# Another beautiful integral (Part 2)

One of the ways of calculating the integral in closed form is to think of crafitly using the geometric series, but even so it seems evil enough.

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

Maybe you can guide me, bless me with another precious hints, clues. Thanks MSE users!

Supplementary question: How about the generalization?

$$\int_0^1\int_0^1\cdots\int_0^1\frac{1}{(1+x_1) (1+x_2)\cdots (1+x_n)(1+ x_1 x_2 \cdots x_n)} \ dx_1 \ dx_2 \cdots \ dx_n$$

• maple cannot go further than two integrations, at that point giving a complicated expression involving dilogarithms. It just returns the third integration attempt as a symbolic restatement of the integration (having the last variable inside it as a remaining parameter). Sep 30, 2015 at 14:31
• This is equivalent to $\displaystyle \sum_{n=0}^{\infty} (-1)^n (\log 2 -H_{n^{-}})^4$ where $\displaystyle H_{n^{-}}=\sum_{k=1}^{n} \frac{(-1)^{k+1}}{k}=\log 2+(-1)^{n+1}\int_0^1 \frac{x^n}{1+x}dx$ are skew (alternating) harmonic numbers. the squared case was discussed here. Sep 30, 2015 at 14:43
• @nospoon: I agree with you. The computation of such a series looks like just a tedious exercise in summation by parts. Sep 30, 2015 at 14:48
• @Chris'ssistheartist in which way this integral looks beautiful? For me it just looks like a total mess ;-) Sep 30, 2015 at 17:30
• Meanwhile i reduced it to: $$\small \frac52\zeta(3)\ln2 - \frac{11\pi^4}{576} -\frac{\pi^2}{12}\ln^2 2 + \frac{\ln^4 2}{16} + \frac32 \operatorname{Li_4}\left(\frac12\right)-\int_0^1 \frac{ \operatorname{Li}_3(x) }{1+x} dx + \frac12 \int_0^1\frac{\ln(1-x^2) \operatorname{Li_2} \left(\frac{1-x}{2}\right)}{x}dx$$ If there indeed a very nice and short way to do it, alot of stuff should cancel here.. (Thanks, Chris.) Oct 1, 2015 at 13:35

This response will only address the $n=4$ case,

$$I_{4}:=\int_{[0,1]^{4}}\frac{\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z\,\mathrm{d}w}{\left(1+x\right)\left(1+y\right)\left(1+z\right)\left(1+w\right)\left(1+xyzw\right)}.\tag{1}$$

According to WolframAlpha, the multiple integral $(1)$ above has the approximate numerical value $I_{4}\approx0.223076.$

Starting with the substitution $w=\frac{1-t}{1+xyzt}$, we can whittle the multiple integral down to the following double integral:

