Integral involving Dilogarithm $\int_{1/2}^{1} \frac{\mathrm{Li}_{2}\left(x\right)\ln\left(2x - 1\right)}{x}\,\mathrm{d}x$

Question

I need your help in evaluating the following integral in closed form. $$\displaystyle\int\limits_{0.5}^{1} \frac{\mathrm{Li}_{2}\left(x\right)\ln\left(2x - 1\right)}{x}\,\mathrm{d}x$$

Since the function is singular at $x = 0.5$, we are looking for Principal Value. The integral is finite and was evaluated numerically.

I expect the closed form result to contain $\,\mathrm{Li}_{3}$ and $\,\mathrm{Li}_{2}$.

Thanks

Answer 1

This provisional answer is an horribly inelegant conjecture with no proof attached

$$\int\limits_{0.5}^{1}\frac{Li_2(x) \ln(2x-1)}{x} dx =-\frac{1}{2}\sum_{n=1}^{\infty}\frac{\sum_{k=0}^{n-1} \frac{\binom{n-1}{k}}{(k+1)^2}}{ \sum_{k=0}^n (k (2 k-1)) \binom{n}{k}}\tag1$$

I have checked this solution using $m=100$ approximation

$$\int_{\frac{1}{2}}^1 \frac{ \left(\sum _{k=1}^{m} \frac{x^k}{k^2}\right)\log (2 x-1)}{x} \, dx =-\frac{1}{2}\sum_{n=1}^{m}\frac{\sum_{k=0}^{n-1} \frac{\binom{n-1}{k}}{(k+1)^2}}{ \sum_{k=0}^n (k (2 k-1)) \binom{n}{k}}$$

The numerator summation arises from the number sequence given here (binomial transform of $1/(k+1)^2$: that is 1, 5/4, 29/18, 103/48, 887/300, 1517/360, etc.) and the denominator summation arises from the number sequence given here (the binomial transform of the hexagonal numbers)

Hopefully someone can make a little more sense of this than I have.

Later Edit: In working to understand user90369's answer to this question I found these identities using Mathematica

$$Li_n(\frac{2}{m})-Li_n(\frac{1}{m})=\sum _{k=1}^{\infty } \frac{1}{m^k k^{n-1}}\sum _{v=1}^k \binom{k-1}{v-1}\frac{1}{v}$$

in the case of $m=2$ $$\zeta(n)-Li_n(\frac{1}{2})=\sum _{k=1}^{\infty } \frac{1}{2^k k^{n-1}}\sum _{v=1}^k \binom{k-1}{v-1}\frac{1}{v}$$

Added Later Still: More Trivial Identities

$$Li_2(\frac{1}{2})=\sum\limits_{k=1}^\infty \frac{1}{k^3 2^k}\sum\limits_{v=1}^1 {\binom k v} \frac{1}{v}$$

$$Li_4(\frac{1}{2})=\sum\limits_{k=1}^\infty \frac{1}{k^3 2^k}\sum\limits_{v=k}^k {\binom k v} \frac{1}{v}$$

$$\int\limits_{0.5}^{1}\frac{Li_2(x) \ln(2x-1)}{x} dx=-Li_2(\frac{1}{2})-Li_4(\frac{1}{2})-\sum\limits_{k=1}^\infty \frac{1}{k^3 2^k}\sum\limits_{v=2}^{k-1} {\binom k v} \frac{1}{v}$$

Since $$\sum\limits_{k=1}^{\infty} \frac{1}{k^3 2^k}\sum\limits_{v=1}^{k} {\binom k v} \frac{1}{v}=\sum _{v=1}^{\infty} \frac{1}{v}\sum _{k=1}^{\infty} \frac{1}{k^{3} 2^k}\binom{k}{v}$$

There is a brute force pattern matching approach that I have found, using

$$S_a=\sum\limits_{k=1}^{\infty} \frac{1}{k^3 2^k}\sum\limits_{v=a}^{a} {\binom k v} \frac{1}{v}=\sum _{v=a}^a \frac{1}{v}\sum _{k=1}^{\infty} \frac{1}{k^{3} 2^k}\binom{k}{v}$$

where $$\int\limits_{0.5}^{1}\frac{Li_2(x) \ln(2x-1)}{x} dx= -\sum_{a=1}^{\infty} S_a$$

