TITLE: Class of integrals: $I(a)=\int_0^\infty \frac{dx}{e^x+ax}$
QUESTION [23 upvotes]: I'm investigating integrals in the form
$$I(a):=\int_0^\infty \frac{dx}{e^x+ax}$$
So far, I haven't been able to find any special values other than $I(0)=1$, and I've only managed to evaluate these similar indefinite integrals:
$$\int \frac{x-1}{e^x+ax}dx=-\frac{\ln(1+axe^{-x})}{a}+C$$
$$\int \frac{xdx}{e^x+x+1}=-\ln(1+e^{-x}(x+1))+C$$
I've also found the following series representation for $I(a)$:
$$I(a)=\sum_{n=0}^\infty \frac{(-a)^n n!}{(n+1)^{n+1}}$$
...which looks remarkably similar to the Maclaurin series for the Lambert-W function.
QUESTION: Can anyone find any non-trivial special values of this integral? I find this unlikely because of the weird series representation of $I(a)$, so if this isn't feasible, can anyone find any interesting properties or functional/differential equations for $I(a)$?
UPDATE: I've managed to show that
$$\lim_{a\to\infty }\frac{aI(a)}{\ln(a)}=1$$

REPLY [2 votes]: Starting from the series that you already got
$$
\eqalign{
  & I(a) = \int_0^\infty  {{{dx} \over {e^{\,x}  + ax}}}  = \int_0^\infty  {{{e^{\, - x} dx} \over {\left( {1 + axe^{\, - x} } \right)}}}  =   \cr 
  &  = \int_0^\infty  {\sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( {a^{\,k} x^{\,k} e^{\, - \,\left( {k + 1} \right)\,x} } \right)} \;dx} 
  = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} {{k!} \over {\left( {k + 1} \right)^{k + 1} }}a^{\,k} }  \cr} 
$$
and which converges for
$$
\left| a \right|x/e^{\,x}  < \left| a \right|1/e < 1\quad  \Rightarrow \quad \left| a \right| < e
$$
From this related post we get
$$
\sum\limits_{1\, \le \,\,n} {{1 \over {n^{\,n} }}x^{\,n} }
  = x\sum\limits_{0\, \le \,\,n} {{1 \over {\left( {n + 1} \right)^{\,n + 1} }}x^{\,n} }
  = x\int_{\,0}^{\,1} {t^{\, - \,x\,t} dt} 
$$
and since
$$
A(z) = \sum\limits_{0\, \le \,n} {a_n \,z^n } \quad  \Leftrightarrow \quad \int_{\;t\, = \,0}^\infty  {e^{\, - \,t} A(z\,t)\,d\,t}
  = \sum\limits_{0\, \le \,n} {n!a_n z^{\,n} } 
$$
we get another integral representation
$$ \bbox[lightyellow] {  
\eqalign{
  & I( - x) = \sum\limits_{0\, \le \,\,n} {{{n!} \over {\left( {n + 1} \right)^{\,n + 1} }}x^{\,n} }
  = \int_0^\infty  {{{e^{\, - u} du} \over {1 - x\,u\,e^{\, - u} }}}  =   \cr 
  &  = \int_{\,u\, = \,0}^{\,\infty } {e^{\, - \,u} \int_{\,t\, = \,0}^{\,\,1} {t^{\, - \,x\,u\,t} dt\,} du}
  = \int_{\,t\, = \,0}^{\,1} {\int_{\,u\, = \,0}^{\,\infty } {e^{ - \,u\left( {1 + x\,t\ln t} \right)}\, dt\,} du}  =   \cr 
  &  = \int_{\,t\, = \,0}^{\,1} {{{dt} \over {\left( {1 + x\,t\ln t} \right)}}}  \cr} 
}$$
Now the second line tells us that
$$
I( - 1/s) = \int_{\,u\, = \,0}^{\,\infty } {e^{\, - \,u} \int_{\,t\, = \,0}^{\,\,1} {t^{\, - \,\,\left( {u/s} \right)\,t} dt\,} du} 
$$
i.e.
$$ \bbox[lightyellow] {  
\eqalign{
  & {1 \over s}I( - 1/s)
 = \int_{\,\alpha \, = \,0}^{\,\infty } {e^{\, - \,s\,\alpha } \left( {\int_{\,t\, = \,0}^{\,\,1} {t^{\, - \,\,\alpha \,t} dt\,} } \right)d\alpha }  =   \cr 
  &  = \int_0^\infty  {{{e^{\, - u} } \over {s - \,u\,e^{\, - u} }}du}  = \int_{\,t\, = \,0}^{\,1} {{{dt} \over {\left( {s + \,t\ln t} \right)}}}  \cr} 
}$$
so that our integral is tied to the Laplace transform of the interesting function
$\int_{0}^1 {t^{-xt}}dt=\sum_{n=1}^\infty \frac{x^{n-1}}{n^n} =  $ Sphd$(-x;1)$
cited by JJacquelin in his answer to the already cited post.