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

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.

• 'The site is classified'? – Steven Stadnicki Aug 6 '14 at 14:58
• @StevenStadnicki Yes Sir, it is classified but I don't think that will be the issue. I use that term to avoid the question from other users here: "where did you get this problem?" and sort of. Now, can we back to the main issue, calculate the integral (((o(ﾟ▽ﾟ)o))) – Anastasiya-Romanova 秀 Aug 6 '14 at 15:09
• The reasons that people ask 'where did you get this problem?' are (a) understanding what level solutions should be aimed at and what the initial framework is, and (b) making sure they aren't answering problems that come from e.g. contest sites or other sites that have specific policies against looking for outside help. – Steven Stadnicki Aug 6 '14 at 15:24
• @StevenStadnicki Well, I can assure you that this problem is NOT taken from contest sites nor also my homework – Anastasiya-Romanova 秀 Aug 6 '14 at 15:29
• Anyway I welcome any level solutions and I don't think you need to worry about point (a) because of this one: meta.math.stackexchange.com/questions/11419/… – Anastasiya-Romanova 秀 Aug 6 '14 at 15:33


\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}

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

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

• Sir, please read the OP carefully (ﾉ≧∀≦)ﾉ – Anastasiya-Romanova 秀 Aug 6 '14 at 15:44
• @V-Moy $0$ k. I'll rethink the answer. I'm sorry I didn't read carefully your question. Thanks. – Felix Marin Aug 6 '14 at 16:30
• even though the OP is not satisfied with this answer, I like it. I saw this question just a few minutes after it was posted and my first thought was: "I'm sure this could be solved using that "limit of the derivative" approach that Felix often uses" :) – etothepitimesi Aug 6 '14 at 16:35
• @el.Salvador I didn't like the first answer because I also used the same approach as Mr. Felix Marin did but I do like the ADDENDA. – Anastasiya-Romanova 秀 Aug 6 '14 at 17:30
• Mr. Felix Marin: To be honest, your solution is good but it would be better if you add some extra explanations i.e: you use by parts method multiple times and also use property of logarithmic integral function. I've checked your answer and I accept it. Thank you very much (∩_∩) – Anastasiya-Romanova 秀 Aug 6 '14 at 17:36


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}

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

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

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 $$\zeta(5,\frac{1}{3})=\frac{2\pi^5}{3\sqrt3}+121\zeta(5).$$

• The last step is what worries the OP. – Felix Marin Aug 6 '14 at 16:35
• @FelixMarin, I added the calculation of $\zeta(5,\frac{1}{6})$ and $\zeta(5,\frac{1}{3})$ in my answer. – xpaul Aug 7 '14 at 13:18
• I just saw your calculation. It was pretty fine. – Felix Marin Aug 11 '14 at 18:43
• @FelixMarin, it took me some time to get it. – xpaul Aug 11 '14 at 23:52

Here's my solution to this problem.