Prove that $\int_0^1\frac{1-x}{1-x^6}\ln^4x\,dx=\frac{16\sqrt{3}}{729}\pi^5+\frac{605}{54}\zeta(5)$

Solution 1:

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}{1 - x \over 1 - x^{6}}\,\ln^{4}\pars{x}\,\dd x:\ {\large ?}}$

\begin{align}&\color{#c00000}{\int_{0}^{1}% {1 - x \over 1 - x^{6}}\,\ln^{4}\pars{x}\,\dd x} =\lim_{\mu \to 0}\partiald[4]{}{\mu}\int_{0}^{1} {1 - x \over 1 - x^{6}}\,x^{\mu}\,\dd x \\[3mm]&=\lim_{\mu \to 0}\partiald[4]{}{\mu}\int_{0}^{1} {x^{\mu/6} - x^{\pars{\mu + 1}/6} \over 1 - x}\,{1 \over 6}\,x^{-5/6}\,\dd x ={1 \over 6}\,\lim_{\mu \to 0}\partiald[4]{}{\mu}\int_{0}^{1} {x^{\pars{\mu - 5}/6} - x^{\pars{\mu - 4}/6} \over 1 - x^{6}}\,\dd x \\[3mm]&={1 \over 6}\,\lim_{\mu \to 0}\partiald[4]{}{\mu}\bracks{% \int_{0}^{1}{1 - x^{\pars{\mu - 4}/6} \over 1 - x^{6}}\,\dd x -\int_{0}^{1}{1 - x^{\pars{\mu - 5}/6} \over 1 - x^{6}}\,\dd x} \\[3mm]&={1 \over 6}\,\lim_{\mu \to 0}\partiald[4]{}{\mu}\bracks{% \Psi\pars{\mu + 2 \over 6} - \Psi\pars{\mu + 1 \over 6}} \end{align}

$$ \color{#66f}{\large\int_{0}^{1}% {1 - x \over 1 - x^{6}}\,\ln^{4}\pars{x}\,\dd x ={1 \over 7776}\,\bracks{% \Psi^{\tt\pars{IV}}\pars{1 \over 3} - \Psi^{\tt\pars{IV}}\pars{1 \over 6}}} \approx {\tt 23.2507} $$

ADDENDA

\begin{align} &\color{#00f}{\int_{0}^{1}{\ln^{4}\pars{x} \over x - a}\,\dd x} =-\int_{0}^{1}{\ln^{4}\pars{a\bracks{x/a}} \over 1 - x/a}\,{\dd x \over a} =-\int_{0}^{1/a}{\ln^{4}\pars{ax} \over 1 - x}\,\dd x \\[3mm]&=-\int_{0}^{1/a}\ln\pars{1 - x}\,4\ln^{3}\pars{ax}\,{1 \over x}\,\dd x =4\int_{0}^{1/a}{\rm Li}_{2}'\pars{x}\ln^{3}\pars{ax}\,\dd x \\[3mm]&=-4\int_{0}^{1/a}{\rm Li}_{2}\pars{x}\,3\ln^{2}\pars{ax}\,{1 \over x} \,\dd x \\[3mm]&=-12\int_{0}^{1/a}{\rm Li}_{3}'\pars{x}\ln^{2}\pars{ax}\,\dd x =12\int_{0}^{1/a}{\rm Li}_{3}\pars{x}2\ln\pars{ax}\,{1 \over x}\,\dd x \\[3mm]&=24\int_{0}^{1/a}{\rm Li}_{4}'\pars{x}\ln\pars{ax}\,\dd x =-24\int_{0}^{1/a}{\rm Li}_{4}\pars{x}\,{1 \over x}\,\dd x =-24\int_{0}^{1/a}{\rm Li}_{5}'\pars{x}\,\dd x \\[3mm]&=\color{#00f}{-24\,{\rm Li}_{5}\pars{1 \over a}} \end{align}

Now, you can use partial fractions. For instance: $$ \int_{0}^{1}{\ln^{4}\pars{x} \over 3\pars{x + 1}}= -8\,{\rm Li}_{5}\pars{-1} ={15 \over 2}\,\zeta\pars{5} $$

