Asymptotic expansion for $\frac{1}{2\zeta(3)}\int_x^\infty \frac{u^2}{e^u - 1} du$? Is there an asymptotic expansion for the function:
\begin{equation}
g(x)=\frac{1}{2\zeta(3)}\int_x^\infty \frac{u^2}{e^u - 1} du,
\end{equation}
over the domain $x\in [0,\infty)$ in terms of elementary functions? Here $\zeta$ is the Riemann zeta function and $1/2\zeta(3)$ is a normalization factor included to ensure $g(0)=1$.
We have $$g(x) = \dfrac1{2 \zeta(3)}\underbrace{\int_x^{\infty} \dfrac{u^2}{e^u-1} du}_{I(x)}$$
We will now obtain a series expansion for $I(x)$. We have $$I(x) = \int_x^{\infty} \dfrac{u^2}{e^u-1} du = \int_x^{\infty} \dfrac{u^2 e^{-u}}{1-e^{-u}} du = \int_x^{\infty} \sum_{k=0}^{\infty} u^2 e^{-(k+1)u} du = \sum_{k=1}^{\infty} \int_x^{\infty} u^2 e^{-ku} du$$
We now have that
$$\int_x^{\infty} u^2 e^{-ku} du = \dfrac{e^{-kx} \left(k^2 x^2 + 2kx + 2\right)}{k^3}$$
We have have
$$I(x) = \sum_{k=1}^{\infty} \dfrac{e^{-kx} \left(k^2 x^2 + 2kx + 2\right)}{k^3}\tag{$\star$}$$
For a given $x$, truncating $(\star)$ will give you an exponentially converging approximation. Note that $$I(0) = \displaystyle \sum_{k=1}^{\infty} \dfrac2{k^3} = 2 \zeta(3)$$
yes is equal to the incomplete gamma function http://mathworld.wolfram.com/IncompleteGammaFunction.html

$$ \Gamma (3,x) $$