Compute the integral in a closed form : $\int_0^{1}\operatorname{Li}_2(1-x)dx$

Question

How I can find the closed form of the following integration :

$I=\int_0^{1}\operatorname{Li}_2(1-x)dx$

$J=\int_0^{1}\operatorname{Li}_2(1-x)\operatorname{Li}_2(1-\frac{1}{x})dx$

$K=\int_0^{1}\ln(x)\operatorname{Li}_2(1-\frac{1}{x})dx$

Wolfram alpha give me for $I=$ see here

For $J=$ see here

For $K=$ see here

My attempt first integral is :

$\operatorname{Li}_2(1-x)=\sum_{n=1}^{\infty}\frac{(1-x)^n}{n^2}$

So :

$I=\sum_{n=1}^{\infty}\frac{\int_0^{1}(1-x)^{n}dx}{n^2}$

$=\sum_{n=1}^{\infty}\frac{\beta(1,n+1)}{n^2}$

$=\sum_{n=1}^{\infty}\frac{1}{(n+1)n^2}=\zeta (2)-1$

But what about from second and last integral ?

We can use same method ?

Answer 1

For $I$, the antiderivative is "simple".

$$I=\int\text{Li}_2(1-x) \,dx$$ Let $$u=\text{Li}_2(1-x)\implies du= \frac{\log (x)}{1-x}\,dx$$ $$I=x\text{Li}_2(1-x) -\int \frac{x\log (x)}{1-x}\,dx=x\text{Li}_2(1-x)-\Big(\text{Li}_2(1-x)-x (\log (x)-1) \Big)$$ $$I=(x-1) \text{Li}_2(1-x)+x (\log (x)-1)$$

For $J$ and $K$, I really do not know if closed for would exist. Computing with high accuracy, none of the numbers are recognized by inverse symbolic calculators.

If you do not want numerical integration, you could use series expansions to $O(x^{n+1})$ $$\text{Li}_2(1-x)=\frac{\pi ^2}{6}+x (\log (x)-1)+x^2 \left(\frac{\log (x)}{2}-\frac{1}{4}\right)+x^3 \left(\frac{\log (x)}{3}-\frac{1}{9}\right)+x^4 \left(\frac{\log (x)}{4}-\frac{1}{16}\right)+O\left(x^5\right)$$ $$\text{Li}_2\left(1-\frac{1}{x}\right)=\left(-\frac{1}{2} \log ^2(x)-\frac{\pi ^2}{6}\right)+x (1-\log (x))+x^2 \left(\frac{1}{4}-\frac{\log (x)}{2}\right)+x^3 \left(\frac{1}{9}-\frac{\log (x)}{3}\right)+x^4 \left(\frac{1}{16}-\frac{\log (x)}{4}\right)+O\left(x^5\right)$$ and integrate termwise.

As a function of $n$, for $J$ the numerical results would be $$\left( \begin{array}{ccc} n & J_{(n)} & \approx \\ 1 & \frac{5}{16}+\frac{\pi ^2}{12}-\frac{\pi ^4}{36} & -1.57084 \\ 2 & -\frac{125}{432}+\frac{7 \pi ^2}{54}-\frac{\pi ^4}{36} & -1.71577 \\ 3 & -\frac{3689}{6912}+\frac{7 \pi ^2}{48}-\frac{\pi ^4}{36} & -1.8002 \\ 4 & -\frac{2804441}{4320000}+\frac{23 \pi ^2}{150}-\frac{\pi ^4}{36} & -1.84164 \\ 5 & -\frac{9219923}{12960000}+\frac{17 \pi ^2}{108}-\frac{\pi ^4}{36} & -1.86367 \\ 6 & -\frac{7762159241}{10372320000}+\frac{47 \pi ^2}{294}-\frac{\pi ^4}{36} & -1.87637 \end{array} \right)$$ which would converge very slowly.