\begin{align} I_{4} &=\small{\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}z\int_{0}^{1}\frac{\mathrm{d}w}{\left(1+x\right)\left(1+y\right)\left(1+z\right)\left(1+w\right)\left(1+xyzw\right)}}\\ &=\small{\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}z\int_{0}^{1}\frac{\mathrm{d}t}{\left(1+x\right)\left(1+y\right)\left(1+z\right)\left(2-t+xyzt\right)}}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}z\,\frac{\ln{(2)}-\ln{\left(1+xyz\right)}}{\left(1+x\right)\left(1+y\right)\left(1+z\right)\left(1-xyz\right)}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\int_{0}^{xy}\mathrm{d}v\,\frac{\ln{\left(\frac{2}{1+v}\right)}}{\left(1+x\right)\left(1+y\right)\left(xy+v\right)\left(1-v\right)};~~~\small{\left[xyz=v\right]}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,\frac{\ln{\left(\frac{2}{1+v}\right)}}{\left(1+x\right)\left(x+u\right)\left(u+v\right)\left(1-v\right)};~~~\small{\left[xy=u\right]}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}v\int_{v}^{x}\mathrm{d}u\,\frac{\ln{\left(\frac{2}{1+v}\right)}}{\left(1+x\right)\left(x+u\right)\left(u+v\right)\left(1-v\right)}\\ &=\int_{0}^{1}\mathrm{d}v\int_{v}^{1}\mathrm{d}x\int_{v}^{x}\mathrm{d}u\,\frac{\ln{\left(\frac{2}{1+v}\right)}}{\left(1+x\right)\left(x+u\right)\left(u+v\right)\left(1-v\right)}\\ &=\int_{0}^{1}\mathrm{d}v\int_{v}^{1}\mathrm{d}u\int_{u}^{1}\mathrm{d}x\,\frac{\ln{\left(\frac{2}{1+v}\right)}}{\left(1+x\right)\left(x+u\right)\left(u+v\right)\left(1-v\right)}\\ &=\int_{0}^{1}\mathrm{d}v\int_{v}^{1}\mathrm{d}u\,\frac{\ln{\left(\frac{(1+u)^2}{4u}\right)}\ln{\left(\frac{2}{1+v}\right)}}{\left(1-u\right)\left(u+v\right)\left(1-v\right)}\\ &=\int_{0}^{1}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,\frac{\ln{\left(\frac{(1+u)^2}{4u}\right)}\ln{\left(\frac{2}{1+v}\right)}}{\left(1-u\right)\left(u+v\right)\left(1-v\right)}.\tag{2}\\ \end{align}

WolframAlpha's numerical approximation of the iterated integral obtained in the last line of $(2)$ is consistent with the original approximation stated above, so I am reasonably confident that I haven't made any errors so far.

Continuing, transforming variables and changing the order of integration yields the following equivalent double integral representation of $I_{4}$:

\begin{align} I_{4} &=\int_{0}^{1}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,\frac{\ln{\left(\frac{(1+u)^2}{4u}\right)}\ln{\left(\frac{2}{1+v}\right)}}{\left(1-u\right)\left(u+v\right)\left(1-v\right)}\\ &=\int_{0}^{1}\mathrm{d}u\int_{\frac{1-u}{1+u}}^{1}\mathrm{d}y\,\frac{\ln{\left(\frac{(1+u)^2}{4u}\right)}\ln{\left(1+y\right)}}{\left(1-u\right)\left(u+\frac{1-y}{1+y}\right)y\left(1+y\right)};~~~\small{\left[\frac{1-v}{1+v}=y\right]}\\ &=-\frac12\int_{0}^{1}\mathrm{d}x\int_{x}^{1}\mathrm{d}y\,\frac{\ln{\left(1-x^2\right)}\ln{\left(1+y\right)}}{xy\left(1-xy\right)};~~~\small{\left[\frac{1-u}{1+u}=x\right]}\\ &=-\frac12\int_{0}^{1}\mathrm{d}y\int_{0}^{y}\mathrm{d}x\,\frac{\ln{\left(1-x^2\right)}\ln{\left(1+y\right)}}{xy\left(1-xy\right)}.\tag{3}\\ \end{align}

Now, the dilogarithm function $\operatorname{Li}_{2}{\left(z\right)}$ for complex argument is traditionally defined via the integral representation

$$\operatorname{Li}_{2}{\left(z\right)}:=-\int_{0}^{z}\frac{\ln{\left(1-t\right)}}{t}\,\mathrm{d}t;~~~\small{z\in\mathbb{C}\setminus(1,\infty)}.\tag{4}$$

The following indefinite integral may then be confirmed by differentiated both sides of the equation:

$$\small{\int\frac{\ln{\left(c+dx\right)}}{a+bx}\,\mathrm{d}x=\frac{\operatorname{Li}_{2}{\left(\frac{b\left(c+dx\right)}{bc-ad}\right)}+\ln{\left(c+dx\right)}\ln{\left(\frac{d\left(a+bx\right)}{ad-bc}\right)}}{b}+\color{grey}{constant}.}\tag{5}$$

