# What's about $-\int_0^1\frac{\log(1+x^{10})\log x}{x}dx$?

Playing with Wolfram Alpha and inspired in  (I refers it if someone know how relates my problem with some of problems involving the Apéry constant in this reference, but the relation doesn't seem explicit), defining $$I_n:=-\int_0^1\frac{\log(1+x^{2n})\log x}{x}dx$$ for integers $n\geq 1$, I can calculate, as I am saying with Wolfram Alpha (but I don't know how get the indefinite integrals) $I_1$, $I_2$, $I_3$ and $I_4$. And as a conjecture $$I_8=\frac{6\zeta(3)}{8\cdot 16^2}.$$

Motivation. I would like to do a comparison with the sequence $I_1$, $I_2$, $I_3$, $I_4$ and $I_8$.

Question. If do you know that this problem was solved in the literature please add a comment: can you evaluate in a closed-form $I_5$? Many thanks.

## References:

 Walther Janous , Around's Apéry's constant, J. Ineq. Pure and Appl. Math. 7(1) Art. 35 (2006).

• For instance, if you type Walther Janous , Around's Apéry's constant in Google you find the paper from the European Mathematical Information Service.
– user243301
Mar 14, 2017 at 11:12

By Taylor series expansion ,

For $0<x<1$ and $n\geq 1$,

$\displaystyle -\ln(1+x^{2n})=\sum_{k=1}^{+\infty} \dfrac{(-1)^kx^{2kn}}{k}$

For $k \geq 0$,

$\displaystyle \int_0^1 x^k\ln x dx=-\dfrac{1}{(k+1)^2}$

(integration by parts)

Therefore,

\begin{align} I_n&=\int_0^1 \left(\sum_{k=1}^{+\infty} \dfrac{(-1)^kx^{2kn-1}\ln x}{k}\right) dx\\ &=\sum_{k=1}^{+\infty} \left(\int_0^1 \dfrac{(-1)^kx^{2kn-1}\ln x}{k} dx\right)\\ &=-\sum_{k=1}^{+\infty} \dfrac{(-1)^k}{k(2kn)^2}\\ &=-\dfrac{1}{4n^2}\sum_{k=1}^{+\infty} \dfrac{(-1)^k}{k^3}\\ &=-\dfrac{1}{4n^2}\left(\sum_{k=1}^{+\infty} \dfrac{1}{(2k)^3}-\sum_{k=0}^{+\infty}\dfrac{1}{(2k+1)^3}\right)\\ &=-\dfrac{1}{4n^2}\left(\dfrac{1}{8}\zeta(3)-\left(\zeta(3)-\dfrac{1}{8}\zeta(3)\right)\right)\tag{1}\\ &=-\dfrac{1}{4n^2}\times -\dfrac{3}{4}\zeta(3)\\ &=\boxed{\dfrac{3\zeta(3)}{16n^2}} \end{align}

For (1) observe that,

\begin{align}\sum_{k=0}^{+\infty}\dfrac{1}{(2k+1)^3}&=\sum_{k=1}^{+\infty}\dfrac{1}{k^3}-\sum_{k=1}^{+\infty}\dfrac{1}{(2k)^3}\\ &=\zeta(3)-\dfrac{1}{8}\zeta(3) \end{align}

• Many thanks to you and @PaulEnta you are generous and nice. It can encourage to me and young students to read and study your answers.
– user243301
Mar 14, 2017 at 14:04
• In my genuine calculations I tried your approach but I had a mistake a step.
– user243301
Mar 14, 2017 at 14:15

Integration by part leads to$$I_n=-I_n+2n\int_0^1\frac{x^{2n-1}\log^2 x}{1+x^{2n}}\,dx$$ Then $$I_n=n\int_0^1\frac{x^{2n-1}\log^2 x}{1+x^{2n}}\,dx$$ and Maple integrates:$$I_n=\frac{3\zeta(3)}{16n^2}$$

• Many thanks for your attention, now I am going to do the calculation to see it. I was breaking my head and spirit in my attempt to get the closed-form. Thanks!
– user243301
Mar 14, 2017 at 11:37

$\newcommand{\bbx}{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}{\mathcal{#1}} \newcommand{\mrm}{\mathrm{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$ \begin{align} &-\int_{0}^{1}{\ln\pars{1 + x^{2n}}\ln\pars{x} \over x}\,\dd x \,\,\,\,\stackrel{x^{2n}\ \mapsto\ x}{=}\,\, -\,{1 \over 4n^{2}}\int_{0}^{1}{\ln\pars{1 + x}\ln\pars{x} \over x}\,\dd x \\[5mm]= &\ -\,{1 \over 4n^{2}}\int_{0}^{-1}{\ln\pars{1 - x}\ln\pars{-x} \over x}\,\dd x = {1 \over 4n^{2}}\int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\ln\pars{-x}\,\dd x = -\,{1 \over 4n^{2}}\int_{0}^{-1}{\mrm{Li}_{2}\pars{x} \over x}\,\dd x \\[5mm] = &\ -\,{1 \over 4n^{2}}\int_{0}^{-1}\mrm{Li}_{3}'\pars{x}\,\dd x = -\,{1 \over 4n^{2}}\,\mrm{Li}_{3}\pars{-1} = -\,{1 \over 4n^{2}}\sum_{k = 1}^{\infty}{\pars{-1}^{k} \over k^{3}} \\[5mm] = &\ -\,{1 \over 4n^{2}}\pars{\sum_{k = 1\ \mrm{even}}^{\infty}{1 \over k^{3}} - \sum_{k = 1\ \mrm{odd}}^{\infty}{1 \over k^{3}}} = -\,{1 \over 4n^{2}}\pars{2\sum_{k = 1}^{\infty}{1 \over \pars{2k}^{3}} - \sum_{n = 1}^{\infty}{1 \over n^{3}}} = \bbx{\ds{{3 \over 16n^{2}}\,\zeta\pars{3}}} \end{align}

• I've understand all your steps, the change of variables to get $-\int_0^{-1}\frac{\log(1+u)\frac{1}{2n}\log u}{u^{1/(2n)}}\frac{1}{2n}\frac{u^{1/(2n)}}{u}du$, after other change of variables and you are introducing the derivative of the dilogarihtm, after the integration by parts with $\operatorname{Li}_2'(x)\log(-1)|_0^{-1}=0$ and after the direct integration of $\operatorname{Li}_3'(x)$ and final statement. It was very nice, Many thanks.
– user243301
Mar 14, 2017 at 19:08
• @user243301 Yes. You're right. Thanks. Mar 14, 2017 at 20:28