Lambert transform of monomials

Question

The Lambert transform of a function $f(x)$, is given by: $$\int_{0}^{\infty}\frac{f(x)}{e^{xt}-1}dx\;\;\;\;(\Re(t)>0)$$

We wish for a closed form of the transform : $$\int_{0}^{\infty}\frac{x^{n}}{e^{xt}-1}dx\;\;\;\;\;\;(n\in\mathbb{Z}^{+})$$

Answer 1

$$ \frac{1}{e^{xt} - 1} = e^{-xt} \frac{1}{1 - e^{-xt}} = e^{-xt}(1+e^{-xt} + e^{-2xt}+ \dots ). $$ Thus, $$ \int_0^\infty \frac{x^n}{e^{xt} - 1} dx = \int_0^\infty x^n e^{-xt} \left(\sum_{k=0}^\infty e^{-k xt} \right)dx = \sum_{k=1}^\infty \int_0^\infty x^n e^{-kxt} dx. $$ Now $$ \int_0^\infty x^n e^{-kxt} dx = \frac 1 {(kt)^{n+1}} \int_0^\infty y^n e^{-y} dy = \frac{n!}{(kt)^{n+1}}. $$ So $$ \int_0^\infty \frac{x^n}{e^{xt} - 1} dx = \frac{n!}{t^{n+1}} \sum_{k=1}^\infty \frac{1}{k^{n+1}} =\frac{n!}{t^{n+1}} \ \zeta(n+1). $$