Next, splitting up the logarithm function of $x$ in the numerator and applying partial fraction decomposition to the rational part, we find

\begin{align} I_{4} &=-\frac12\int_{0}^{1}\mathrm{d}y\int_{0}^{y}\mathrm{d}x\,\frac{\ln{\left(1-x^2\right)}\ln{\left(1+y\right)}}{xy\left(1-xy\right)}\\ &=-\frac12\int_{0}^{1}\mathrm{d}y\int_{0}^{y}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}\ln{\left(1+y\right)}}{xy\left(1-xy\right)}\\ &~~~~~-\frac12\int_{0}^{1}\mathrm{d}y\int_{0}^{y}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}\ln{\left(1+y\right)}}{xy\left(1-xy\right)}\\ &=-\frac12\int_{0}^{1}\mathrm{d}y\,\ln{\left(1+y\right)}\int_{0}^{y}\mathrm{d}x\,\left[\frac{1}{1-xy}+\frac{1}{xy}\right]\ln{\left(1+x\right)}\\ &~~~~~-\frac12\int_{0}^{1}\mathrm{d}y\,\ln{\left(1+y\right)}\int_{0}^{y}\mathrm{d}x\,\left[\frac{1}{1-xy}+\frac{1}{xy}\right]\ln{\left(1-x\right)}\\ &=-\frac12\int_{0}^{1}\mathrm{d}y\,\ln{\left(1+y\right)}\int_{0}^{y}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}}{1-xy}\\ &~~~~~-\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\int_{0}^{y}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}}{x}\\ &~~~~~-\frac12\int_{0}^{1}\mathrm{d}y\,\ln{\left(1+y\right)}\int_{0}^{y}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{1-xy}\\ &~~~~~-\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\int_{0}^{y}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{x}\\ &=\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\left[-\int_{0}^{y}\mathrm{d}x\,\frac{y\ln{\left(1+x\right)}}{1-xy}\right]\\ &~~~~~+\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(-y\right)}}{y}\\ &~~~~~-\frac12\int_{0}^{1}\mathrm{d}y\,\ln{\left(1+y\right)}\int_{1-y}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}}{1-y\left(1-t\right)};~~~\small{\left[1-x=t\right]}\\ &~~~~~+\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(y\right)}}{y}\\ &=\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\left[\operatorname{Li}_{2}{\left(y\right)}+\ln{\left(1-y\right)}\ln{\left(1+y\right)}-\operatorname{Li}_{2}{\left(\frac{y}{1+y}\right)}\right]\\ &~~~~~-\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\int_{1-y}^{1}\mathrm{d}t\,\frac{\left(\frac{y}{1-y}\right)\ln{\left(t\right)}}{1+\left(\frac{y}{1-y}\right)t}\\ &~~~~~+\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(-y\right)}}{y}+\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(y\right)}}{y}\\ &=\small{\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\left[\operatorname{Li}_{2}{\left(y\right)}+\ln{\left(1-y\right)}\ln{\left(1+y\right)}+\operatorname{Li}_{2}{\left(-y\right)}+\frac12\ln^{2}{\left(1+y\right)}\right]}\\ &~~~~~\small{-\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\left[\operatorname{Li}_{2}{\left(\frac{y}{y-1}\right)}-\operatorname{Li}_{2}{\left(-y\right)}-\ln{\left(1-y\right)}\ln{\left(1+y\right)}\right]}\\ &~~~~~+\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(-y\right)}}{y}+\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(y\right)}}{y}\\ &=\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\left[\operatorname{Li}_{2}{\left(y\right)}+\operatorname{Li}_{2}{\left(-y\right)}+\ln{\left(1-y\right)}\ln{\left(1+y\right)}\right]\\ &~~~~~+\frac14\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1+y\right)}}{y}\\ &~~~~~\small{+\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\left[\operatorname{Li}_{2}{\left(y\right)}+\frac12\ln^{2}{\left(1-y\right)}+\operatorname{Li}_{2}{\left(-y\right)}+\ln{\left(1-y\right)}\ln{\left(1+y\right)}\right]}\\ &~~~~~+\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(-y\right)}}{y}+\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(y\right)}}{y}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}}{y}\left[\operatorname{Li}_{2}{\left(y\right)}+\operatorname{Li}_{2}{\left(-y\right)}+\ln{\left(1-y\right)}\ln{\left(1+y\right)}\right]\\ &~~~~~+\frac14\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1+y\right)}}{y}+\frac14\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{2}{\left(1-y\right)}\ln{\left(1+y\right)}}{y}\\ &~~~~~+\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(-y\right)}}{y}+\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(y\right)}}{y}\\ &=\frac32\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(y\right)}}{y}+\frac32\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(-y\right)}}{y}\\ &~~~~~+\frac14\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1+y\right)}}{y}+\frac14\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{2}{\left(1-y\right)}\ln{\left(1+y\right)}}{y}\\ &~~~~~+\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1-y\right)}\ln^{2}{\left(1+y\right)}}{y}.\tag{6}\\ \end{align}