From Mathematica $$S_1=\frac{\pi ^2}{12}-\frac{\log ^2(2)}{2}=Li_2(\frac{1}{2})$$ $$S_2=1/48 \left(-\pi^2 + 12 \log(2) + 6 \log^2(2) \right)$$ $$S_3=\frac{1}{108} \left(6+\pi ^2-18 \log (2)-6 \log ^2(2)\right) $$ $$S_4=\frac{1}{192} \left(-8-\pi ^2+22 \log (2)+6 \log ^2(2)\right)$$

and so on. From $S_2$ onwards the general term as far as I have been able to determine is

$$S_a=(-1)^{k-1}C_a+(-1)^{k-1}\frac{\pi^2}{12a^2}+(-1)^{k}\frac{H_{a-1}\log(2)}{a^2}+(-1)^{k}\frac{\log^2(2)}{2a^2}$$

where $H_a$ is the Harmonic Number. I haven't been able to determine the pattern for the rational term, $C_a$ [$0$,$\frac{6}{(12\times3^2)}$,$\frac{8}{(12\times4^2)}$,$\frac{21}{2(12\times5^2)}$,$\frac{119}{10(12\times6^2)}$,$\frac{202}{15(12\times7^2)}$,$\frac{1525}{105(12\times8^2)}$,...]. Since the last three sum up to known closed forms it would be unfortunate if the rational term summation didn't do the same.

Once again I hope someone can make a little more sense of this than I have.

Answer 2

$\displaystyle \int\limits_{0.5}^1 \frac{Li_2(x)\ln(2x-1)}{x}dx=$

$\displaystyle =\sum\limits_{k=1}^\infty \frac{1}{k^2 2^k}\sum\limits_{v=0}^{k-1} {\binom {k-1} v} \lim\limits_{h\to 0}\frac{1}{h}\left(\frac{(2x-1)^{v+h+1}}{v+h+1}-\frac{(2x-1)^{v+1}}{v+1}\right)|_{0.5}^1$

$\displaystyle =-\sum\limits_{k=1}^\infty \frac{1}{k^3 2^k}\sum\limits_{v=1}^k {\binom k v} \frac{1}{v} = -\int\limits_0^1 \frac{Li_3(\frac{x+1}{2})-Li_3(\frac{1}{2})}{x}dx $

First note:

Be $\,\displaystyle H_k(x):=x\int\limits_0^1 \frac{(xt)^k-1}{xt-1}dt=\sum\limits_{v=1}^k \frac{x^v}{v}$ . $\,$ It's $\enspace\displaystyle \sum\limits_{v=1}^k {\binom k v} \frac{1}{v}=H_k(2)-H_k(1)$ .

Second note:

We can define e.g. $\,\displaystyle Fi_n(x):=\int\limits_0^{1-x}\frac{Li_n(t+x)-Li_n(x)}{t}dt\,$ for $\,|x|\leq 1\,$ .

Then it's $\,\displaystyle \int\limits_{0.5}^1 \frac{Li_2(x)\ln(2x-1)}{x}dx=-Fi_3(\frac{1}{2})\,$ .

Answer 3

Note that

\begin{align} I_1=&\int_0^1 \frac{\ln^2(1-x)\ln(1+x)}{1+x}dx =\int_0^1 \frac{\ln^2x \ln(2-x)}{2-x}dx\\ =& \>\ln2 \int_0^1 \frac{\ln^2x }{2-x}dx +\int_0^1 \frac{\ln^2x }{2-x} \left( -\int_0^1 \frac x{2-xy}dy\right) dx\\ =&\>2\ln2 Li_3(\frac12)+\int_0^1 \int_0^1 \frac{\ln^2x }{1-y}\left(\frac1{2-xy}-\frac1{2-x}\right)dy\>dx\\ = &\>2\ln2 Li_3(\frac12)+ 2\int_0^1\frac{Li_3(\frac y2)}ydy +2 \int_0^1 \frac{Li_3(\frac y2)-Li_3(\frac12)}{1-y}\>\overset{ibp}{dy}\\ = & \> 2\ln2 Li_3(\frac12)+2Li_4(\frac12)+ 2\int_0^{\frac12}\underset{=J}{\frac{\ln(1-2y)Li_2(y)}y}dy\tag1 \end{align} A similar procedure establishes \begin{align} I_2=\int_0^1 \frac{\ln^2(1-x)\ln x}{1+x}dx = \>2Li_4(1)+ 2J + 2\int^1_{\frac12}\frac{\ln(2y-1)Li_2(y)}ydy\tag2 \end{align} Combine (1) and (2) to express the original integral as \begin{align} \int^1_{\frac12}\frac{\ln(2y-1)Li_2(y)}ydy =\frac12 (I_2-I_1)+\ln2 Li_3(\frac12)+Li_4(\frac12)-Li_4(1)\tag3 \end{align} where the integrals $I_1$ and $I_2$ are known, given by \begin{align} &I_1 =-\frac{\pi^4}{360} +2\ln2 \zeta(3)-\frac{\pi^2}6\ln^22+\frac14 \ln^42\\ &I_2 = -6Li_4(\frac12)+\frac{11\pi^4}{360}-\frac14\ln^42 \end{align}

Substitute into (3) to obtain the close-form

\begin{align}\int^1_{\frac12}\frac{\ln(2y-1)Li_2(y)}ydy =& -2{Li}_4\left(\frac{1}{2}\right)-\frac{1}{8}\ln2\zeta(3) + \frac{\pi^4}{180}- \frac{1}{12}\ln^42 \end{align}