Is it possible to further simplify the following improper integral? $$ \int_{0}^{\infty}\frac{\ln(e^{x+s}+1)-\ln(e^{x}+1)}{se^{s}}ds,\;x>0 $$
The denominator leads me to think about Gamma function, however, $\Gamma(0)=+\infty$. I also tried integration by parts, but I couldn't further simplify it. Does any one know how to calculate this integral? Any hints, references or help would be appreciated. Thank you very much.
Write $$f(x)=\log \left(a+e^{s+x}\right)-\log \left(a+e^x\right)$$ and expand as a series around $a=0$ which gives $$f(x)=s+\sum_{n=1}^\infty (-1)^n\,\, \frac {e^{-n x}-e^{-n (s+x)} }n $$
$$\frac{\log(e^{x+s}+1)-\log(e^{x}+1)}{se^{s}}=e^{-s}+\sum_{n=1}^\infty (-1)^n\,\,\frac 1{ns}\left(e^{-s-nx}-e^{-(n+1) s-n x} \right)$$ So, the definite integral is $$\large\color{blue}{I=1+\sum_{n=1}^\infty (-1)^n\,\,\frac{\log (n+1)}{n}\, e^{-n x} }$$
Edit
Writing $$\log(n+1)=\log(n)-\sum_{k=1}^\infty \frac{(-1)^k }{k \,n^{k}}$$ we also have $$\large\color{blue}{I=1-\text{Li}_{1}^{(1,0)}\left(-e^{-x}\right)-\sum_{k=1}^\infty \frac{(-1)^k }{k}\text{Li}_{k+1}\left(-e^{-x}\right)}$$
