Is there a closed form function for the integral $\int_{0}^{z} \frac{1}{x}\log(1-x)\log(x)\log(1+x)\,dx$?

Question

Recently, in connection with the problem of calculating generating functions of the antisymmetric harmonic number (https://math.stackexchange.com/a/3526006/198592, and What's the generating function for $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}x^n\ ?$) I stumbled on the beautiful integral

$$f(z) = \int_{0}^z \frac{\log(1-x)\log(x)\log(1+x)}{x}\,dx\tag{1}$$

which seems to be hard.

I tried the common procedure of partial integrations, variable transformations and antiderivatives hunting with Mathematica which generated a multitude of different variants of the integral but finally I could not solve it.

Question Can you calculate the integral $(1)$?

Notice that we are looking here for the integral as a function of the upper limit $z$, or equvalently, for an antiderivative. The problem counts as solved once $f(z)$ is expressed through known functions, we also say that $f(z)$ has a "closed functional form".

On the other hand there are myriads of integrations problems in this forum which are similar but have fixed limits, i.e. they are definite integrals which define a constant, and the question is then if this constant is expressible by known constants - has a "closed form".

Our problem also has a compagnion in the constant species ( Evaluating $\int^1_0 \frac{\log(1+x)\log(1-x) \log(x)}{x}\, \mathrm dx$) which provided the closed form for $f(1)$.

Answer 1

EDIT 05.02.2020

Please see my full solution here https://math.stackexchange.com/a/3535943/198592

Original post 01.02.2020

This extended comment has now become an answer: I find a solution in terms of the expected polylogs and as derivatives of Hypergeometric functions.

Solution

We have to solve the integral

$$i(z) =\int_0^z \frac{\log(1- t)\log( t)\log(1+ t)}{t}\,dt \tag{1} $$

First we transfer the dependence on $z$ to the integrand thereby restricting the integration limits to $0$ and $1$.

Letting $x=t z$ the integral becomes

$$i(z)=\int_0^1 \frac{\log(1-z t)\log(z t)\log(1+z t)}{t}\,dt \\= \underbrace{\log(z) \int_0^1 \frac{\log(1-z t)\log(1+z t)}{t}\,dt}_{i_1}+\underbrace{\int_0^1 \frac{\log(1-z t)\log(t)\log(1+z t)}{t}\,dt}_{i_2}\tag{2}$$

It splits up into two integrals since $\log(x z)=\log(z)+\log(x)$.

The first integral $i_1$ can be solved immediately (using Mathematica, see appendix) so we continue with $i_2$.

Now the integrand of $i_2$ can be simplified using the well-known trick of splitting a product of two $\log$s according to

$$\log(1+w) \log(1-w) =\frac{1}{2} \log ^2\left(1-w^2\right)-\frac{1}{2} \log ^2(1-w)-\frac{1}{2} \log ^2(w+1)\tag{3a}$$

where we have used that for $|w|<1$

$$\log(1+w) + \log(1-w) = \log(1-w^2)\tag{3b}$$

With $w = z x$ the integral is split into three integrals

$$i_2 =A+B_{-}+B_{+}\tag{4a}$$

where

$$B_{\pm} = -\int \frac{\log (x) \log ^2(1\pm x z)}{2 x} \, dx\tag{4b}$$

can be solved immediately (results not provided here for the sake of brevity) but

$$A = \int \frac{\log (x) \log ^2\left(1-x^2 z^2\right)}{2 x} \, dx\tag{5}$$

escapes the antiderivative hunt with Mathematica.

Hence, to proceed, with $A$ we adopt the differentiating procedure described in section and generate the term with the complicated argument by

$$ \log ^2\left(1-x^2 z^2\right) =\frac{\partial ^2\left(1-x^2 z^2\right)^a}{\partial a^2}|_{a\to0}\tag{6} $$

The integral (actually the antiderivative because at this stage we ignore the integration limits) can now be done under the derivative leading to the "kernel"

$$\kappa(z,a) = \frac{1}{2} \int \frac{\log (x) \left(1-x^2 z^2\right)^a}{x} \, dx\\ =-\frac{1}{8 a^2} \left(1-\frac{1}{x^2 z^2}\right)^{-a} \left(1-x^2 z^2\right)^a \left(\, _3F_2\left(-a,-a,-a;1-a,1-a;\frac{1}{x^2 z^2}\right)\\ -2 a \log (x) \, _2F_1\left(-a,-a;1-a;\frac{1}{x^2 z^2}\right)\right)\tag{7}$$

Hence we get

$$A = \frac{\partial ^2\kappa(z,a)}{\partial a^2}|_{a\to0}\tag{8}$$

Here $ _3F_2$ and $_2F_1$ are Hypergeometric functions which can be represented as single infinite sums and hence are simpler than the double infinite series of the Appell-function we have obtained in the section The derivative method.

