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How to prove that

$$4\int_0^1\frac{\chi_2(x)\operatorname{Li}_2(x)}{x}\ dx+\int_0^1\frac{\log(1-x)\log^2(x)\log(1+x)}{x}\ dx=\frac{29}4\zeta(2)\zeta(3)-\frac{91}8\zeta(5)$$

Where $\chi_2(x)=\sum_{n=1}^\infty\frac{x^{2n-1}}{(2n-1)^2}$ is the Legendre Chi function and $ \operatorname{Li}_2(x)=\sum_{n=1}^\infty\frac{x^n}{n^2}$ is the Dilogarithm function.

This integral was proposed by Cornel.

Ali Shadhar
  • 25,498

2 Answers2

7

Using the relation between the Chi function and Dilogarithm we can rewrite the first integral as: $$4\int_0^1\frac{\chi_2(x)\operatorname{Li}_2(x)}{x}dx=2\int_0^1\frac{\operatorname{Li}^2_2(x)}{x} dx-2\int_0^1\frac{\operatorname{Li}_2(x)\operatorname{Li}_2(-x)}{x} dx$$ You solved the first part here. $$\int_0^1\frac{\operatorname{Li}_2^2(x)}{x}dx=2\zeta(2)\zeta(3)-3\zeta(5)$$ And the second one is found here: $$\int_0^1\frac{\operatorname{Li}_2(x){\operatorname{Li}_2(-x)}}{x}dx =-\frac54\zeta(2)\zeta(3)+\frac{59}{32}\zeta(5)$$ Combinging the two results from above yields: $$\boxed{4\int_0^1\frac{\chi_2(x)\operatorname{Li}_2(x)}{x}dx=\frac{13}{2}\zeta(2)\zeta(3)-\frac{155}{16}\zeta(5)}$$ The second integral is solved here. $$\boxed{\int_0^1\frac{\ln(1-x)\ln^2 x\ln(1+x)}{x}dx=\frac34 \zeta(2)\zeta(3)-\frac{27}{16}\zeta(5)}$$ Combining the two boxed results gives: $$4\int_0^1\frac{\chi_2(x)\operatorname{Li}_2(x)}{x} dx+\int_0^1\frac{\ln(1-x)\ln^2(x)\ln(1+x)}{x} dx=\frac{29}4\zeta(2)\zeta(3)-\frac{91}8\zeta(5)$$

Remark.

We know from above that: $$\int_0^1\frac{\chi_2(x)\operatorname{Li}_2(x)}{x}dx=\frac{13}{8}\zeta(2)\zeta(3)-\frac{155}{64}\zeta(5)$$ But integating by parts also gives us: $$\sum_{n=0}^\infty \frac{1}{(2n+1)^2}\int_0^1 x^{2n}\operatorname{Li}_2 (x)dx$$$$\overset{IBP}=\sum_{n=0}^\infty \frac{\operatorname{Li}_2(1)}{(2n+1)^3}+\sum_{n=0}^\infty \frac{1}{(2n+1)^3}\int_0^1 x^{2n}\ln(1-x)dx$$ $$=\frac{7}{8}\zeta(2)\zeta(3) +\sum_{n=0}^\infty \frac{H_{2n+1}}{(2n+1)^4}$$ Which results in: $$\sum_{n=0}^\infty \frac{H_{2n+1}}{(2n+1)^4}=\frac34\zeta(2)\zeta(3)-\frac{155}{64}\zeta(5)$$ Alteratively one can compute that sum in a different way to find the value of the first integral.

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Zacky
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3

This approach is pretty identical to Cornel's solution posted on his FB page.

using the fact that $\quad\displaystyle \sum_{n=1}^\infty a_{2n}=\frac12\left(\sum_{n=1}^\infty a_n+\sum_{n=1}^\infty (-1)^na_n\right),\ $ we have \begin{align} \sum_{n=1}^\infty\frac{x^{2n-1}}{(2n-1)^2}&=\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)^2}=\frac12\left(\sum_{n=0}^\infty\frac{x^{n+1}}{(n+1)^2}+\sum_{n=0}^\infty(-1)^n\frac{x^{n+1}}{(n+1)^2}\right)\\ &=\frac12\left(\sum_{n=1}^\infty\frac{x^n}{n^2}-\sum_{n=1}^\infty(-1)^n\frac{x^n}{n^2}\right)=\frac12\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)\right) \end{align}

then, the first integral: \begin{align} I_1&=4\int_0^1\left(\sum_{n=1}^\infty\frac{x^{2n-1}}{(2n-1)^2}\right)\frac{\operatorname{Li}_2(x)}{x}\ dx\\ &=2\sum_{n=1}^\infty\left(\frac1{n^2}-\frac{(-1)^n}{n^2}\right)\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx\\ &=2\sum_{n=1}^\infty\left(\frac1{n^2}-\frac{(-1)^n}{n^2}\right)\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)\\ &=\zeta(2)\zeta(3)-2\zeta(2)\operatorname{Li}_3(-1)-2\sum_{n=1}^\infty\frac{H_n}{n^4}+2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}\\ &\boxed{=\frac72\zeta(2)\zeta(3)-2\sum_{n=1}^\infty\frac{H_n}{n^4}+2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}} \end{align} and the second integral:

using the following identity proved by Cornel and can be found in his book, (Almost) Impossible Integrals, Sums and Series. $\quad\displaystyle\ln(1-x)\ln(1+x)=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)x^{2n}$.

multiply both sides by $\displaystyle\frac{\ln^2x}{x}$ then integrate from $0$ to $1$, we get \begin{align} I_2&=\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\int_0^1x^{2n-1}\ln^2x\ dx\\ &=\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\left(\frac{2}{(2n)^3}\right)\\ &=-4\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}+\frac14\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac18\zeta(5)\\ &=-2\sum_{n=1}^\infty\frac{H_n}{n^4}-2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}+\frac14\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac18\zeta(5)\\ &\boxed{=-2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}-\frac74\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac18\zeta(5)} \end{align} Finally \begin{align} I&=I_1+I_2\\ &=\frac72\zeta(2)\zeta(3)-\frac18\zeta(5)-\frac{15}4\sum_{n=1}^\infty\frac{H_n}{n^4}\\ &=\frac72\zeta(2)\zeta(3)-\frac18\zeta(5)-\frac{15}4\left(3\zeta(5)-\zeta(2)\zeta(3)\right)\\ &\boxed{=\frac{29}{4}\zeta(2)\zeta(3)-\frac{91}{8}\zeta(5)} \end{align}

Ali Shadhar
  • 25,498