}^{+\infty}\frac{\text{d}x}{1+xe^x}$

Question

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 ?

Answer 1

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.

Answer 2

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

$$I=\int_{0}^{+\infty}\frac{dx}{1+x\,e^x}=0.7672292594\cdots$$