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

Question

This integral comes from a well-known site (I am sorry, the site is classified due to regarding the OP.)

$$\int_0^1\frac{1-x}{1-x^6}\ln^4x\,dx$$

I can calculate the integral using the help of geometric series and I get the answer \begin{align} \sum_{n=0}^\infty\left(\frac{24}{(6n+1)^5}-\frac{24}{(6n+2)^5}\right) &=\frac{1}{6^5}\left(\Psi^{(4)}\left(\frac{1}{3}\right)-\Psi^{(4)}\left(\frac{1}{6}\right)\right)\\ &=\frac{16\sqrt{3}}{729}\pi^5+\frac{605}{54}\zeta(5) \end{align} To be honest, I use Wolfram Alpha to calculate the sum of series. The problem is I don't think this is the correct way to calculate the integral because I use a machine to help me. I tried another way, I used partial fraction to decompose the integrand as $$\frac{\ln^4x}{3(x+1)}+\frac{\ln^4x}{2(x^2+x+1)}-\frac{2x-1}{6(x^2-x+1)}\ln^4x$$ but none of them seemed easy to calculate. Could anyone here please help me to calculate the integral preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.

Answer 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: $$ \int_{0}^{1}{\ln^{4}\pars{x} \over 3\pars{x + 1}}= -8\,{\rm Li}_{5}\pars{-1} ={15 \over 2}\,\zeta\pars{5} $$

Answer 2

You can use the geometric series to evaluate. In fact, \begin{eqnarray} I&=&\int_0^1\frac{(1-x)\ln^4x}{1-x^6}dx\\ &=&\int_0^1\sum_{n=0}^\infty(1-x)x^{6n}\ln^4xdx\\ &=&\sum_{n=0}^\infty(1-x)x^{6n}\ln^4xdx\\ &=&\sum_{n=0}^\infty\left(\frac{24}{(6n+1)^5}-\frac{24}{(6n+2)^5}\right)\\ &=&\frac{1}{324}\left(\sum_{n=0}^\infty\frac{1}{(n+1/6)^5}-\sum_{n=0}^\infty\frac{1}{(n+1/3)^5}\right)\\ &=&\frac{1}{324}(\zeta(5,\frac{1}{6})-\zeta(5,\frac{1}{3}))\\ &=&\frac{16\pi^5}{243\sqrt3}+\frac{605}{54}\zeta(5). \end{eqnarray} The (tedious) calculation of $\zeta(5,\frac{1}{6})$ and $\zeta(5,\frac{1}{3})$ is derived from two results from this (in Pages 1628-1629). First we have $$ \zeta(5,\frac{1}{6})-\zeta(5,\frac{5}{6})=\frac{44\pi^5}{\sqrt3}, \zeta(5,\frac{1}{6})+\zeta(5,\frac{5}{6})=(2^5-1)(2^5-1)\zeta(5), $$ from which we obtain $$ \zeta(5,\frac{1}{6})=\frac{22\pi^5}{\sqrt3}+3751\zeta(5). $$ Since
$$ \zeta(5,\frac{1}{6})+\zeta(5,\frac{1}{3})+\zeta(5,\frac{1}{2})+\zeta(5,\frac{2}{3})+\zeta(5,\frac{5}{6})+\zeta(5)=6^5\zeta(5),\zeta(5,\frac{1}{2})=31\zeta(5) $$ we have $$ \zeta(5,\frac{1}{3})+\zeta(5,\frac{2}{3})=242\zeta(5). $$ Also $$ \zeta(5,\frac{1}{3})-\zeta(5,\frac{2}{3})=\sum_{n=-\infty}^\infty\frac1{(n+\frac{1}{3})^5}=\frac{4\pi^5}{3\sqrt3}.$$ From this, we obtain $$\zeta(5,\frac{1}{3})=\frac{2\pi^5}{3\sqrt3}+121\zeta(5). $$

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

Answer 4

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

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}

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

Answer 5

Here's my solution to this problem.