# More on the integral $\int_0^1\int_0^1\int_0^1\int_0^1\frac{1}{(1+x) (1+y) (1+z)(1+w) (1+ x y z w)} \ dx \ dy \ dz \ dw$

$$I = \int_0^1\int_0^1\int_0^1\int_0^1\frac{1}{(1+x) (1+y) (1+z)(1+w) (1+ x y z w)} \ dx \ dy \ dz \ dw$$

I. User DavidH gave a beautiful (albeit long) answer in terms of the Nielsen generalized polylogarithm,

$$S_{n,p}(z) = \frac{(-1)^{n+p-1}}{(n-1)!\,p!}\int_0^1\frac{(\ln t)^{n-1}\big(\ln(1-z\,t)\big)^p}{t}dt$$

namely,

$$I = \tfrac32 S_{2,2}(-1)+\tfrac{11}{8} S_{1,3}(1)-S_{1,3}(-1) + \tfrac32 S_{3,1}(-1) \approx 0.223076$$

with the last addend tweaked by yours truly. A session with Mathematica shows that these explicitly are,

$$S_{3,1}(-1) = -\tfrac78\zeta(4) \\ S_{1,3}(1) = \zeta(4) \\ S_{2,2}(-1) = 2S_{1,3}(-1)-\tfrac18\zeta(4)$$

and,

$$S_{1,3}(-1) = \tfrac18\ln^3(2)\,\rm{Li}_1\big(\tfrac12\big)+\tfrac12\ln^2(2)\,\rm{Li}_2\big(\tfrac12\big)+\ln(2)\,\rm{Li}_3\big(\tfrac12\big)+\rm{Li}_4\big(\tfrac12\big)-\zeta(4)$$

Since $$S_{1,3}(-1)$$ and $$S_{2,2}(-1)$$ have a linear relation, then the integral can be simplified as,

$$\color{blue}{I = 2S_{1,3}(-1)+\tfrac14\zeta(4)}$$

Note that $$\rm{Li}_n\big(\tfrac12\big)$$ for $$n=1,2,3$$ have closed-forms.

II. User nospoon gave an equal but alternative form as,

$$I=\tfrac52\ln(2)\zeta(3)-\tfrac{11}{576}\pi^4-\tfrac1{2}\ln^2(2)\zeta(2)+\tfrac1{16}\ln^4(2)+\tfrac32\rm{Li}_4\big(\tfrac12\big)-A+\tfrac12B\\ \approx 0.223076$$

where

$$A = \int_0^1\frac{\rm{Li}_3(x)}{1+x}dx$$ $$B= \int_0^1\frac{\ln(1-x^2)\,\rm{Li}_2\big(\tfrac{1-x}2\big)}{x}dx$$

III. Question

After guessing on various candidate variables, is it true that the closed-forms of $$A$$ and $$B$$ are,

$$A = -4S_{2,2}(-1)+6S_{1,3}(-1) +\ln(2)\zeta(3) = 0.339545\dots$$ $$B = -\tfrac12S_{2,2}(-1)-2S_{1,3}(-1)-\tfrac38\ln(2)\zeta(3)+\tfrac14\ln^2(2)\zeta(2) = -0.1112606\dots$$

IV. Solution $$A=\int_0^1 \frac{\text{Li}_3(x)}{x+1} \, dx=-2 \text{Li}_4\left(\frac{1}{2}\right)-\frac{3}{4} \zeta (3) \log (2)+\frac{\pi ^4}{60}-\frac{1}{12} \log ^4(2)+\frac{1}{12} \pi ^2 \log ^2(2)$$ $$B=\int_0^1 \frac{\text{Li}_2\left(\frac{1-x}{2}\right) \log \left(1-x^2\right)}{x} \, dx=-3 \text{Li}_4\left(\frac{1}{2}\right)-3 \zeta (3) \log (2)+\frac{47 \pi ^4}{1440}-\frac{1}{8} \log ^4(2)+\frac{1}{6} \pi ^2 \log ^2(2)$$ $$I=2 \text{Li}_4\left(\frac{1}{2}\right)+\frac{7}{4} \zeta (3) \log (2)-\frac{7 \pi ^4}{360}+\frac{\log ^4(2)}{12}-\frac{1}{12} \pi ^2 \log ^2(2)$$