# A somewhat intriguing dilogarithmic integral

The following problem has been recently proposed by C.I. Valean, $$\int_0^1 \frac{\displaystyle \left(\operatorname{Li}_2\left(\frac{1-x}{2}\right)\right)^2}{1+x} \textrm{d}x$$ $$=\frac{3}{16}\zeta(5)-\frac{1}{4}\zeta(2)\zeta(3)+\frac{5}{8}\log(2)\zeta(4)-\frac{1}{6}\log^3(2)\zeta(2)+\frac{1}{20}\log^5(2).$$

One of the author's solutions makes use of the following (very useful) Cauchy product, $$\frac{(\operatorname{Li_2}(x))^2}{1-x}=\sum_{n=1}^{\infty}x^n\left((H_n^{(2)})^2-5 H_n^{(4)}+4\sum_{k=1}^n\frac{H_k}{k^3}\right),$$ which appears in More (Almost) Impossible Integrals, Sums, and Series: A New Collection of Fiendish Problems and Surprising Solutions (2023), in Sect. 4.5, page 398 (see the top of the page), which is the sequel of (Almost) Impossible Integrals, Sums, and Series (2019).

We observe the main integral, after the variable change $$x\mapsto 1-x$$, may be put in the form
$$\frac{1}{2}\int_0^1 \frac{\displaystyle \left(\operatorname{Li}_2\left(\frac{x}{2}\right)\right)^2}{1-x/2} \textrm{d}x,$$ and from here we get manageable results, all easily derived with results from the mentioned books.

For example, at this point we might find helpful to know and use generating functions like $$\sum_{n=1}^{\infty} x^n (H_n^{(2)})^2$$ $$=\frac{1}{1-x}\biggr(-4\zeta(4)-4\zeta(3)\log(1-x)-2\zeta(2)\log^2(1-x)-\frac{1}{6}\log^4(1-x)$$ $$+\frac{2}{3}\log(x)\log^3(1-x)+(\operatorname{Li_2}(x))^2+4\log(1-x)\operatorname{Li_3}(x)-3\operatorname{Li_4}(x)$$ $$+4\operatorname{Li_4}(1-x)-4\operatorname{Li_4}\left(\frac{x}{x-1}\right)\biggr),$$

which appears in the first-mentioned book (and visiting further its Sect. 3.51 is definitely illuminating for the next steps to do).

Another solution idea: we could start with letting $$x\mapsto 2x-1$$ and then exploiting the Dilogarithm reflection formula, in an effort to reduce everything to manageable (known) results.

The interesting question: Now, as previously seen, the path described above involves the use of harmonic series. May we design ways that avoid the use of harmonic series and then elegantly get the desired closed form?

• Nice problem with nice closed form +1 Jul 24, 2023 at 23:58
• @AliShadhar Thank you. Jul 25, 2023 at 7:55
• Substituting $\dfrac{1-x}2\mapsto x$ and integrating by parts once/twice leads to simpler-looking integrals, $$\newcommand{dilog}[1]{\operatorname{Li}_2\left(#1\right)} \\ \newcommand{trilog}[1]{\operatorname{Li}_3\left(#1\right)} \\ \ln2 \dilog{\frac12}^2 - 2 \int_0^{\tfrac12} \frac{\ln^2(1-x) \dilog x}{x} \, dx,$$ $$\ln 2 \dilog{\frac12}^2 - 2 \ln^22 \trilog{\frac12} - 4 \int_0^{\tfrac12} \frac{\ln(1-x) \trilog x}{1-x} \, dx$$though I wouldn't know how to proceed with either... Jul 26, 2023 at 19:43
• @user170231 Thanks for sharing. Jul 27, 2023 at 10:31

This is NOT a complete answer, but suggests some remarks.