Summarizing very briefly, we have expressed the integral $i_2$ in terms of the second derivative of Hypergeometric functions with respect to a parameter.

This means that the question has been answered (inlcuding the polylog functions $\operatorname{Li}_{1,2,3,4}(z)$ which for brevity we did not write down explcitly) provided we agree that Hypergeometric functions and their derivatives are acceptable as "closed form functions".

The derivative method

@clathratus suggested in a comment that our integral "looks almost like a derivative of $_2F_1$".

Let us have a closer look. Writing $x^a = e^{a \log{x}}$ we can generate a factor $\log(x)$ by differentiating with respect to $a$ and then letting $a\to0$, or, letting $a\to (-1)$ we obtain $\frac{\log(t)}{t}$ etc.

Hence defining

$$d = t^{b-1} (1-t z)^a (1+t z)^c\tag{d.1}$$

we can generate the integrand of $i_2$ thus

$$\frac{\partial ^3d}{\partial a\, \partial b\, \partial c}|_{a\to 0,b\to 0,c\to 0}\\=\frac{\log (t) \log (1-t z) \log (t z+1)}{t}\tag{d.2}$$

Now we carry out the integration which gives

$$\int_0^1 d \, dt=\frac{1}{b} F_1(b;-a,-c;b+1;z,-z)\tag{d.3}$$

here

$$F_1\left(\alpha ,\beta ,\beta ',\gamma ,x,y\right)=\sum _{m=0}^{\infty } \sum _{n=0}^{\infty } \frac{(\alpha )_{m+n} (\beta )_m \left(\beta '\right)_n}{ (\gamma )_{m+n}}\frac{x^m y^n }{m! n!}\tag{d.4}$$

is the AppellF1-Hypergeometric function (http://mathworld.wolfram.com/AppellHypergeometricFunction.html), a two-dimensional generalization of the common hypergeometric function and

$$(\alpha )_{n}=\frac{\Gamma(\alpha+n)}{\Gamma(\alpha)}$$

is the Pochhammer symbol.

Hence our integral $i_2$ can be expressed as derivatives of the AppellF1 function in the origin of the parameters.

There are indeed cases in which specifiy relations between the parameters hold in which $F_1$ reduces to $_2F_1$ (see reference cited above) but, unfortunately, these are not met in our case $(d.4)$. Hence the word "almost" of @clathratus is sigificant.

Appendix

Mathematica gives for $i_1$

$$i_1 = \log (z) \left(-\text{Li}_3\left(\frac{t z+1}{t z-1}\right)-\text{Li}_3(t z+1)+\text{Li}_3\left(\frac{t z+1}{1-t z}\right)-\text{Li}_3(1-t z)+\left(\text{Li}_2\left(\frac{t z+1}{t z-1}\right)-\text{Li}_2\left(\frac{t z+1}{1-t z}\right)\right) \log \left(\frac{t z+1}{1-t z}\right)+\text{Li}_2(t z+1) \left(\log \left(\frac{t z+1}{1-t z}\right)+\log (1-t z)\right)+\text{Li}_2(1-t z) \left(\log (t z+1)-\log \left(\frac{t z+1}{1-t z}\right)\right)+\frac{1}{2} \left(\log (t z)-\log \left(\frac{2 t z}{t z-1}\right)+\log \left(\frac{2}{1-t z}\right)\right) \log ^2\left(\frac{t z+1}{1-t z}\right)+(\log (-t z)-\log (t z)) \log (t z+1) \log \left(\frac{t z+1}{1-t z}\right)+\frac{1}{2} (\log (t z)-\log (-t z)) \log (t z+1) (\log (t z+1)-2 \log (1-t z))+\log (t z) \log (t z+1) \log (1-t z)\right)$$

Answer 2

Introduction

Heureka! I have eventually solved the indefinite integral

$$a(x) = \int \frac{\log(1-x)\log(x)\log(1+x)}{x}\,dx\tag{1}$$

in other words, I have found the antidervative of the integrand

$$i(x) = \frac{\log(1-x)\log(x)\log(1+x)}{x}\tag{2}$$

This means that we have also the integral

$$f(z) = \int_{0}^z \frac{\log(1-x)\log(x)\log(1+x)}{x}\,dx\tag{3}$$

since it is convergent in the vicinity of $x=0$.