And so we have reduced our multiple integral to a sum of five single-variable polylogarithmic integrals. Instead of attempting to evaluate each of these in turn, we'll save much energy if we make a few rearrangements first.

\begin{align} I_{4} &=\frac32\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(y\right)}}{y}+\frac32\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(-y\right)}}{y}\\ &~~~~~+\frac14\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1+y\right)}}{y}+\frac14\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{2}{\left(1-y\right)}\ln{\left(1+y\right)}}{y}\\ &~~~~~+\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1-y\right)}\ln^{2}{\left(1+y\right)}}{y}\\ &=\frac32\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(-y\right)}}{y}+\frac32\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(y\right)}}{y}\\ &~~~~~+\frac14\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1+y\right)}}{y}+\frac14\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{2}{\left(1-y\right)}\ln{\left(1+y\right)}}{y}\\ &~~~~~\small{+\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1-y^2\right)}-\ln^{3}{\left(1-y\right)}-\ln^{3}{\left(1+y\right)}-3\ln^{2}{\left(1-y\right)}\ln{\left(1+y\right)}}{3y}}\\ &=-\frac34\int_{0}^{1}\mathrm{d}y\,\frac{(-2)\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(-y\right)}}{y}+\frac32\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(y\right)}}{y}\\ &~~~~~-\frac13\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1-y\right)}}{y}-\frac{1}{12}\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1+y\right)}}{y}\\ &~~~~~+\frac13\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1-y^2\right)}}{y}-\frac34\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{2}{\left(1-y\right)}\ln{\left(1+y\right)}}{y}\\ &=-\frac34\left[\operatorname{Li}_{2}{\left(-y\right)}^{2}\right]_{0}^{1}+\frac32\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(y\right)}}{y}\\ &~~~~~\small{-\frac13\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1-y\right)}}{y}-\frac{1}{12}\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1+y\right)}}{y}+\frac13\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1-y^2\right)}}{y}}\\ &~~~~~-\frac18\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1-y^2\right)}-\ln^{3}{\left(\frac{1-y}{1+y}\right)}-2\ln^{3}{\left(1+y\right)}}{y}\\ &=-\frac34\left[\operatorname{Li}_{2}{\left(-1\right)}\right]^{2}+\frac32\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(y\right)}}{y}\\ &~~~~~-\frac13\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1-y\right)}}{y}+\frac16\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1+y\right)}}{y}\\ &~~~~~+\frac{5}{24}\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1-y^2\right)}}{y}+\frac18\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(\frac{1-y}{1+y}\right)}}{y}\\ &=-\frac34\left[\operatorname{Li}_{2}{\left(-1\right)}\right]^{2}+\frac32\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+y\right)}\operatorname{Li}_{2}{\left(y\right)}}{y}\\ &~~~~~-\frac13\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1-y\right)}}{y}+\frac16\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1+y\right)}}{y}\\ &~~~~~+\frac{5}{48}\int_{0}^{1}\mathrm{d}z\,\frac{\ln^{3}{\left(1-z\right)}}{z};~~~\small{\left[y=\sqrt{z}\right]}\\ &~~~~~-\int_{0}^{1}\mathrm{d}y\,\frac{\left[\frac12\ln{\left(\frac{1+y}{1-y}\right)}\right]^{3}}{y}\\ &=-\frac34\left[\operatorname{Li}_{2}{\left(-1\right)}\right]^{2}-\frac32\operatorname{Li}_{2}{\left(1\right)}\operatorname{Li}_{2}{\left(-1\right)}-\frac32\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1-y\right)}\operatorname{Li}_{2}{\left(-y\right)}}{y}\\ &~~~~~-\frac{11}{48}\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1-y\right)}}{y}+\frac16\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1+y\right)}}{y}-\int_{0}^{1}\mathrm{d}y\,\frac{\left[\operatorname{arctanh}{\left(y\right)}\right]^{3}}{y}.\tag{7}\\ \end{align}