We wish to find $$\int_0^\frac{1}{2}\frac{\text{Li}^2(x)}{1-x}dx$$. A few steps of applying IBP yields

$$\int \frac{\text{Li}_2^2(x)}{1-x}dx {= -\text{Li}_2^2(x)\ln(1-x)-\int \frac{2\text{Li}_2(x)\ln^2(1-x)}{x}dx \\= -\text{Li}_2^2(x)\ln(1-x)-2\text{Li}_2(x)\ln^2(1-x)\ln x \\-4\int \frac{\text{Li}_2(x)\ln(1-x)\ln x}{1-x}dx-2\int \frac{\ln^3(1-x)\ln x}{x}dx \\= -\text{Li}_2^2(x)\ln(1-x)-2\text{Li}_2(x)\ln^2(1-x)\ln x-2\int \frac{\ln^3(1-x)\ln x}{x}dx \\-4\int \frac{[\zeta(2)-\ln x\ln(1-x)-\text{Li}_2(1-x)]\ln(1-x)\ln x}{1-x}dx \\= -\text{Li}_2^2(x)\ln(1-x)-2\text{Li}_2(x)\ln^2(1-x)\ln x-4\zeta(2)\text{Li}_2(1-x)\ln(1-x) \\+4\zeta(2)\text{Li}_3(1-x) -2\int \frac{\ln^3(1-x)\ln x}{x}dx \\+4\int \frac{\ln^2 x\ln^2(1-x)+\text{Li}_2(1-x)\ln(1-x)\ln x}{1-x}dx \\= -\text{Li}_2^2(x)\ln(1-x)-2\text{Li}_2(x)\ln^2(1-x)\ln x-4\zeta(2)\text{Li}_2(1-x)\ln(1-x) \\+4\zeta(2)\text{Li}_3(1-x) -\ln^3(1-x)\ln^2x \\+\int \frac{\ln^2 x\ln^2(1-x)+4\text{Li}_2(1-x)\ln(1-x)\ln x}{1-x}dx. }$$ Therefore, $$\int_0^\frac{1}{2} \frac{\text{Li}_2^2(x)}{1-x}dx {= \text{Li}_2^2\left(\frac{1}{2}\right)\ln2+2\text{Li}_2\left(\frac{1}{2}\right)\ln^32+4\zeta(2)\text{Li}_2\left(\frac{1}{2}\right)\ln2 \\+4\zeta(2)\text{Li}_3\left(\frac{1}{2}\right)+\ln^52-4\zeta(2)\zeta(3) \\+\int_0^\frac{1}{2} \frac{\ln^2 x\ln^2(1-x)+4\text{Li}_2(1-x)\ln(1-x)\ln x}{1-x}dx, }$$ where we need to find $$\int_0^\frac{1}{2} \frac{\ln^2 x\ln^2(1-x)+4\text{Li}_2(1-x)\ln(1-x)\ln x}{1-x}dx$$. At this step, we exploit the series representation of the integrand. Therefore, $${ \int_0^\frac{1}{2} \frac{\ln^2 x\ln^2(1-x)+4\text{Li}_2(1-x)\ln(1-x)\ln x}{1-x}dx \\= \int_\frac{1}{2}^1 \frac{\ln^2 x\ln^2(1-x)+4\text{Li}_2(x)\ln(1-x)\ln x}{x}dx \\= \sum_{m,n} \frac{x^{m+n}}{mn} \left[\frac{\ln^2 x}{m+n}-\frac{2\ln x}{(m+n)^2}+\frac{2}{(m+n)^3}\right]_\frac{1}{2}^1 - \frac{4x^{m+n}}{m^2n} \left[\frac{\ln x}{m+n}-\frac{1}{(m+n)^2}\right]_\frac{1}{2}^1 \\= \sum_{m,n} \frac{2}{mn(m+n)^3}+\frac{4}{m^2n(m+n)^2} \\- \sum_{m,n} \frac{\left(\frac{1}{2}\right)^{m+n}}{mn} \left[\frac{\ln^2 2}{m+n}+\frac{2\ln 2}{(m+n)^2}+\frac{2}{(m+n)^3}\right] + \frac{\left(\frac{1}{2}\right)^{m+n}}{m^2n} \left[\frac{4\ln 2}{m+n}+\frac{4}{(m+n)^2}\right] \\= \underbrace{\sum_{m,n} \frac{2}{mn(m+n)^3}+\frac{4}{m^2n(m+n)^2}}_{A} - \ln^22\underbrace{\sum_{m,n} \frac{\left(\frac{1}{2}\right)^{m+n}}{mn(m+n)}}_{B} \\- 