Integral $I=\int_0^1 \frac{\log x \log (1+x) \log(1-x) \log(1+x^2)\log(1-x^2)}{x^{3/2}}dx$

Question

Hi I am trying to integrate and obtain a closed form result for $$ I:=\int_0^1 \frac{\log x \log (1+x) \log(1-x) \log(1+x^2)\log(1-x^2)}{x^{3/2}}dx. $$ Here is what I tried (but I do not think this is a smart way because of all the sums): writing $$ I=\int_0^1 \frac{dx\log x}{x^{3/2}} \sum_{n=1}^\infty\frac{ (-1)^{n}x^n}{n}\sum_{m=1}^\infty \frac{x^m}{m}\sum_{l=1}^\infty \frac{(-1)^lx^{2l}}{l}\sum_{p=1}^\infty \frac{(-1)^p(-x^2)^{p}}{p}. $$ Now we can write $$ \sum_{n=1}^\infty\frac{ (-1)^{n}}{n}\sum_{m=1}^\infty \frac{1}{m}\sum_{l=1}^\infty \frac{(-1)^l}{l}\sum_{p=1}^\infty \frac{1}{p} \int_0^1 x^{n+m+2l+2p-3/2}\log x \, dx. $$ We can simplify this as $$ \sum_{n=1}^\infty \frac{ (-1)^{n}}{n}v\sum_{m=1}^\infty \frac{1}{m}\sum_{l=1}^\infty \frac{(-1)^l}{l}\sum_{p=1}^\infty \frac{1}{p}\left[ \frac{-1}{\big( n+m+2l+2p -\frac{1}{2} \big)^2}\right]. $$ I am stuck as to what to do From here, note above I used $$ -\sum_{n=1}^\infty \frac{x^n}{n}=\log(1-x),\quad \log(1+x)=-\sum_{n=1}^\infty \frac{(-1)^{n}x^n}{n},\quad \int_0^1 x^n \log x \, dx =\frac{-1}{(n+1)^2}. $$

Answer 1

Using $$\log (1+x) \log(1-x)= -\sum_{n=1}^\infty \frac{a_n}{b_n} \, x^{2n}$$ where the $a_n$ and $b_n$ correspond to sequences $A049281$ and $A069685$ in $OEIS$ So $$\log (1+x) \log(1-x)\log (1+x^2) \log(1-x^2)=\Bigg(\sum_{n=1}^\infty \frac{a_n}{b_n} \, x^{2n} \Bigg)\Bigg(\sum_{n=1}^\infty \frac{a_n}{b_n} \, x^{4n} \Bigg)$$ Make the Cauchy product to face $$\log (1+x) \log(1-x)\log (1+x^2) \log(1-x^2)=x^6\,\sum_{n=0}^\infty {c_n} \, x^{2n}$$ The $c_n$ are all positive and decreasing So $$I=\sum_{n=0}^\infty c_n\, \int_0^1 x^{2 n+\frac{9}{2}}\, \log (x)\,dx=-4\sum_{n=0}^\infty\frac{c_n}{(4 n+11)^2}$$ Converting to decimale, this converges quite slowly to $-0.061442$.

Inverse symbolic calculators do not find anything.