The first two logarithmic integrals can immediately be written as Nielsen generalized polylogarithms. It's also not difficult to reduce the third logarithmic integral to Nielsen polylogarithms:

\begin{align} \int_{0}^{1}\mathrm{d}y\,\frac{\left[\operatorname{arctanh}{\left(y\right)}\right]^{3}}{y} &=\int_{0}^{1}\mathrm{d}y\,\frac{\left[\frac12\ln{\left(\frac{1+y}{1-y}\right)}\right]^{3}}{y}\\ &=-\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(\frac{1-y}{1+y}\right)}}{8y}\\ &=-\frac14\int_{0}^{1}\mathrm{d}x\,\frac{\ln^{3}{\left(x\right)}}{1-x^2};~~~\small{\left[\frac{1-y}{1+y}=x\right]}\\ &=-\frac18\int_{0}^{1}\mathrm{d}x\,\frac{\ln^{3}{\left(x\right)}}{1-x}-\frac18\int_{0}^{1}\mathrm{d}x\,\frac{\ln^{3}{\left(x\right)}}{1+x}\\ &=-\frac38\int_{0}^{1}\mathrm{d}x\,\frac{\ln^{2}{\left(x\right)}\ln{\left(1-x\right)}}{x}+\frac38\int_{0}^{1}\mathrm{d}x\,\frac{\ln^{2}{\left(x\right)}\ln{\left(1+x\right)}}{x}\\ &=\frac34\,S_{3,1}{\left(1\right)}-\frac34\,S_{3,1}{\left(-1\right)}.\tag{8}\\ \end{align}

This just leaves the dilogarithmic integral to evaluate.

\begin{align} \int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1-y\right)}\operatorname{Li}_{2}{\left(-y\right)}}{y} &=-\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1-y\right)}}{y}\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+yx\right)}}{x}\\ &=-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1-y\right)}\ln{\left(1+xy\right)}}{xy}\\ &=:-\int_{0}^{1}\mathrm{d}x\,\frac{J{\left(-x\right)}}{x}\\ &=-\int_{0}^{1}\mathrm{d}x\,\frac{S_{1,2}{\left(-x\right)}}{x}-\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{3}{\left(-x\right)}}{x}\\ &=-S_{2,2}{\left(-1\right)}-\operatorname{Li}_{4}{\left(-1\right)}.\tag{9}\\ \end{align}

(See Appendix 2 for definition and evaluation of the auxiliary function $J{(a)}$ used above.)

