Closed form of $\int_{0}^{1} \frac{\log(1+x)\log(2+x) \log(3+x)}{1+x}\,dx$

Question

In https://math.stackexchange.com/a/3414337/198592 I have mentioned some integrals which I could not solve.

One of them is

$$i_{1}=\int_{0}^{1} \frac{\log(1+x)\log(2+x)\log(3+x)}{1+x}\,dx \simeq 0.295123\tag{1}$$

I have tried several approaches like partial integration and series expansion but with no avail.

The substiution

$$\log(x) = \int_0^{\infty } \frac{e^{-t}-e^{-t\; x}}{t} \, dt\tag{2}$$

did allow doing the $x$-integral but already the first integration of the triple integral failed.

Interestingly, with "one $\log$ less" or without the denominator $1+x$ the integration leads to a closed expression.

Problem: find a closed expression for $i_1$ or, equivalently, for

$$i_{1s}=\int_{0}^{1} \frac{\log(1+x)\log(1+\frac{x}{2})\log(1+\frac{x}{3})}{1+x}\,dx \simeq 0.0130713\tag{3}$$

Answer 1

The process flows elegantly if we simply make use of the variable change $x\mapsto (1-x)/(1+x)$. I'll develop it.

So, we have that

$$\int_0^1\frac{\log(1+x)\log(2+x)\log(3+x)}{1+x}\textrm{d}x$$ $$=\frac{1}{2}\log^2(2)\int_0^1 \frac{\log(3+x)}{1+x}\textrm{d}x-\frac{1}{2}\log^2(2)\int_0^1 \frac{\log(1+x)}{1+x}\textrm{d}x+\log(2)\int_0^1 \frac{\log^2(1+x)}{1+x}\textrm{d}x$$ $$-\frac{1}{2}\int_0^1\frac{\log^3(1+x)}{1+x}\textrm{d}x-\frac{\log(2)}{2}\int_0^1 \frac{\log(1+x)\log(2+x)}{1+x}\textrm{d}x$$ $$+\frac{\log(2)}{2}\int_0^1 \frac{\log(2+x)\log(3+x)}{1+x}\textrm{d}x-\log(2)\int_0^1 \frac{\log(1+x)\log(3+x)}{1+x}\textrm{d}x$$ $$+\frac{1}{2}\int_0^1 \frac{\log^2(1+x)\log(2+x)}{1+x}\textrm{d}x+\frac{1}{2}\int_0^1 \frac{\log^2(1+x)\log(3+x)}{1+x}\textrm{d}x.$$

What's next? We look at these integral with a trained eye and see all are reducible immediately to known, trivial integrals (mainly with integration by parts and usual knowledge of polylogarithms).

End of story.

Answer 2

Approach 1: Series expansion

Having studied approach 2 in more detail I found that it corresponds just to a direct series expansion of the integrand of the integral.

Using

$$\frac{\log(1+x)}{1+x} = \sum _{i=1}^{\infty } (-1)^{i+1} H_i x^i$$

where

$$H_i = \sum _{n=1}^i \frac{1}{n}$$

is the harmonic number, and

$$\log \left(1+\frac{x}{2}\right)=\sum _{j=1}^{\infty } \frac{(-1)^{j+1} 2^{-j} x^j}{j}$$

and

$$\log \left(1+\frac{x}{3}\right)= \sum _{k=1}^{\infty } \frac{(-1)^{k+1} 3^{-k} x^k}{k}$$

we can write the integrand of $i_{1s}$ as

$$\frac{\log(1+x)\log(1+\frac{x}{2})\log(1+\frac{x}{3})}{1+x}=\sum_{m=1}^{\infty} (-1)^{m+1} c(m) x^m$$

where the coefficients are defined as

$$c(m)=\sum _{i=1}^{\infty} \sum _{j=1}^{\infty} \sum _{k=1}^{\infty} \frac{ H_i}{ 2^{j}\;j\;3^{k}\; k} \delta _{m,i+j+k}$$

Here $\delta _{n,m}$ is Kronecker's delta ($=1$ if $n=m$, $= 0$ else).

Notice that the triple sum consists effectively only of $\left\lceil \frac{1}{2} (m-3) \left(m+\frac{1}{2}-3\right)\right\rceil$ summands.

