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In this post, the OP asks about the integral,

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

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,

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

Rodrigo de Azevedo
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Tito Piezas III
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1 Answers1

7

First solution: Modulo constants, the Nielsen polylog $S_{1,3}(-1)$ is equivalent to a special case of the celebrated Nielsen-Ramanujan integral:

  • $\int_0^1 \frac{\log ^3(t+1)}{t} \, dt=-6 \text{Li}_4\left(\frac{1}{2}\right)-\frac{21}{4} \zeta (3) \log (2)+\frac{\pi ^4}{15}-\frac{1}{4} \log ^4(2)+\frac{1}{4} \pi ^2 \log ^2(2)$

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

  • $I=2 \text{Li}_4\left(\frac{1}{2}\right)+\frac{7}{4} \zeta (3) \log (2)-\frac{7 \pi ^4}{360}+\frac{\log ^4(2)}{12}-\frac{1}{12} \pi ^2 \log ^2(2)$

Also one have $S_{2,2}(-1)$ equivalent to

  • $\int_0^1 \frac{\log (x) \log ^2(x+1)}{x} \, dx=-4 \text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{2} \zeta (3) \log (2)+\frac{\pi ^4}{24}-\frac{1}{6} \log ^4(2)+\frac{1}{6} \pi ^2 \log ^2(2)$

Which leads to the

Second solution: All integrals proposed by OP are certain 4-admissible integrals (for reference, see here) which can be calculated via Multiple Zeta Values. The result is:

  • $A=\int_0^1 \frac{\text{Li}_3(x)}{x+1} \, dx=-2 \text{Li}_4\left(\frac{1}{2}\right)-\frac{3}{4} \zeta (3) \log (2)+\frac{\pi ^4}{60}-\frac{1}{12} \log ^4(2)+\frac{1}{12} \pi ^2 \log ^2(2)$

  • $B=\int_0^1 \frac{\text{Li}_2\left(\frac{1-x}{2}\right) \log \left(1-x^2\right)}{x} \, dx=-3 \text{Li}_4\left(\frac{1}{2}\right)-3 \zeta (3) \log (2)+\frac{47 \pi ^4}{1440}-\frac{1}{8} \log ^4(2)+\frac{1}{6} \pi ^2 \log ^2(2)$

So based on relations given by @nospoon it is proved again. Also here is an elementary solution of numerous integrals including $A,B$. Moreover this answers the OP's question i.e. verifying the correctness of $2$ formulas connecting $A, B, S_{p,q}$. In fact for these types of integrals a more powerful method exists, that is the

Third solution: Evidently, modulo a trivial $\log^n(2)$ term, the generalized integral $$I(n)=\int_{(0,1)^n} \frac{ \prod_1^n dx_i}{(1+\prod_1^n x_i)\prod_1^n (1+x_i)}$$ is equivalent to $$J(n)=\int_{(0,1)^n} \frac{\prod_1^n x_i \prod_1^n dx_i}{(1+\prod_1^n x_i)\prod_1^n (1+x_i)}$$ then tailed Euler sums $$\sum_{k=1}^\infty (\log(2)-\widetilde{H_k})^n (-1)^{(n+1)k}$$ then ordinary Euler sums (by Abel summation calculating partial sums of $(-1)^{(n+1)k}$ and taking difference on $(\log(2)-\widetilde{H_k})^n$, for at most $2$ times), then alternating (level $2$) MZVs via stuffle relations. Plugging in known special values of MZVs completes the evaluation of $I(4)$.

One obtain high-weight results similarly, say

  • $ I(6)=-33\zeta(\bar5,1)+60 \text{Li}_6\left(\frac{1}{2}\right)+30 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)+60 \text{Li}_5\left(\frac{1}{2}\right) \log (2)+\frac{771 \zeta (3)^2}{64}+\frac{35}{4} \zeta (3) \log ^3(2)-\frac{29 \pi ^6}{360}+\frac{5 \log ^6(2)}{6}-\frac{5}{8} \pi ^2 \log ^4(2)$

  • $ I(7)=1729\zeta(\bar5,1)+\frac{35}{3} \pi ^2 \text{Li}_4\left(\frac{1}{2}\right)-3360 \text{Li}_6\left(\frac{1}{2}\right)-420 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)-1680 \text{Li}_5\left(\frac{1}{2}\right) \log (2)-\frac{5397 \zeta (3)^2}{8}-\frac{315}{4} \zeta (3) \log ^3(2)+7 \pi ^2 \zeta (3) \log (2)-\frac{50813}{32} \zeta (5) \log (2)+\frac{1589281 \pi ^6}{362880}-\frac{14}{3} \log ^6(2)+\frac{175}{36} \pi ^2 \log ^4(2)+\frac{4739 \pi ^4 \log ^2(2)}{1440}$

Note that both $I(2k)$ and $I(2k+1)$ are generated by weight $2k$ constants, rather than of different weights (as expected). One may carry out Abel partial summation for both cases to see the reason. I cannot resist to give the following

Bonus: We have an associated magnificent series (have a try):

  • $\scriptsize \sum _{n=1}^{\infty } \left(\sum _{j=2 n+1}^{\infty } \frac{(-1)^{j-1}}{j}\right)^7=\frac{1729}{2} \zeta(\bar5,1)+\frac{209}{4} \zeta(\bar5,1,1)+\frac{253}{8} \zeta(5,\bar1,1)-\frac{253}{8} \log (2) \zeta(\bar5,1)+\frac{815}{16} \text{Li}_4\left(\frac{1}{2}\right) \zeta (3)+\frac{35}{6} \pi ^2 \text{Li}_4\left(\frac{1}{2}\right)-1680 \text{Li}_6\left(\frac{1}{2}\right)+210 \text{Li}_7\left(\frac{1}{2}\right)+35 \text{Li}_4\left(\frac{1}{2}\right) \log ^3(2)-210 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)+105 \text{Li}_5\left(\frac{1}{2}\right) \log ^2(2)-840 \text{Li}_5\left(\frac{1}{2}\right) \log (2)+210 \text{Li}_6\left(\frac{1}{2}\right) \log (2)-\frac{5397 \zeta (3)^2}{16}-\frac{53679 \zeta (7)}{256}-\frac{1243 \pi ^2 \zeta (5)}{1024}-\frac{26207 \pi ^4 \zeta (3)}{46080}+\frac{3755}{384} \zeta (3) \log ^4(2)-\frac{315}{8} \zeta (3) \log ^3(2)-\frac{815}{384} \pi ^2 \zeta (3) \log ^2(2)-\frac{7843}{256} \zeta (5) \log ^2(2)+\frac{9133}{256} \zeta (3)^2 \log (2)+\frac{7}{2} \pi ^2 \zeta (3) \log (2)-\frac{50813}{64} \zeta (5) \log (2)+\frac{1589281 \pi ^6}{725760}-\frac{1}{6} \log ^7(2)-\frac{7 \log ^6(2)}{3}-\frac{7}{12} \pi ^2 \log ^5(2)+\frac{175}{72} \pi ^2 \log ^4(2)+\frac{4739 \pi ^4 \log ^2(2)}{2880}+\frac{9361 \pi ^6 \log (2)}{161280}$
Infiniticism
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