Putting everything together, we arrive at

\begin{align} I_{4} &=-\frac34\left[\operatorname{Li}_{2}{\left(-1\right)}\right]^{2}-\frac32\operatorname{Li}_{2}{\left(1\right)}\operatorname{Li}_{2}{\left(-1\right)}\\ &~~~~~-\frac32\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1-y\right)}\operatorname{Li}_{2}{\left(-y\right)}}{y}\\ &~~~~~-\frac{11}{48}\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1-y\right)}}{y}+\frac16\int_{0}^{1}\mathrm{d}y\,\frac{\ln^{3}{\left(1+y\right)}}{y}\\ &~~~~~-\int_{0}^{1}\mathrm{d}y\,\frac{\left[\operatorname{arctanh}{\left(y\right)}\right]^{3}}{y}\\ &=-\frac34\left[\operatorname{Li}_{2}{\left(-1\right)}\right]^{2}-\frac32\operatorname{Li}_{2}{\left(1\right)}\operatorname{Li}_{2}{\left(-1\right)}\\ &~~~~~+\frac32\,S_{2,2}{\left(-1\right)}+\frac32\operatorname{Li}_{4}{\left(-1\right)}\\ &~~~~~+\frac{11}{8}\,S_{1,3}{\left(1\right)}-S_{1,3}{\left(-1\right)}\\ &~~~~~-\frac34\,S_{3,1}{\left(1\right)}+\frac34\,S_{3,1}{\left(-1\right)}\\ &=\frac32\,S_{2,2}{\left(-1\right)}+\frac{11}{8}\,S_{1,3}{\left(1\right)}-S_{1,3}{\left(-1\right)}-\frac{7\pi^4}{480}.\\ \end{align}

Appendix 1.

The Nielsen generalized polylogarithm may be defined for positive integer indices via the integral representation

$$S_{n,p}{\left(z\right)}:=\frac{\left(-1\right)^{n+p-1}n}{n!\,p!}\int_{0}^{1}\frac{\ln^{n-1}{\left(t\right)}\ln^{p}{\left(1-zt\right)}}{t}\,\mathrm{d}t;~~~\small{n,p\in\mathbb{N}^{+}}.$$

Setting $n=1$,

$$S_{1,p}{\left(z\right)}:=\frac{\left(-1\right)^{p}}{p!}\int_{0}^{1}\frac{\ln^{p}{\left(1-zt\right)}}{t}\,\mathrm{d}t;~~~\small{p\in\mathbb{N}^{+}}.$$

Setting $p=1$,

$$S_{n,1}{\left(z\right)}=\frac{\left(-1\right)^{n}n}{n!}\int_{0}^{1}\frac{\ln^{n-1}{\left(t\right)}\ln{\left(1-zt\right)}}{t}\,\mathrm{d}t;~~~\small{n\in\mathbb{N}^{+}}.$$

Appendix 2.

Define the real function $J:(-\infty,1]\to\mathbb{R}$ via the integral representation

$$J{\left(a\right)}:=\int_{0}^{1}\frac{\ln{\left(1-y\right)}\ln{\left(1-ay\right)}}{y}\,\mathrm{d}y;~~~\small{a\le1}.$$

Then, for $a\le1$ we have

\begin{align} J{\left(a\right)} &=\int_{0}^{1}\frac{\ln{\left(1-y\right)}\ln{\left(1-ay\right)}}{y}\,\mathrm{d}y\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1-y\right)}}{y}\int_{0}^{1}\mathrm{d}x\,\frac{ay}{ayx-1}\\ &=-a\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-y\right)}}{1-ayx}\\ &=-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{a\ln{\left(1-y\right)}}{1-axy}\\ &=-\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{2}{\left(\frac{ax}{ax-1}\right)}}{x}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\frac12\ln^{2}{\left(1-ax\right)}+\operatorname{Li}_{2}{\left(ax\right)}}{x}\\ &=\frac12\int_{0}^{1}\mathrm{d}x\,\frac{\ln^{2}{\left(1-ax\right)}}{x}+\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{2}{\left(ax\right)}}{x}\\ &=S_{1,2}{\left(a\right)}+\operatorname{Li}_{3}{\left(a\right)}.\\ \end{align}

