Cómo mostramos que $\int_{0}^{1}{\ln^k(x)\over x}\ln\left(1-\sqrt[n]{x}\right)\mathrm dx=(-n)^{k+1}k!\zeta(k+2)?$

Question

Propuesto:

Una forma cerrada simple

$$\int_{0}^{1}{\ln^k(x)\over x}\ln\left(1-\sqrt[n]{x}\right)\mathrm dx=(-n)^{k+1}k!\zeta(k+2)\tag1$$

Donde $n,k=1,2,3,\cdots$

Mi intento:

$u=\sqrt[n]{x}\implies {nx\over u}du=dx$, $x=u^n$, entonces $(1)$ se convierte en

$$n^{k+1}\int_{0}^{1}{\ln^k(u)\over u}\ln(1-u)\mathrm du\tag2$$

$$-n^{k+1}\sum_{v\ge1}{1\over v}\int_{0}^{1}u^v\ln^k(u)\mathrm du\tag3$$

$$-n^{k+1}\sum_{v\ge1}{1\over v}\cdot{(-1)^kk!\over (v+1)^{k+1}}\tag4$$

$$(-n)^{k+1}k!\sum_{v\ge1}{1\over v(v+1)^{k+1}}\tag5$$

Cómo podemos probar $(1)?$

Answer 1

Con el cambio de la variable $x=z^n$ se quedan con algo que claramente depende de los derivados de una función Beta:

$$ \int_{0}^{1}\log(1-u)\log^k(u)\frac{du}{u} = -\lim_{\substack{\alpha\to 0^+\\ \beta\to 0^+}}\frac{d^k}{d\alpha^k}\frac{d}{d\beta}\int_{0}^{1}u^{\alpha-1}(1-u)^{\beta}\,du$ $ y el lado derecho es simplemente el $$ \lim_{\alpha\to 0^+}\frac{d^k}{d\alpha^k}\left(\frac{H_\alpha}{\alpha}\right) =(-1)^k k!\,\zeta(k+2)$ $ por un simple cálculo a través de la función Digamma y sus derivados.

Como alternativa, permite para calcular el % de OP $$\begin{eqnarray*} \sum_{n\geq 1}\frac{1}{n(n+1)^{k+1}}&=&\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+1}\right)\frac{1}{(n+1)^k}\\&\stackrel{SBP}{=}&\sum_{n\geq 1}\left(1-\frac{1}{n+1}\right)\frac{1}{(n+1)^k}-\sum_{n\geq 1}\left(1-\frac{1}{n+1}\right)\frac{1}{(n+2)^k}\\&=&1-\zeta(k+1)+\sum_{n\geq 1}\frac{1}{n(n+1)^k} \end{eqnarray*}$por inducción $(5)$ $: $$ \sum_{n\geq 1}\frac{1}{n(n+1)^k} = k-\zeta(2)-\zeta(3)-\ldots-\zeta(k).$ $

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}{\ln^{k}\pars{x} \over x}\ln\pars{1 - \root[\large n]{x}}\,\dd x & \,\,\,\stackrel{\root[\large n]{x}\ \mapsto\ x}{=}\,\,\, n^{k + 1}\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\ln^{k}\pars{x}\,\dd x \\[5mm] & = -n^{k + 1}\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{k}\pars{x}\,\dd x \label{1}\tag{1} \end{align}

Tenga en cuenta que

\begin{align} &\left.\int_{0}^{1}\mrm{Li}_{s}'\pars{x}\ln^{\ell}\pars{x}\,\dd x\, \right\vert_{\ \ell\ >\ 0} = -\ell\int_{0}^{1}\mrm{Li}_{s + 1}'\pars{x}\ln^{\ell - 1}\pars{x}\,\dd x \\[5mm] = &\ \ell\pars{\ell - 1}\int_{0}^{1}\mrm{Li}_{s + 2}'\pars{x} \ln^{\ell - 2}\pars{x}\,\dd x = \cdots = \pars{-1}^{\ell}\ell\pars{\ell - 1}\cdots 1 \int_{0}^{1}\mrm{Li}_{s + \ell}'\pars{x}\,\dd x \\[5mm] = &\ \pars{-1}^{\ell}\,\ell!\,\,\mrm{Li}_{s + \ell}\pars{1} = \bbx{\pars{-1}^{\ell}\,\ell!\,\zeta\pars{s + \ell}} \end{align}

La expresión \eqref{1} se convierte en

$$ \int_{0}^{1}{\ln^{k}\pars{x} \over x}\ln\pars{1 - \root[\gran n]{x}}\,\dd x = -n^{k + 1}\bracks{\pars{-1}^{k}\,k!\,\zeta\pars{2 + k}} = \bbx{\pars{-n}^{k + 1}\,k!\,\zeta\pars{k + 2}} $$