# Hard Definite integral involving the Zeta function

Prove that: $$\displaystyle \int_{0}^{1}\frac{1-x}{1-x^{6}}{\ln^4{x}} \ {dx} = \frac{16{{\pi}^{5}}}{243\sqrt[]{{3}}}+\frac{605\zeta(5)}{54}$$

I was able to simplify it a bit by substituting ${y = -\ln{x}}$ and some further mathematical manipulation but was not able to get the correct form.

#### Solutions Collecting From Web of "Hard Definite integral involving the Zeta function"

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$\ds{\int_{0}^{1}{1 – x \over 1 – x^{6}}\,\ln^{4}\pars{x}\,\dd x= {16 \over 243\root{3}}\,\pi^{5} + {605 \over 54}\,\zeta\pars{5}}$

Lets $\ds{\quad x_{n} = \expo{n\pi\ic/3}\,,\quad n = 0,1,2,3,4,5\quad}$ such that
\begin{align}
{1 – x \over 1 – x^{6}}&
=\pars{x – 1}\sum_{n = 0}^{5}{x_{n}/6 \over x – x_{n}}
={1 \over 6}\sum_{n = 0}^{5}x_{n}
\pars{{x – x_{n} \over x – x_{n}} + {x_{n} – 1 \over x – x_{n}}}
\\[3mm]&={1 \over 6}\,\underbrace{\sum_{n = 0}^{5}x_{n}}_{\ds{=\ 0}}
+{1 \over 6}\sum_{n = 0}^{5}{x_{n}\pars{x_{n} – 1} \over x – x_{n}}
={1 \over 6}\sum_{n = 1}^{5}{x_{n}\pars{x_{n} – 1} \over x – x_{n}}
={1 \over 6}\sum_{n = -2}^{2}{x_{n + 3}\pars{x_{n + 3} – 1} \over x – x_{n + 3}}
\\[3mm]&={1 \over 6}\sum_{n = -2}^{2}{x_{n}\pars{x_{n} + 1} \over x + x_{n}}
\end{align}

Then,
$$\color{#c00000}{\int_{0}^{1}{1 – x \over 1 – x^{6}}\,\ln^{4}\pars{x}\,\dd x} ={1 \over 6}\sum_{n = -2}^{2}x_{n}\pars{x_{n} + 1} \color{#00f}{\int_{0}^{1}{\ln^{4}\pars{x} \over x + x_{n}}\,\dd x}\tag{1}$$

Lets evaluate the integral:
\begin{align}
&\color{#00f}{\int_{0}^{1}{\ln^{k}\pars{x} \over x – a}\,\dd x}
=-\int_{0}^{1}{\ln^{k}\pars{a\bracks{x/a}} \over 1 – x/a}
\,{\dd x \over a}
=-\int_{0}^{1/a}{\ln^{k}\pars{ax} \over 1 – x}\,\dd x
\\[3mm]&=-\int_{0}^{1/a}\ln\pars{1 – x}k\ln^{k – 1}\pars{ax}\,{1 \over x}\,\dd x
=k\int_{0}^{1/a}{{\rm Li}_{1}\pars{x} \over x}\,\ln^{k – 1}\pars{ax}\,\dd x
\\[3mm]&=-k\pars{k – 1}\int_{0}^{1/a}
{{\rm Li}_{2}\pars{x} \over x}\,\ln^{k – 2}\pars{ax}\,\dd x
=\cdots
\\[3mm]&=\pars{-1}^{r}\,{k! \over \pars{k – r – 1}!}
\int_{0}^{1/a}
{{\rm Li}_{1 + r}\pars{x} \over x}\,\ln^{k – r – 1}\pars{ax}\,\dd x
=\cdots
\\[3mm]&=\pars{-1}^{k – 1}k!
\int_{0}^{1/a}{{\rm Li}_{k}\pars{x} \over x}\,\dd x
=\pars{-1}^{k + 1}k!\,{\rm Li}_{k + 1}\pars{1 \over a}
\end{align}

such that
$$\color{#00f}{\int_{0}^{1}{\ln^{4}\pars{x} \over x – x_{n}}\,\dd x} =-24\,{\rm Li}_{5}\pars{-\,{1 \over x_{n}}} =-24\,{\rm Li}_{5}\pars{-x_{-n}}$$