I was surprised that the use of "standard" integrations methods turned out to be sufficient (i.e. no (Euler-)series, no parametric derivatives were used).

For evaulating and checking the various integrals I made extensive use of Mathematica. The advantage of having indefinite integrals is that we can verify the result over a given range of the integration variable by calculating the derivative and comparing it to the integrand. This is in contrast to the case of constants (closed forms) obtained by assigning fixed values to the integration limits where the final result can only be checked numerically (most of the problems here are of the latter type).

Although an antiderivative was found it is available first of all as a Mathematica expression, and because the number of terms in it exceeds 1.000 printing it out is not really useful.

Solution

Step 1:

We simplify the integrand by the well-known decomposion of the product of two logarithms with different arguments into a sum of independent logarithmic terms

$$\log (1-x) \log (x+1)=\frac{1}{2} \log ^2\left(1-x^2\right)-\frac{1}{2} \log ^2(1-x)-\frac{1}{2} \log ^2(x+1)\tag{4}$$

The integral then becomes

$$a(x) = A(x) + B(x)$$

with

$$A=-\frac{1}{2} \int \frac{\log (x) \left(\log ^2(1-x)+\log ^2(x+1)\right)}{x} \, dx\tag{5a}$$

$$B= \frac{1}{2} \int \frac{\log (x) \log ^2\left(1-x^2\right)}{x} \, dx\tag{5b}$$

Step 2:

Doing the integrals.

$A(x)$ is immediately done by Mathematica producing a huge number of terms (leafcount=709) which consist of $\log$s and polylogs $\operatorname{Li}_2$ through $\operatorname{Li}_4$.

We normalize the antiderivative by subtracting a constant (obatined by calculating the limit for $x\to+0$ from above, i.e. from inside the integration interval) so that $A(0)=0$.

$B(x)$ is refused by Mathematica, so we have to prepare it manually. First we change the variable $x\to\sqrt{t}$:

$$\frac{\log (x) \log ^2\left(1-x^2\right)}{2 x}dx =\frac{ \log ^2(1-t) \log \left(\sqrt{t}\right)}{4 t}dt= \frac{ \log ^2(1-t) \log \left({t}\right)}{8 t}dt\tag{6}$$

Notice that the logarithm simplifies nicely.

Now the antiderivative is found by Mathematica, and in the end we have to reverse the variable transform by letting $t\to x^2$. It produces an expression with leafcount of 404 and consists of the same functions as $A(x)$. In the end we renormalize the result as above to $B(0)=0$.

The antiderivatives encountered here produced by Mathematica frequently contain complex terms. They occur typically through logarithms and polylogarithms, as e.g. in $\log(x)$ for $x<0$ and in $\operatorname{Li_n}(x)$ for $x\gt1$. When the integral is real by definition in some interval we should seek to transform all terms into real terms in this interval. For the polylog functions with natural index we have employed systematically the transformation formula from https://en.wikipedia.org/wiki/Polylogarithm thus making all terms real in the interval $0\lt x lt 1$.

$$\operatorname{Li}_n(x)\to (-1)^{n+1} \operatorname{Li}_n\left(\frac{1}{x}\right)-\frac{(2 \pi i)^n}{n!} B_n\left(\frac{\log (x)}{2 \pi i}+1\right)\text{ for x>1}\tag{7}$$

where $B_n(z)$ are the Bernoulli polynomials of order $n$ (https://en.wikipedia.org/wiki/Bernoulli_polynomials).

In total the antiderivative consists of the appreciable amount of about 1.100 terms, and it is not useful to print them all out here. But I can provide the Mathematica code of it on demand.

It is remarkable that all polylogs except $\operatorname{Li_1}(x) = - \log(1-x)$ appear in first power and linearly, i.e. there are no products of these functions appearing.

Evaluations

Here is the plot of the exact function $f(z)$

This picture as such could have been produced, of course, also by numerical integration.

The knowledge of $f(z)$ permits us to calculate any definite integral we wish in the interval $0\le z \le 1$.

