Prove $\int_0^\infty \frac{\ln^2x\ln(1+x)}{x(1+x)} dx=7\zeta(4)$

Question

How to prove, without using beta function that

$$\int_0^\infty\frac{\ln^2x\ln(1+x)}{x(1+x)}\ dx=7\zeta(4)$$

where $\zeta$ is the Riemann zeta function.

Also, can we take advantage of this result to solve some harmonic series?

Answer 1

Denote the integral by $I$, split up into two parts in the point $1$ and let $x\to \frac{1}{x}$ in the second part to get: $$I=\color{blue}{\int_0^1\frac{\ln^2x\ln(1+x)}{x(1+x)} dx}+\color{red}{\int_0^1 \frac{\ln^2 x \ln(1+1/x)}{1+x}dx}$$ $$\require{cancel}=\color{blue}{\int_0^1 \frac{\ln^2 x\ln(1+x)}{x}dx-\cancel{\int_0^1\frac{\ln^2 x\ln(1+x)}{1+x}dx}}+\color{red}{\cancel{\int_0^1 \frac{\ln^2 x\ln(1+x)}{1+x}dx}-\int_0^1\frac{\ln^3 x}{1+x}dx}$$ $$\overset{\color{blue}{IBP}}=\color{blue}{-\frac13\int_0^1 \frac{\ln^3 x}{1+x}dx}\color{red}{-\int_0^1 \frac{\ln^3 x}{1+x}dx}=\color{purple}{-\frac43\int_0^1\frac{\ln^3 x}{1+x}dx}$$ $$=-\frac43\sum_{n=1}^\infty (-1)^{n-1} \int_0^1 x^{n-1} \ln^3 xdx=8\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^4}=7\zeta(4)$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[10px,#ffd]{\int_{0}^{\infty}{\ln^{2}\pars{x} \ln\pars{1 + x} \over x\pars{1 + x}}\,\dd x = 7\zeta\pars{4}}}:\ {\Large ?}.\qquad$ $\ds{\zeta}$ is the Riemann Zeta Function.

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\begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}{\ln^{2}\pars{x}\ln\pars{1 + x} \over x\pars{1 + x}}\,\dd x} = \int_{1}^{\infty}{\ln^{2}\pars{x - 1}\ln\pars{x} \over \pars{x - 1}x}\,\dd x \\[5mm] = &\ \int_{1}^{0}{\ln^{2}\pars{1/x - 1}\ln\pars{1/x} \over \pars{1/x - 1}\pars{1/x}}\,\pars{-\,{\dd x \over x^{2}}} \\[5mm] = &\ -\int_{0}^{1}{\bracks{\ln\pars{1 - x} - \ln\pars{x}}^{\, 2}\ln\pars{x} \over 1 - x}\,\dd x \\[5mm] = &\ -\int_{0}^{1}{\bracks{\ln\pars{x} - \ln\pars{1 - x}}^{\, 2}\ln\pars{1 - x} \over x}\,\dd x \\[1cm] = &\ -\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\ln^{2}\pars{x}\,\dd x + 2\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x \\[1mm] & \phantom{=\!} -\int_{0}^{1}{\ln^{3}\pars{1 - x} \over x}\,\dd x \label{1}\tag{1} \end{align} Integrating by parts the last integral in (\ref{1}),

$\ds{\int_{0}^{1}{\ln^{3}\pars{1 - x} \over x}\,\dd x = \int_{0}^{1}{\ln^{3}\pars{x} \over 1 - x}\,\dd x = 3\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\ln^{2}\pars{x}\,\dd x}$.

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(\ref{1}) is reduced to \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}{\ln^{2}\pars{x}\ln\pars{1 + x} \over x\pars{1 + x}}\,\dd x} \\[5mm] = &\ -4\int_{0}^{1}\ \underbrace{\ln\pars{1 - x} \over x} _{\ds{-\,\mrm{Li}_{2}'\pars{x}}}\ \ln^{2}\pars{x}\,\dd x + 2\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x \label{2}\tag{2} \end{align}

$\ds{\mrm{Li}_{\large s}}$ is the Polylogarithm.

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$\ds{\large\mbox{Evaluation of}\ \int_{0}^{1}{\ln\pars{1 - x} \over x}\,\ln^{2}\pars{x}\,\dd x}$: \begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\ln^{2}\pars{x}\,\dd x} = -\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\,\ln^{2}\pars{x}\,\dd x \\[5mm] \,\,\,\stackrel{\mrm{IBP}}{=}\,\,\,& 2\int_{0}^{1}{\mrm{Li}_{2}\pars{x} \over x}\,\ln\pars{x}\,\dd x = 2\int_{0}^{1}\mrm{Li}_{3}'\pars{x}\ln\pars{x}\,\dd x = \\[5mm] = &\ -2\int_{0}^{1}{\mrm{Li}_{3}\pars{x} \over x}\,\dd x = -2\int_{0}^{1}\mrm{Li}_{4}'\pars{x}\,\dd x = -2\,\mrm{Li}_{4}\pars{1} = \bbx{-2\zeta\pars{4}} \label{3}\tag{3} \end{align}

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$\ds{\large\mbox{Evaluation of}\ \int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x}$: \begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x} = \left.{\partial^{3} \over \partial\mu\,\partial\nu^{2}}\int_{0}^{1}x^{\mu - 1} \bracks{\pars{1 - x}^{\nu} - 1}\,\dd x\, \right\vert_{\ds{\mu\ =\ 0^{+} \atop \nu\ =\ 0}} \\[5mm] = &\ {\partial^{3} \over \partial\mu\,\partial\nu^{2}} \bracks{{\Gamma\pars{\mu}\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu + 1}} - {1 \over \mu}} _{\ds{\mu\ =\ 0^{+} \atop \nu\ =\ 0}} \\[5mm] = &\ \partiald[2]{}{\nu}\partiald{}{\mu}\braces{ {1 \over \mu}\bracks{{\Gamma\pars{\mu + 1}\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu + 1}} - 1}} _{\ds{\mu\ =\ 0^{+} \atop \nu\ =\ 0}} = \bbx{-\,{1 \over 2}\,\zeta\pars{4}} \label{4}\tag{4} \end{align}

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Lets replace (\ref{3}) and (\ref{4}) in \ref{2}: $$ \bbox[10px,#ffd]{\int_{0}^{\infty}{\ln^{2}\pars{x}\ln\pars{1 + x} \over x\pars{1 + x}}\,\dd x} = -4\bracks{-2\zeta\pars{4}} + 2\bracks{-\,{1 \over 2}\,\zeta\pars{4}} = \bbx{7\zeta\pars{4}} $$

Answer 3

Following the same approach here, we can get the generalization

$$\int_0^\infty\frac{\ln^{2a}(x)\ln(1+x)}{x(1+x)}dx=(2a)!(2a+2)\left(1-2^{-2a-1}\right)\zeta(2a+2)$$