Integral of the function $S(x)=\ln\left(1-\frac{x}{\exp(x)}\right)$ I have to check if the following series:
$$S(x)=\sum_{k=1}^{\infty}\frac{x^k}{k\exp(kx)}$$ gives a function of $x$
$$S(x)=-\ln\left(1-\frac{x}{\exp(x)}\right)$$
for which:
$$J=\left|\int_{0}^{+\infty}S(x)dx\right|\lt\infty$$
I used Maple and Mathematica to solve the integral without any result. Does anyone have an idea how to calculate $J$? Thanks
 A: For $x \geqslant 0$, all terms in
$$S(x) = \sum_{k=1}^\infty \frac{x^k}{x\exp(kx)}$$
are non-negative. Hence we have
$$\begin{align}
J &= \int_0^\infty S(x)\,dx\\
&= \sum_{k=1}^\infty \frac{1}{k}\underbrace{\int_0^\infty x^ke^{-kx}\,dx}_{t = kx}\\
&= \sum_{k=1}^\infty \frac1k \int_0^\infty \left(\frac tk\right)^ke^{-t}\,d\left(\frac tk\right)\\
&= \sum_{k=1}^\infty \frac{1}{k^{k+2}} \int_0^\infty t^ke^{-t}\,dt\\
&= \sum_{k=1}^\infty \frac{k!}{k^{k+2}},
\end{align}$$
using $\int_0^\infty t^{\alpha-1}e^{-t}\,dt = \Gamma(\alpha)$ for $\Re \alpha > 0$. By Stirling's approximation,
$$k! \sim \sqrt{2\pi k}k^ke^{-k} \Rightarrow \frac{k!}{k^{k+2}} \sim \frac{\sqrt{2\pi}}{k^{3/2}e^k},$$
so the sum converges. For a decent approximation of the value, one needs not compute many terms, but I don't know if there is a known closed form for the sum.
A: Differentiating the power series with respect to $x$ gives
$$ S(x)=\sum_{k=1}^{\infty}\frac{x^k}{ke^{kx}} \implies S'(x)=\sum _{k=1}^{\infty }\left({\frac {{x}^{k}}{x{{\rm e}^{kx}}}}-{\frac {{x}^{k
}}{{{\rm e}^{kx}}}}\right)=\frac{1-x}{e^{x}-x}. $$
Now, to get $S(x)$, you need to integrate the last expression to get

$$ S(x)=x-\ln(x-e^x)+c. $$

Now, you need to determine the constat of integration.
Note:
We used the identity

$$ \sum_{k=1}^{\infty}t^k=\frac{t}{1-t}. $$

A: First, the sum of the series is correct because
$$S(x)=\sum_{k=1}^\infty \frac{x^k}{k\exp(kx)}=\sum_{k=1}^\infty\frac1k\left(\frac x{\mathrm{e}^x}\right)^k=-\ln\left(1-\frac x{\mathrm{e}^x}\right).$$
We note that $x\,\mathrm{e}^{-x}$ takes its values between $0$ and $\mathrm{e}^{-1}$ for real positive values of $x$, hence $S(x)\geq 0$ for all $x\geq 0$.
Second the integral:


*

*when $x\to0$, we have $S(x)=-\ln(1-x)+o(x)=x+o(x)$

*when $x\to\infty$, $S(x)=-\ln(1-x\,\mathrm e^{-x})=x\,\mathrm{e}^{-x}+o(\mathrm{e}^{-x})$


from what we conclude that the integral $J=\int_0^\infty S(x)\mathrm{d}x$ converges absolutely (and therefore is finite).