Answer 2

The integrals $J$ and $K$ can be computed using the polylogarithm identity $$ \operatorname{Li}_2 \left(1 - \frac{1}{x}\right) = - \operatorname{Li}_2(1-x) - \frac{1}{2} \log^2(x) \, , \, x > 0 \, , \tag{1} $$ and the integrals $$ \int \limits_0^1 \frac{[-\log(x)]^n}{1-x} \, \mathrm{d} x = n! \zeta(n+1) \, , \, n \in \mathbb{N} \, . \tag{2}$$ We have \begin{align} K &\stackrel{(1)}{=} \int \limits_0^1 \left[\frac{- \log^3(x)}{2} - \log(x) \operatorname{Li}_2(1-x)\right]\, \mathrm{d} x \\ &= 3 + \left[\left(x - 1 - x \log(x)\right)\operatorname{Li}_2(1-x)\right]_{x = 0}^{x=1} + \int \limits_0^1 \frac{[1 - x + x \log(x)] \log(x)}{1-x} \, \mathrm{d} x \\ &= 3 + \zeta(2) - 1 - 2 + \int \limits_0^1 \frac{\log^2(x)}{1-x} \, \mathrm{d} x \stackrel{(2)}{=} \zeta(2) + 2 \zeta(3) = \frac{\pi^2}{6} + 2\zeta(3) \, . \end{align} Similarly, we can write $$ - J \stackrel{(1)}{=} \int \limits_0^1 \operatorname{Li}_2^2(1-x) \, \mathrm{d} x + \frac{1}{2} \int \limits_0^1 \log^2(x) \operatorname{Li}_2(1-x) \, \mathrm{d} x \equiv M + \frac{1}{2} N \, .$$ The remaining integrals can be evaluated using integration by parts: \begin{align} M &= \zeta^2(2) - 2 \int \limits_0^1 - \log(x) \operatorname{Li}_2(1-x) \, \mathrm{d} x \\ &= \zeta^2(2) - 2 \zeta(2) + 2 \int \limits_0^1 \frac{\log(x)[x - 1 - x \log(x)]}{1-x} \, \mathrm{d} x \\ &= -\zeta(2)(2-\zeta(2)) + 2 + 4 - 2 \int \limits_0^1 \frac{\log^2(x)}{1-x} \, \mathrm{d} x \stackrel{(2)}{=} 6 - \zeta(2)(2-\zeta(2)) - 4 \zeta(3) \end{align} and \begin{align} N &= \int \limits_0^1 \frac{- x \log(x) [\log^2(x) - 2 \log(x) + 2]}{1-x} \, \mathrm{d} x \stackrel{(2)}{=} 6 \zeta(4) + 4 \zeta(3) + 2 \zeta(2) - 12 \, . \end{align} Therefore, $$ - J = 3 \zeta(4) - 2 \zeta(3) + \zeta(2)(\zeta(2)-1) = \frac{11 \pi^4}{180} - \frac{\pi^2}{6} - 2 \zeta(3) \, . $$

Answer 3

\begin{align} I&=\int_0^1\operatorname{Li}_2(1-x)\ dx=\int_0^1\operatorname{Li}_2(x)\ dx=\sum_{n=1}^\infty\frac1{n^2}\int_0^1x^n\ dx=\sum_{n=1}^\infty\frac{1}{n^2(n+1)}\\ &=\sum_{n=1}^\infty\frac{1}{n^2}-\sum_{n=1}^\infty\left(\frac{1}{n}+\frac1{n+1}\right)=\zeta(2)-H_1=\zeta(2)-1 \end{align}

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\begin{align} J&=\int_0^1\operatorname{Li}_2(1-x)\operatorname{Li}_2(1-1/x)\ dx=\int_0^1\operatorname{Li}_2(1-x)\left(-\operatorname{Li}_2(1-x)-\frac12\ln^2x\right)\ dx\\ &=-\int_0^1\operatorname{Li}^2_2(1-x)\ dx-\frac12\int_0^1\operatorname{Li}_2(1-x)\ln^2x\ dx\\ &=-\int_0^1\operatorname{Li}^2_2(1-x)\ dx-\frac12\int_0^1\operatorname{Li}_2(1-x)\ln^2x\ dx\\ &=-J_1-\frac12J_2\\ J_1&=\int_0^1\operatorname{Li}^2_2(1-x)\ dx=\int_0^1\operatorname{Li}^2_2(x)\ dx=\sum_{n=1}^\infty\frac1{n^2}\int_0^1x^n\operatorname{Li}_2(x)\ dx\\ &=\sum_{n=1}^\infty\frac1{n^2}\left(\frac{\zeta(2)}{n+1}-\frac{H_{n+1}}{(n+1)^2}\right)\\ &=\zeta(2)(I)-\sum_{n=1}^\infty\frac{H_{n+1}}{n^2(n+1)^2}\\ &\boxed{J_1=\zeta(2)(\zeta(2)-1)-\sum_{n=1}^\infty\frac{H_{n}}{n^2(n+1)^2}-\sum_{n=1}^\infty\frac{1}{n^2(n+1)^3}}\\ J_2&=\int_0^1\operatorname{Li}_2(1-x)\ln^2x\ dx\\ &=\int_0^1\left(\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)\right)\ln^2x\ dx\\ &=\zeta(2)\int_0^1\ln^2x\ dx-\int_0^1\ln^3x\ln(1-x)\ dx-\int_0^1\operatorname{Li}_2(x)\ln^2x\ dx\\ &=\zeta(2)(2)+\sum_{n=1}^\infty\frac1{n}\int_0^1x^n\ln^3x\ dx-\sum_{n=1}^\infty\frac1{n^2}\int_0^1x^n\ln^2x\ dx\\ &\boxed{J_2=2\zeta(2)-6\sum_{n=1}^\infty\frac{1}{n(n+1)^4}-2\sum_{n=1}^\infty\frac{1}{n^2(n+1)^3}}\\ \end{align} Thus \begin{align} J&=-J_1-\frac12J_2=-\zeta^2(2)+\sum_{n=1}^\infty\frac{H_n}{n^2(n+1)^2}+2\sum_{n=1}^\infty\frac{1}{n^2(n+1)^3}+3\sum_{n=1}^\infty\frac{1}{n(n+1)^4}\\ &=-\frac52\zeta(4)+(3\zeta(3)-2\zeta(2))+2(\zeta(3)+3\zeta(2)-6)+3(-\zeta(4)-\zeta(3)-\zeta(2)+4)\\ &=\zeta(2)+2\zeta(3)-\frac{11}2\zeta(4) \end{align}