With expression $\pars{1}$:
\begin{align}
&\color{#c00000}{\int_{0}^{1}{1 – x \over 1 – x^{6}}\,\ln^{4}\pars{x}\,\dd x}
=-4\sum_{n = -2}^{2}x_{n}\pars{x_{n} + 1}{\rm Li}_{5}\pars{-x_{-n}}
\\[3mm]&=-8\,{\rm Li}_{5}\pars{-1}
-8\,\Re\sum_{n = 1}^{2}x_{n}\pars{x_{n} + 1}{\rm Li}_{5}\pars{-x_{-n}}
\\[3mm]&=-8\,{\rm Li}_{5}\pars{-1}
-8\,\Re\sum_{n = 1}^{2}\expo{n\pi\ic/3}\pars{\expo{n\pi\ic/3} + 1}
{\rm Li}_{5}\pars{\expo{\bracks{3 – n}\pi\ic/3}}
\\[3mm]&=-8\,{\rm Li}_{5}\pars{-1}
-8\,\Re\sum_{n = 1}^{2}\expo{n\pi\ic/2}\pars{\expo{n\pi\ic/6} + \expo{-n\pi\ic/6}}
{\rm Li}_{5}\pars{\expo{\bracks{3 – n}\pi\ic/3}}
\\[3mm]&=-8\,{\rm Li}_{5}\pars{-1}
-16\,\Re\sum_{n = 1}^{2}\expo{n\pi\ic/2}\cos\pars{n\pi \over 6}
{\rm Li}_{5}\pars{\expo{\bracks{3 – n}\pi\ic/3}}
\\[3mm]&=-8\ \underbrace{{\rm Li}_{5}\pars{-1}}
_{\ds{\color{#c00000}{-\,{15 \over 16}\,\zeta\pars{5}}}}\ +\
8\root{3}\ \underbrace{\Im{\rm Li}_{5}\pars{\expo{2\pi\ic/3}}}
_{\ds{\color{#c00000}{2\pi^{5} \over 729}}}\ +\
8\ \underbrace{\Re{\rm Li}_{5}\pars{\expo{\pi\ic/3}}}
_{\ds{\color{#c00000}{{25 \over 54}\,\zeta\pars{5}}}}
\end{align}

So,
$$\color{#66f}{\large% \int_{0}^{1}{1 – x \over 1 – x^{6}}\,\ln^{4}\pars{x}\,\dd x = {16 \over 243\root{3}}\,\pi^{5} + {605 \over 54}\,\zeta\pars{5}}$$

Note that $\frac{1-x}{1-x^6}=\sum_{k=0}^\infty (x^{6k}-x^{6k+1})$. And the integration $\int_0^1 x^n \ln{x}^4=\partial_n^4 \int_0^1 x^n dx=\frac{24}{(n+1)^5}$. We have
$$LHS = 24\sum_{k=0}^\infty \left(\frac{1}{(6k+1)^5}-\frac{1}{(6k+2)^5}\right)$$
Use the discrete Fourier, and denote $\xi=\exp(i\frac{\pi}{3})$, $\xi_i =\xi^i$. Then
$$LHS=24\sum_{i=1}^6a_i\sum_{k=1}^\infty \frac{\xi_i^k}{k^5}=24\sum_{i=1}^6 a_i Li_5(\xi_i).$$
where $a_1=a_5=\frac{1}{6}$,$a_2=-\frac{i}{2\sqrt{3}}=-a_4$,$a_6=0$,$a_3=-\frac{1}{3}$.Thus we use the summation formula for polylogarithm
$$Li_5(\xi_2)+(-1)^5 Li_5(\xi_4)=-\frac{(2\pi i)^5}{5!}B_5(\frac{1}{3})=\frac{4i\pi^5}{729}.$$
$B$ is the Bernoulli polynomial. Also, $$Li_5(\xi_1)+Li_5(\xi_5)=(1-\frac{2}{2^5}-\frac{3}{3^5}+\frac{6}{6^5})\sum_{k=1}^\infty\frac{1}{k^5}=\frac{25}{27}\sum_{k=1}^\infty\frac{1}{k^5}.$$
Also note that $L_5(-1)=-(1-\frac{2}{2^5})\zeta(5).$ We conclude that
$$24\sum_{i=1}^6 a_i Li_5(\xi_i)=24\frac{-i}{2\sqrt{3}}\cdot\frac{4i\pi^5}{729}+24(\frac{1}{6}\frac{25}{27}+\frac{1}{3}\cdot\frac{15}{16})\zeta(5)=\frac{16\pi^5}{243\sqrt{3}}+\frac{605}{54}\zeta(5)$$

The series
$$24\sum_{k=0}^\infty \left(\frac{1}{(6k+1)^5}-\frac{1}{(6k+2)^5}\right)$$
$$=\frac{24}{6^5}\sum_{k=0}^\infty \left(\frac{1}{(k+1/6)^5}-\frac{1}{(k+2/6)^5}\right)$$
Can be evaluated by the polygamma function $$=\frac{24}{6^5} \left(\frac{-\psi^4(1/6)}{24} – \frac{-\psi^4(1/3)}{24} \right)$$
$$=\frac{1}{6^5} \left({\psi^4(1/3)} – {\psi^4(1/6)} \right)$$
$$= {\frac{16{{\pi}^{5}}}{243\sqrt[]{{3}}}+\frac{605\zeta(5)}{54}}$$