Writing $c(m) = p(m)/q(m)$ as an irreducible fraction we have for the first few terms

$$\{p\}_{m=1}^{m=10}=\{0,0,0,1,23,283,2725,46261,1821713,4554217\}$$ $$\{q\}_{m=1}^{m=10} = \{1,1,1,6,72,648,5184,77760,2799360,6531840\}$$

none of which is contained in https://oeis.org/.

The integral itself is then given by

$$i_{1s} = \sum_{m=1}^{\infty} \frac{c(m)}{m+1}$$

The coefficients look suffciently complicated so that a closed expression seems to be out of reach.

However, still complicated looking cases do have closed expressions, as for instance

$$\int_0^1 \frac{\log (x+1) \log \left(\frac{x}{2}+1\right)}{x+1} \, dx=\operatorname{Li}_3(-2)-\operatorname{Li}_2(-2) \log (2)+\frac{3 \zeta (3)}{4}-\frac{1}{2} \log ^3(2)$$

Hence we don't give up but consider this sequence of integrals

$$i(k) = (-1)^{k+1} \int_0^1 \frac{x^k \log (x+1) \log \left(\frac{x}{2}+1\right)}{x+1} \, dx$$

from which we find the integral by forming the series of $\log(1+x/3)$ thus

$$i_{1s} = \sum_{k=1}^{\infty} \frac{ i(k)}{k \;3^k}$$

We have

$$i(1) = -\operatorname{Li}_2\left(-\frac{1}{2}\right)-\operatorname{Li}_3\left(-\frac{1}{2}\right)-\operatorname{Li}_2\left(-\frac{1}{2}\right) \log (2)-\frac{3 \zeta (3)}{4}-\frac{\pi ^2}{12}+2+\frac{1}{6}\log ^3(2)\\-\frac{5 \log ^2(2)}{2}+3 \log (3) \log (2)-\log \left(\frac{27}{2}\right)$$

$$i(2) = -\frac{5 \operatorname{Li}_2\left(-\frac{1}{2}\right)}{2}-\operatorname{Li}_3\left(-\frac{1}{2}\right)-\operatorname{Li}_2\left(-\frac{1}{2}\right) \log (2)-\frac{3 \zeta (3)}{4}-\frac{5 \pi ^2}{24}+4+\frac{1}{6} \log ^3(2)\\-\frac{5 \log ^2(2)}{4}+\frac{9}{2} \log (3) \log (2)+\frac{5 \log (2)}{4}-\frac{21 \log (3)}{4}-\frac{1}{12} \log (8) \log (256)$$

$$i(3) = -\frac{29 \operatorname{Li}_2\left(-\frac{1}{2}\right)}{6}-\operatorname{Li}_3\left(-\frac{1}{2}\right)-\operatorname{Li}_2\left(-\frac{1}{2}\right) \log (2)-\frac{3 \zeta (3)}{4}-\frac{29 \pi ^2}{72}+\frac{707}{108}+\frac{1}{6}\log ^3(2)\\-\frac{61 \log ^2(2)}{12}+\frac{15}{2} \log (3) \log (2)+\frac{31 \log (2)}{36}-\frac{31 \log (3)}{4}$$

$$i(4) =-\frac{103 \operatorname{Li}_2\left(-\frac{1}{2}\right)}{12}-\operatorname{Li}_3\left(-\frac{1}{2}\right)-\operatorname{Li}_2\left(-\frac{1}{2}\right) \log (2)-\frac{3 \zeta (3)}{4}-\frac{103 \pi ^2}{144}+\frac{2179}{216}-\frac{1}{6} \log ^3(2)\\-\frac{167 \log ^2(2)}{24}+\frac{45}{4} \log (3) \log (2)-\frac{95 \log (2)}{144}-\frac{169 \log (3)}{16}$$

It seems that the expressions have the form

$$i(k) = a_{1,k} + a_{2,k}\;\pi^2 + a_{3,k}\; \log (2)+ a_{4,k}\;\log ^2(2)+\\+ a_{5,k}\;\log (3)+ a_{6,k}\;\log (2)\log (3)+a_{7,k} \operatorname{Li}_2\left(-\frac{1}{2}\right)\\ + \left(\frac{1}{6} \log ^3(2)- \operatorname{Li}_3\left(-\frac{1}{2}\right)-\operatorname{Li}_2\left(-\frac{1}{2}\right) \log (2)-\frac{3 \zeta (3)}{4}\right)$$

where the $a_{i,k}$ are rational numbers.