Example 1: $f(1)$

We obtain

$$f(1) = 2 \text{Li}_4\left(\frac{1}{2}\right)+\frac{7}{4} \zeta (3) \log (2)-\frac{3 \pi ^4}{160}+\frac{\log ^4(2)}{12}-\frac{1}{12} \pi ^2 \log ^2(2)\simeq 0.290721\tag{8}$$

in agreement with the expression for $f(1)$ calculated some time ago by others (Evaluating $\int^1_0 \frac{\log(1+x)\log(1-x) \log(x)}{x}\, \mathrm dx$).

Example 2: $f(\frac{1}{2})$

This will be given in full length.

$$f(\frac{1}{2})= -\operatorname{Li}_4\left(\frac{1}{2}\right)+\operatorname{Li}_4\left(\frac{1}{3}\right)+\operatorname{Li}_4\left(\frac{2}{3}\right)-\frac{\operatorname{Li}_4\left(-\frac{1}{3}\right)}{4}-\frac{\operatorname{Li}_4\left(\frac{1}{4}\right)}{8}+\frac{\operatorname{Li}_4\left(\frac{3}{4}\right)}{4}+\frac{1}{2} \operatorname{Li}_2\left(-\frac{1}{2}\right) \log ^2(2)+\frac{3}{4} \operatorname{Li}_2\left(\frac{3}{4}\right) \log ^2(2)+\frac{1}{2} \operatorname{Li}_2\left(-\frac{1}{2}\right) \log ^2(3)+\frac{1}{2} \operatorname{Li}_2\left(\frac{1}{3}\right) \log ^2(3)-\frac{1}{6} \pi ^2 \left(3 \operatorname{Li}_2\left(-\frac{1}{2}\right)+3 \operatorname{Li}_2\left(\frac{1}{3}\right)-4 \log ^2(2)+5 \log ^2(3)-\log (3) \log (2)+4 (\log (3)-\log (2))^2-(\log (3)-\log (2)) (2 \log (2)+7 \log (3))\right)-\frac{1}{8} \operatorname{Li}_2\left(-\frac{1}{3}\right) \log ^2(3)+\frac{1}{8} \operatorname{Li}_2\left(\frac{3}{4}\right) \log ^2(3)-\operatorname{Li}_2\left(-\frac{1}{2}\right) \log (3) \log (2)-\frac{1}{2} \operatorname{Li}_2\left(\frac{3}{4}\right) \log (3) \log (2)+\operatorname{Li}_3\left(-\frac{1}{2}\right) \log (2)+\operatorname{Li}_3\left(\frac{1}{3}\right) \log (2)-\frac{1}{4} \operatorname{Li}_3\left(\frac{1}{4}\right) \log (2)+\frac{1}{2} \operatorname{Li}_3\left(\frac{3}{4}\right) \log (2)-\operatorname{Li}_3\left(-\frac{1}{2}\right) \log (3)-\frac{1}{4} \operatorname{Li}_3\left(-\frac{1}{3}\right) \log (3)-\frac{1}{4} \operatorname{Li}_3\left(\frac{3}{4}\right) \log (3)-\frac{15}{8} \zeta (3) \log (2)+\zeta (3) \log (3)-\frac{\pi ^4}{80}+\frac{4 \log ^4(2)}{3}+\frac{3 \log ^4(3)}{32}-\log (3) \log ^3(2)+\frac{5}{24} (\log (3)-\log (2))^4\simeq 0.155958\tag{9}$$

Discussion

There are several topics to be discussed in the wake of this type of calculations.

1. Generalization: solution of a whole class of trilog integrals

See Evaluating the indefinite integral $\int x^k \log (1-x) \log (x) \log (x+1) \, dx$

I have found that integrals of the type

$$f_k(x) = \int x^{k}\log(1-x)\log(x)\log(1+x)\,dx\tag{10}$$

with integer $k$ can be solved explicitly. For $k=-1$ it was shown here, for other $k$ integration by parts leads to integrable exppressions.

2. Relation to generating function problem

I have mentioned here What's the generating function for $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}x^n\ ?$ that the problem of the generating function of third order for the alternating harmonic series boils down to two integrals, one of which is the integral solved here.

3. Calculations tend to resemble a proof of existence

Because of the large number of terms involved the philosophy approaches more that of a proof of existence of an antiderivative. The procedure typically is: if Mathematica can't solve a given integral then perform manual (mathematical) actions on it leading to a number of integrals which hopefully Mathematica can solve. We could then disregard the actual results and only state that an antiderivative exists.