• A loooonnnnnnnnnnnggggggggggggg answer. Good job though!:-) (+1) Oct 5, 2015 at 14:04
• @Chris'ssistheartist Thanks! I'm going to take a quick break for breakfast and then have a look at the general case. :) Oct 5, 2015 at 14:09
• This is just pure beauty !! Thanks Oct 7, 2015 at 8:22
• your endurance is impressive (+1) Oct 17, 2015 at 11:05
• For a long while, I missed choosing your answer (which I intended to do). Unbeatable answer! Jan 13, 2020 at 10:49

From this paper page $$105$$ we have

$$\overline{H}_n-\ln2=(-1)^{n-1}\int_0^1\frac{x^n}{1+x}dx$$

$$\Longrightarrow (\overline{H}_n-\ln2)^4=\int_{[0,1]^4}\frac{(xyzw)^n}{(1+x)(1+y)(1+z)(1+w)}\ dx\ dy\ dz\ dw$$

now multiply both sides by $$(-1)^n$$ then $$\sum_{n=0}^\infty$$ we get

$$I=\int_{[0,1]^4}\frac{\ dx\ dy\ dz\ dw}{(1+x)(1+y)(1+z)(1+w)(1+xyzw)}=\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^4=S$$

Lets calculate $$S$$

$$S=\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^2\color{blue}{(\overline{H}_n-\ln2)^2}$$

$$=\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^2\left(\color{blue}{\int_0^1\int_0^1\frac{(xy)^n}{(1+x)(1+y)}dx\ dy}\right)$$

$$=\int_0^1\int_0^1\frac{dx\ dy}{(1+x)(1+y)}\left(\sum_{n=0}^\infty(\overline{H}_n-\ln2)^2(-xy)^n\right)$$

In the same paper, page $$97$$ Eq$$(13)$$ we have

$$\sum_{n=0}^\infty(\overline{H}_n-\ln2)^2t^n=\frac{1}{1-t}\left(\operatorname{Li}_2(t)-2\operatorname{Li}_2\left(\frac{1+t}{2}\right)+\operatorname{Li}_2\left(\frac{1}{2}\right)+\ln^22\right)$$

Therefore,

$$S=\int_0^1\int_0^1\frac{\operatorname{Li}_2(-xy)-2\operatorname{Li}_2\left(\frac{1-xy}{2}\right)+\operatorname{Li}_2\left(\frac{1}{2}\right)+\ln^22}{(1+x)(1+y)(1+xy)}\ dx\ dy,\qquad xy=u$$

$$=\int_0^1\int_0^x\frac{\operatorname{Li}_2(-u)-2\operatorname{Li}_2\left(\frac{1-u}{2}\right)+\operatorname{Li}_2\left(\frac{1}{2}\right)+\ln^22}{(1+x)(x+u)(1+u)}\ dx\ du$$

$$=\int_0^1\color{blue}{\int_u^1\frac{1}{(1+x)(x+u)}}\frac{\operatorname{Li}_2(-u)-2\operatorname{Li}_2\left(\frac{1-u}{2}\right)+\operatorname{Li}_2\left(\frac{1}{2}\right)+\ln^22}{1+u}\ dx\ du$$

$$=\int_0^1\color{blue}{\frac{\ln\left(\frac{(1+u)^2}{4u}\right)}{1-u}}\frac{\operatorname{Li}_2(-u)-2\operatorname{Li}_2\left(\frac{1-u}{2}\right)+\operatorname{Li}_2\left(\frac{1}{2}\right)+\ln^22}{1+u}\ du$$

now set $$u=\frac{1-x}{1+x}$$

$$\Longrightarrow S=-\frac12\int_0^1\frac{\ln(1-x^2)}{x}\left[\operatorname{Li}_2\left(-\frac{1-x}{1+x}\right)-2\operatorname{Li}_2\left(\frac{x}{1+x}\right)+2\operatorname{Li}_2\left(\frac{1}{2}\right)+\ln^22\right]\ dx$$

apply integration by parts

$$\Longrightarrow S=\frac14\ln^22\zeta(2)+\frac12\int_0^1\frac{\operatorname{Li}_2(x^2)}{1-x^2}\left(\frac{\ln(1+x)}{x}-\ln2\right)\ dx$$

