16
$\begingroup$

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?

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

$\endgroup$
11
  • 1
    $\begingroup$ 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). $\endgroup$
    – coffeemath
    Sep 30, 2015 at 14:31
  • 2
    $\begingroup$ 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. $\endgroup$ Sep 30, 2015 at 14:43
  • 1
    $\begingroup$ @nospoon: I agree with you. The computation of such a series looks like just a tedious exercise in summation by parts. $\endgroup$ Sep 30, 2015 at 14:48
  • 3
    $\begingroup$ @Chris'ssistheartist in which way this integral looks beautiful? For me it just looks like a total mess ;-) $\endgroup$
    – tired
    Sep 30, 2015 at 17:30
  • 3
    $\begingroup$ 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.) $\endgroup$ Oct 1, 2015 at 13:35

4 Answers 4

26
$\begingroup$

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

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

$\endgroup$
10
  • 4
    $\begingroup$ A loooonnnnnnnnnnnggggggggggggg answer. Good job though!:-) (+1) $\endgroup$ Oct 5, 2015 at 14:04
  • $\begingroup$ @Chris'ssistheartist Thanks! I'm going to take a quick break for breakfast and then have a look at the general case. :) $\endgroup$
    – David H
    Oct 5, 2015 at 14:09
  • $\begingroup$ This is just pure beauty !! Thanks $\endgroup$ Oct 7, 2015 at 8:22
  • 2
    $\begingroup$ your endurance is impressive (+1) $\endgroup$
    – tired
    Oct 17, 2015 at 11:05
  • 1
    $\begingroup$ For a long while, I missed choosing your answer (which I intended to do). Unbeatable answer! $\endgroup$ Jan 13, 2020 at 10:49
8
$\begingroup$

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

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

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

$\endgroup$
4
  • 1
    $\begingroup$ (+1) Wonderful answer. $\endgroup$ Jan 13, 2020 at 10:42
  • $\begingroup$ @ Ali Shather Good idea in your first two lines to go only "halfway" to integrals and leave the rest as a sum. $\endgroup$ Jan 13, 2020 at 15:08
  • $\begingroup$ @user 1591719 Thank you. $\endgroup$ Jan 13, 2020 at 17:19
  • $\begingroup$ @Dr. Wolfgang Hintze Thank you. $\endgroup$ Jan 13, 2020 at 17:19
6
$\begingroup$

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:

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

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

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

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

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

$\endgroup$
2
  • $\begingroup$ (+1) A very good paper. Thank you. $\endgroup$ Jan 13, 2020 at 10:58
  • $\begingroup$ Due to an intentional sign error on my side I also found to your result for 3 dimensions. $\endgroup$ Jan 13, 2020 at 15:01
4
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .