How can we show that $\int_{0}^{1}{\ln^{2k+1}x\ln(1-x)\over x}\mathrm dx=\Gamma(2k+2) \zeta(2k+3)?$

Question

Motivated by this question, by a small change we arrived at a simple closed form

$$\int_{0}^{1}{\ln^{2k+1}x\ln(1-x)\over x}\mathrm dx=\Gamma(2k+2)\zeta(2k+3)\tag1$$ $k\ge0$

How may one prove $(1)?$

Answer 1

Note that we have

$$ \int_0^1\frac{\log^{2k+1}(x) \log(1-x)}{x}\,dx=\left. \left(\frac{d^{2k+1}}{da^{2k+1}}\int_0^1\frac{x^a\log(1-x)}{x}\,dx\right)\right|_{a=0} \tag 1$$

Then, expanding $\log(1-x)$ is its Mclaurin series, we see that

$$\int_0^1 \frac{x^a\log(1-x)}{x}\,dx=-\sum_{n=1}^\infty \frac{1}{n(n+a)}\tag 2$$

Differentiating $2k+1$ times the right-hand side of $(2)$ reveals

$$\frac{d^{2k+1}}{da^{2k+1}}\left (-\sum_{n=1}^\infty \frac{1}{n(n+a)} \right)=(2k+1)!\sum_{n=1}^\infty \frac{1}{n(n+a)^{2k+2}}\tag 3$$

Finally, setting $a=0$ in $(3)$, and using $\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}$, $s>1$, and $\Gamma(n+1)=n!$, we obtain

$$\int_0^1\frac{\log^{2k+1}(x)\log(1-x)}{x}\,dx=\Gamma(2k+2)\zeta(2k+3) $$

as was to be shown!

Answer 2

Making the change of variables $y=-\ln x$ we get $$\begin{align} \int_0^1\frac{\ln^m x\ln(1-x)}{x}\,dx &=-\sum_{n=1}^\infty\frac1n\int_0^1\ln^m(x)\, x^{n-1}\,dx\\ &=-\sum_{n=1}^\infty\frac1n(-1)^m\int_0^\infty y^me^{-ny}\,dy\\ &=(-1)^{m-1}\sum_{n=1}^\infty \frac{m!}{n^{m+2}}=(-1)^{m-1} m!\zeta(m+2). \end{align} $$

Answer 3

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$\ds{\int_{0}^{1}{\ln^{2k + 1}\pars{x}\ln\pars{1 - x} \over x}\,\dd x = \Gamma\pars{2k + 2}\zeta\pars{2k + 3}:\ {\large ?}.\qquad k \geq 0}$.

\begin{align} \mbox{Note that}\quad \int_{0}^{1}{\ln^{2k + 1}\pars{x}\ln\pars{1 - x} \over x}\,\dd x = -\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2k + 1}\pars{x}\,\dd x \end{align}

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Lets $\ds{\,\mc{I}_{sn} \equiv \int_{0}^{1}\mrm{Li}_{s}'\pars{x}\ln^{n}\pars{x}\,\dd x}$. Then, \begin{align} \left.\vphantom{\Large A}\mc{I}_{sn}\right\vert_{\ n\ \in\ \mathbb{N}_{\ \geq\ 1}} & = -\int_{0}^{1}\mrm{Li}_{s}\pars{x}\bracks{n\ln^{n - 1}\pars{x}\,{1 \over x}} \,\dd x = -n\int_{0}^{1}{\mrm{Li}_{s}\pars{x} \over x}\,\ln^{n - 1}\pars{x}\,\dd x \\[5mm] & = -n\int_{0}^{1}\mrm{Li}_{s + 1}'\pars{x}\ln^{n - 1}\pars{x}\,\dd x\implies \left\{\begin{array}{rcl} \ds{\mc{I}_{sn}} & \ds{=} & \ds{-n\,\mc{I}_{s + 1,n - 1}\,,\quad n \in \mathbb{N}_{\ \geq\ 1}} \\[2mm] \ds{\mbox{and}\ \mc{I}_{s0}} & \ds{=} & \ds{\,\mrm{Li}_{s}\pars{1} = \zeta\pars{s}} \end{array}\right. \end{align}

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Then, \begin{align} &\int_{0}^{1}{\ln^{2k + 1}\pars{x}\ln\pars{1 - x} \over x}\,\dd x \\[5mm] = &\ -\,\mc{I}_{2,2k + 1} = \pars{2k + 1}\,\mc{I}_{3,2k} = -\pars{2k + 1}\pars{2k}\,\mc{I}_{4,2k - 1} \\[5mm] = & \cdots = -\pars{2k + 1}\pars{2k}\cdots 1\pars{-\,\mc{I}_{2k + 3,0}} = \pars{2k + 1}!\,\zeta\pars{2k + 3} = \bbx{\Gamma\pars{2k + 2}\zeta\pars{2k + 3}} \end{align}

Moreover,

\begin{align} &\int_{0}^{1}{\ln^{2k + 1}\pars{x}\ln\pars{1 - x} \over x}\,\dd x = -\int_{0}^{1}{\ln^{2k + 2}\pars{x} \over 2k + 2}\, {-1 \over 1 - x}\,\dd x \\[5mm] \stackrel{x\ =\ \exp\pars{-t}}{=}\,\,\,& {1 \over 2k + 2}\int_{\infty}^{0} {\pars{-t}^{2k + 2} \over 1 - \expo{-t}}\pars{-\expo{-t}}\,\dd t = {1 \over 2k + 2}\ \underbrace{% \int_{0}^{\infty}{t^{2k + 2} \over \expo{t} - 1}\,\dd t} _{\ds{\Gamma\pars{2k + 3}\zeta\pars{2s + 3}}}\label{1}\tag{1} \\[5mm] = &\ \bbx{\Gamma\pars{2k + 2}\zeta\pars{2k + 3}} \end{align}

The last integral, in 'line' \eqref{1}, is a well known $\ds{\zeta}$ function integral representation $\ds{\pars{~\mbox{for}\ \Re\pars{k} > - 1~}}$ or/and $\ds{\zeta}$ alternative definition.