The latter integral was nicely calculated by Cornel here

$$\int_0^1\frac{\operatorname{Li}_2(x^2)}{1-x^2}\left(\frac{\ln(1+x)}{x}-\ln2\right)\ dx$$ $$=\frac{1}{6}\ln ^42-\frac{7 }{2}\zeta (4)+\frac{7}{2}\ln2\zeta (3)-\frac{3}{2}\ln ^22\zeta (2)+4 \operatorname{Li}_4\left(\frac{1}{2}\right)$$

$$\Longrightarrow S=\frac{1}{12}\ln ^42-\frac{7 }{4}\zeta (4)+\frac{7}{4}\ln2\zeta (3)-\frac{1}{2}\ln ^22\zeta (2)+2 \operatorname{Li}_4\left(\frac{1}{2}\right)=I$$

• (+1) Wonderful answer. Jan 13, 2020 at 10:42
• @ Ali Shather Good idea in your first two lines to go only "halfway" to integrals and leave the rest as a sum. Jan 13, 2020 at 15:08
• @user 1591719 Thank you. Jan 13, 2020 at 17:19
• @Dr. Wolfgang Hintze Thank you. Jan 13, 2020 at 17:19

See here for explanations.

Let $$I(n)=\int_{(0,1)^n} \frac{ \prod_1^n dx_i}{(1+\prod_1^n x_i)\prod_1^n (1+x_i)}$$ denotes generalized integral that OP mentioned, then:

• $$\small I(1)=\frac{1}{2},\ I(2)=\frac{\pi ^2}{24},\ I(3)=\frac{3 \log ^2(2)}{2}-\frac{\pi ^2}{24}$$

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

• $$\small I(5)=-20 \text{Li}_4\left(\frac{1}{2}\right)-\frac{45}{4} \zeta (3) \log (2)+\frac{259 \pi ^4}{1440}+\frac{5 \log ^4(2)}{3}+\frac{5}{12} \pi ^2 \log ^2(2)$$

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

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

• (+1) A very good paper. Thank you. Jan 13, 2020 at 10:58
• Due to an intentional sign error on my side I also found to your result for 3 dimensions. Jan 13, 2020 at 15:01

Here is a physicist's point of view.

As suggested by OP, i will use the simplest, geometric series approach.

Let us now look at the general case.

$$I_n=\int_0^1...\int_0^1\frac{dx_1...dx_n}{(1+x_1)...(1+x_n)(1+x_1...x_n)}$$

Let's use geometric series

$$\frac{1}{1+x_1...x_n}=1+\sum_{\nu=1}^{\infty}(-1)^\nu(x_1...x_n)^\nu$$

Now let's put the last result into $$I_n$$ and use the following simple result

$$\int_0^1\frac{x^\nu}{1+x}=(-1)^\nu\left [\ln2+\sum_{k=1}^\nu\frac{(-1)^k}{k} \right ]$$ After some simple calculations(I'll skip them) we reach the end result

$$I_n=\ln^n2+\sum_{\nu=1}^\infty(-1)^{\nu(n-1)}\left [\ln2+\sum_{k=1}^\nu\frac{(-1)^k}{k} \right ]^n$$

It is obvious that $$I_n$$ converges asymptotically to $$\ln^n2$$. Already at moderate values of $$n$$, $$\ln^n2$$ gives a good approximation.

For example,in case of $$n=4$$ worked out by David H if we use computed by him value $$I_{4}\approx0.223076$$, absolute error, if we use $$\ln^42$$ instead of $$I_4$$, is about 0.008