Solution 2:

Note that $$ {1 - x \over 1 - x^{6}}=\prod_{n =1}^{5}{1 \over x - x_{n}} =\sum_{n = 1}^{5}{b_{n} \over x - x_{n}}\,,\ \left\lbrace\begin{array}{rcl} x_{n} & = & \expo{n\pi\ic/3} \\[2mm] b_{n} & = &{1 \over 6}\,x_{n}\pars{x_{n} - 1} \end{array}\right. $$

\begin{align}&\color{#c00000}{\int_{0}^{1}% {1 - x \over 1 - x^{6}}\,\ln^{4}\pars{x}\,\dd x} =\sum_{n = 1}^{5}b_{n}\int_{0}^{1}{\ln^{4}\pars{x} \over x - x_{n}}\,\dd x =-24\sum_{n = 1}^{5}b_{n}\,{\rm Li}_{5}\pars{1 \over x_{n}} \\[3mm]&=4{\rm Li}_{5}\pars{\expo{-\pi\ic/3}}+ 4\ic\root{3}{\rm Li}_{5}\pars{\expo{-2\pi\ic/3}}-8{\rm Li}_{5}\pars{-1} -4\ic\root{3}{\rm Li}_{5}\pars{\expo{2\pi\ic/3}} \\[3mm]&+4{\rm Li}_{5}\pars{\expo{\pi\ic/3}} \\[3mm]&=8\bracks{\Re{\rm Li}_{5}\pars{\expo{\pi\ic/3}} +\root{3}\Im{\rm Li}_{5}\pars{\expo{2\pi\ic/3}} - {\rm Li}_{5}\pars{-1}} \end{align} where we used a result of my previous answer: $\ds{\int_{0}^{1}{\ln^{4}\pars{x} \over x - a} =-24\,{\rm Li}_{5}\pars{1 \over a}}$.

The following results, but the last one, are found with Jonquiere Inversion Formula: $$ \Re{\rm Li}_{5}\pars{\expo{\pi\ic/3}}={25 \over 54}\,\zeta\pars{5}\,,\quad \Im{\rm Li}_{5}\pars{\expo{2\pi\ic/3}}={2 \over 729}\,\pi^{5}\,,\quad {\rm Li}_{5}\pars{-1}=-\,{15 \over 16}\,\zeta\pars{5} $$ The last one is found by manipulating the PolyLog serie definition.

\begin{align}&\color{#c00000}{\int_{0}^{1}% {1 - x \over 1 - x^{6}}\,\ln^{4}\pars{x}\,\dd x} =8\braces{{25 \over 54}\,\zeta\pars{5} + \root{3}\,{2 \over 729}\,\pi^{5} -\bracks{-\,{15 \over 16}\,\zeta\pars{5}}} \end{align}

$$ \color{#66f}{\large\int_{0}^{1}% {1 - x \over 1 - x^{6}}\,\ln^{4}\pars{x}\,\dd x ={16\root{3} \over 729}\,\pi^{5} + {605 \over 54}\,\zeta\pars{5}} \approx {\tt 23.2507} $$

Solution 3:

A few identities to consider that may help whittle your solution down to the required form are:

$$\psi^{(4)}(2x)=1/2(\psi^{(4)}(x)+\psi^{(4)}(x+1/2))$$ and

$$\psi^{(4)}(1-x)=\psi^{(4)}(x)+4\pi^{5}(\cos(2\pi x)+5)\cot(\pi x)\csc^{4}(\pi x)$$

i.e. $$\psi^{(4)}(1/3)=1/2(\psi^{(4)}(1/6)+\psi^{(4)}(2/3))$$

and $$\psi^{(4)}(2/3)=\psi^{(4)}(1/3)+4\pi^{5}(\cos(2\pi /3)\cot(\pi /3)\csc^{4}(\pi/3)$$

Prove that $\int_0^1\frac{1-x}{1-x^6}\ln^4x\,dx=\frac{16\sqrt{3}}{729}\pi^5+\frac{605}{54}\zeta(5)$

Solution 1:

ADDENDA

Solution 2:

Solution 3:

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