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\begin{align} K&=\int_0^1\ln x\operatorname{Li}_2(1-1/x)\ dx=\int_0^1\ln x\left(-\operatorname{Li}_2(1-x)-\frac12\ln^2x\right)\ dx\\ &=-\int_0^1\ln x\operatorname{Li}_2(1-x)\ dx-\frac12\int_0^1\ln^3x\ dx\\ &=-\int_0^1\ln x\left(\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)\right)\ dx-\frac12(-6)\\ &=-\zeta(2)\int_0^1\ln x\ dx+\int_0^1\ln^2x\ln(1-x)\ dx+\int_0^1\ln x\operatorname{Li}_2(x)\ dx+3\\ &=-\zeta(2)(-1)-\sum_{n=1}^\infty\frac1n\int_0^1x^n\ln^2x\ dx+\sum_{n=1}^\infty\frac1{n^2}\int_0^1x^n\ln x\ dx+3\\ &=\zeta(2)-2\sum_{n=1}^\infty\frac{1}{n(n+1)^3}-\sum_{n=1}^\infty\frac{1}{n^2(n+1)^2}+3\\ &=\zeta(2)-2(-\zeta(3)-\zeta(2)+3)-(2\zeta(2)-3)+3\\ &=2\zeta(3)+\zeta(2) \end{align}

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Note: \begin{align} S&=\sum_{n=1}^\infty\frac{H_n}{n^2(n+1)^2}\\ &=\sum_{n=1}^\infty\frac{H_n}{n^2}+\sum_{n=1}^\infty\frac{H_n}{(n+1)^2}-2\sum_{n=1}^\infty\frac{H_n}{n(n+1)}\\ &=\sum_{n=1}^\infty\frac{H_n}{n^2}+\sum_{n=1}^\infty\frac{H_{n-1}}{n^2}-2\sum_{n=1}^\infty\frac{H_n}{n(n+1)}\\ &=2\sum_{n=1}^\infty\frac{H_n}{n^2}-\sum_{n=1}^\infty\frac1{n^3}-2\sum_{n=1}^\infty\frac{H_n}{n(n+1)}\\ &=2(2\zeta(3))-\zeta(3)-2\sum_{n=1}^\infty\frac{H_n}{n(n+1)}\\ &=3\zeta(3)-2\sum_{n=1}^\infty\frac{H_n}{n(n+1)}\tag{1}\\ \end{align} We use summation by parts for the remaining sum: \begin{align} S_N&=\sum_{n=1}^N\frac{H_n}{n(n+1)}\\ &=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^N\frac{H_n}{n+1}\\ &=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^{N+1}\frac{H_{n-1}}{n}\\ &=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^{N+1}\frac{H_{n}}{n}+\sum_{n=1}^{N+1}\frac1{n^2}\\ &=\sum_{n=1}^N\frac{H_n}{n}-\left(\sum_{n=1}^{N}\frac{H_{n}}{n}+\frac{H_{N+1}}{N+1}\right)+\sum_{n=1}^{N+1}\frac1{n^2}\\ &=-\frac{H_{N+1}}{N+1}+\sum_{n=1}^{N+1}\frac1{n^2} \end{align} Now, let $N$ approach $\infty$, we get \begin{align} \sum_{n=1}^\infty\frac{H_n}{n(n+1)}&=-\left.\frac{H_{N+1}}{N+1}\right|_{N\ \mapsto\ \infty}+\sum_{n=1}^\infty\frac1{n^2}\\ &=0+\zeta(2) \end{align} Plugging this result in $(1)$, gives $$\sum_{n=1}^\infty\frac{H_n}{n^2(n+1)^2}=3\zeta(3)-2\zeta(2)$$