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

Playing with Wolfram Alpha and inspired in [1] (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:

[1] 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

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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