2\ln2\underbrace{\sum_{m,n} \frac{\left(\frac{1}{2}\right)^{m+n}}{mn(m+n)^2}}_{C} - 2\underbrace{\sum_{m,n} \frac{\left(\frac{1}{2}\right)^{m+n}}{mn(m+n)^3}}_{D} - 4\ln2\underbrace{\sum_{m,n}\frac{\left(\frac{1}{2}\right)^{m+n}}{m^2n(m+n)}}_{E} - 4\underbrace{\sum_{m,n}\frac{\left(\frac{1}{2}\right)^{m+n}}{m^2n(m+n)^2}}_{F}, }$$ where the following information about $$A,B,C,E$$ are given: $$A{= \sum_{m,n} \frac{4}{m(m+n)^4}+\frac{2m^2n+2mn^2}{m^3n^3(m+n)^2} \\= \sum_{m,n} \frac{4}{m(m+n)^4}+\frac{2}{m^2n^2(m+n)} \\= \sum_{m,n} \frac{4}{m(m+n)^4}+\frac{2}{m^3n^2}-\frac{2}{m^4n}+\frac{2}{m^4(m+n)} \\=2\zeta(2)\zeta(3)+\sum_{m,n} \frac{4}{m(m+n)^4}-\frac{2}{m^4n}+\frac{2}{m^4(m+n)} \\=2\zeta(2)\zeta(3)+\sum_{m,n} \frac{4}{m(m+n)^4}-\lim_{x\to 1^-}\left[\sum_{m,n}\frac{2x^n}{m^4n}-\frac{2x^{m+n}}{m^4(m+n)}\right] \\=2\zeta(2)\zeta(3)+\sum_{m,n} \frac{4}{m(m+n)^4}-\lim_{x\to 1^-}\Big[-2\zeta(4)\ln(1-x) \\+2\zeta(4)\ln(1-x)+2\text{Li}_5(x)+\sum_{m,n} \frac{2x^m}{m(m+n)^4}\Big] \\=2\zeta(2)\zeta(3)-2\zeta(5)+\sum_{m,n} \frac{2}{m(m+n)^4}, }$$ $$B=\int_0^\frac{1}{2}\frac{\ln^2(1-x)}{x}dx=2\zeta(3)-2\text{Li}_3\left(\frac{1}{2}\right)-2\text{Li}_2\left(\frac{1}{2}\right)\ln 2-\ln^32,$$ $$C {= \int_0^\frac{1}{2} \frac{2\zeta(3)-2\text{Li}_3(1-x)+2\text{Li}_2(1-x)\ln(1-x)+\ln x\ln^2(1-x)}{x}dx \\= -2\zeta(3)\ln 2-2\text{Li}_2^2\left(\frac{1}{2}\right)+2\text{Li}_3\left(\frac{1}{2}\right)\ln2+\zeta^2(2) \\+\int_0^1 \frac{\text{Li}_2(1-x)\ln(1-x)}{x}dx+\frac{\ln^42}{4}+\frac{1}{2}\int_0^1 \frac{\ln x\ln^2(1-x)}{x}dx \\= -2\zeta(3)\ln 2-2\text{Li}_2^2\left(\frac{1}{2}\right)+2\text{Li}_3\left(\frac{1}{2}\right)\ln2+\zeta^2(2)+\frac{\ln^42}{4} \\-\sum_{m,n}\frac{1}{m^2(m+n)^2}-\sum_{m,n}\frac{1}{2mn(m+n)^2} \\= -2\zeta(3)\ln 2-2\text{Li}_2^2\left(\frac{1}{2}\right)+2\text{Li}_3\left(\frac{1}{2}\right)\ln2+\frac{\ln^42}{4}+\zeta^2(2)-\sum_{m,n}\frac{m+2n}{2m^2n(m+n)^2} \\= -2\zeta(3)\ln 2-2\text{Li}_2^2\left(\frac{1}{2}\right)+2\text{Li}_3\left(\frac{1}{2}\right)\ln2+\frac{\ln^42}{4}+\zeta^2(2)-\sum_{m,n}\frac{1}{4m^2n^2}+\frac{1}{2m^2(m+n)^2} \\= -2\zeta(3)\ln 2-2\text{Li}_2^2\left(\frac{1}{2}\right)+2\text{Li}_3\left(\frac{1}{2}\right)\ln2+\frac{\ln^42}{4}+\frac{2\zeta^2(2)+\zeta(4)}{4} }$$ and $$E=\sum_{m,n}\frac{\left(\frac{1}{2}\right)^{m+n}}{2m^2n^2}=\frac{\text{Li}_2^2\left(\frac{1}{2}\right)}{2}.$$ Therefore, $${ A-B\ln^22-2C\ln2-4E\ln2 = 2\zeta(2)\zeta(3)-2\zeta(5)+\sum_{m,n} \frac{2}{m(m+n)^4} \\- \ln^22\left[2\zeta(3)-2\text{Li}_3\left(\frac{1}{2}\right)-2\text{Li}_2\left(\frac{1}{2}\right)\ln 2-\ln^32\right] \\-2\ln2\left[-2\zeta(3)\ln 2-2\text{Li}_2^2\left(\frac{1}{2}\right)+2\text{Li}_3\left(\frac{1}{2}\right)\ln2+\frac{\ln^42}{4}+\frac{2\zeta^2(2)+\zeta(4)}{4}\right] \\- 4\ln2\frac{\text{Li}_2^2\left(\frac{1}{2}\right)}{2} \\= 2\zeta(2)\zeta(3)-2\zeta(5)+\sum_{m,n} \frac{2}{m(m+n)^4} \\- \left[2\zeta(3)\ln^22-2\text{Li}_3\left(\frac{1}{2}\right)\ln^22-2\text{Li}_2\left(\frac{1}{2}\right)\ln^32-\ln^52\right] \\-\left[-2\zeta(3)2\ln^22-4\text{Li}_2^2\left(\frac{1}{2}\right)\ln2+4\text{Li}_3\left(\frac{1}{2}\right)\ln^22+\frac{\ln^52}{2}+\ln2\frac{2\zeta^2(2)+\zeta(4)}{2}\right] \\- 2\text{Li}_2^2\left(\frac{1}{2}\right)\ln2 }$$ and we obtain $$\int_0^\frac{1}{2} \frac{\text{Li}_2^2(x)}{1-x}dx {= 3\text{Li}_2^2\left(\frac{1}{2}\right)\ln2+4\text{Li}_2\left(\frac{1}{2}\right)\ln^32+4\zeta(2)\text{Li}_2\left(\frac{1}{2}\right)\ln2+4\zeta(2)\text{Li}_3\left(\frac{1}{2}\right) \\+\frac{3\ln^52}{2}+2\zeta(2)\zeta(3)-2\zeta(5)+2\zeta(3)\ln^22-2\text{Li}_3\left(\frac{1}{2}\right)\ln^22 \\-\ln2\frac{2\zeta^2(2)+\zeta(4)}{2}-4\zeta(2)\zeta(3)+\sum_{m,n} \frac{2}{m(m+n)^4}-2D-4F \\= -\frac{\zeta^2(2)\ln2}{4}-\frac{\ln^52}{12}+\frac{\zeta(2)\ln^32}{6}+\frac{3\zeta(2)\zeta(3)}{2}-2\zeta(5) \\+\frac{\zeta(3)\ln^22}{4}-\frac{\zeta(4)\ln2}{2}+\sum_{m,n} \frac{2}{m(m+n)^4} \\- 2\sum_{m,n} \frac{\left(\frac{1}{2}\right)^{m+n}}{mn(m+n)^3} - 4\sum_{m,n}\frac{\left(\frac{1}{2}\right)^{m+n}}{m^2n(m+n)^2}. }$$ The final answer has been numerically verified up to 5 digits and will be updated if any progress was made on the closed form evaluation of any of $$A$$, $$D$$ or $$F$$.

The following Python code confirms the result numerically:

from scipy.special import spence
from numpy import log as ln,pi,arange,meshgrid

def li2(x):
return spence(1-x)

def li3(x,n=1000):
t=arange(1,n+1)
return sum(x**t/t**3)

xi2=pi**2/6
xi3=sum(1/arange(1,2000)**3)
xi4=pi**4/90
xi5=sum(1/arange(1,120)**5)

m,n=meshgrid(arange(1,100),arange(1,100))

s1=sum(sum(1/m/(m+n)**4))
s2=sum(sum(0.5**(m+n)/m/n/(m+n)**3))
s3=sum(sum(0.5**(m+n)/m**2/n/(m+n)**2))