Approach 2: Generating function (original post)

A few minutes after having posted the problem I took up a thread I had considered earlier, and it gives a formal solution in terms of derivatives of known (but not particularly common) functions.

The trick is to generate the $\log$ and the power by

$$\frac{\partial x^a}{\partial a}=x^a \log (x)$$

Consider the function

$$g(x,a,b,c) = (x+1)^a (x+2)^b (x+3)^c\tag{1}$$

from which we can generate the integrand

$$\frac{\log (x+1) \log (x+2) \log (x+3)}{x+1}\tag{2}$$

by a triple derivative and appropriate replacement of the parameters $a,b,c$ as follows

$$\frac{\partial ^3\left((x+1)^a (x+2)^b (x+3)^c\right)}{\partial a\, \partial b\, \partial c}\text{/.}\, \{a\to -1,b\to 0,c\to 0\}\tag{3}$$

Interchanging the order of operations and doing the $x$-integral first we get

$$G(a,b,c) = \int_{0}^{1} g(x,a,b,c)\,dx \\= \frac{2^c}{a+1} \left(2^{a+1} F_1(a+1;-b,-c;a+2;-2,-1)\\-F_1\left(a+1;-b,-c;a+2;-1,-\frac{1}{2}\right)\right)\tag{4}$$

Here

$$F_1(r;s,t;u;x,y)=\sum_{m,n=0}^{\infty} \frac{(r)_{m+n} (s)_{m} (t)_{n}}{(u)_{m+n}}\frac{x^m}{m!} \frac{y^n}{n!}\tag{5}$$

with the notation $(r)_n=\frac{\Gamma (n+r)}{\Gamma (r)}$ is the AppellF1 function which belongs to the class of the hypergeometric functions (http://mathworld.wolfram.com/AppellHypergeometricFunction.html).

The reference also provides an representation as a single integral

$$F_1(r;s,t;u;x,y)=\frac{\Gamma (u)}{\Gamma (r) \Gamma (u-r)} \int_0^1 z^{r-1} (1-z)^{-r+u-1} (1-x z)^{-s} (1-y z)^{-t} \, dz\tag{5a}$$

Hence we have the formal solution

$$i_{1}=\frac{\partial ^3 G(a,b,c)}{\partial a\, \partial b\, \partial c}\text{/.}\, \{a\to -1,b\to 0,c\to 0\}\tag{6}$$

The first steps in trying to evaluate $(6)$ do not support hope for a simple final expression. It seems that we will have replaced a nice integral by a more or less ugly double sum. But let's see ...

Answer 3

We have seen a great elegantly flowing answer of user97357329 (https://math.stackexchange.com/a/3522251/198592), the main step of which consisted in the substitution $x\to\frac{1-t}{1+t}$.

The next steps have been sketched very briefly in the answer, so that I take the freedom to elaborate a bit more and, in the end, provide the explit closed form of our integral.

The transformation has the nice property (which can't be seen in the exposition of that user) that after expanding the resulting logs in the integrand we get one term that is the exact negative of the original term, and all other terms are simpler.

The transformation gives (with $dx\to- 2dt/(t+1)^2$)

$$i_{1}=\int_0^1 \frac{\log (x+1) \log (x+2) \log (x+3))}{x+1}\,dx\to\int_0^1 \frac{\log \left(\frac{2}{t+1}\right) \log \left(\frac{2 (t+2)}{t+1}\right) \log \left(\frac{t+3}{t+1}\right)}{t+1}\,dt$$

Expanding the logs in the integrand yields this list of terms

$$\left\{-\frac{\log ^2(2) \log (t+1)}{t+1},\frac{2 \log (2) \log ^2(t+1)}{t+1},-\frac{\log ^3(t+1)}{t+1},-\frac{\log (2) \log (t+1) \log (t+2)}{t+1},\frac{\log ^2(t+1) \log (t+2)}{t+1},\frac{\log ^2(2) \log (t+3)}{t+1},-\frac{2 \log (2) \log (t+1) \log (t+3)}{t+1},\frac{\log ^2(t+1) \log (t+3)}{t+1},\frac{\log (2) \log (t+2) \log (t+3)}{t+1},-\frac{\log (t+1) \log (t+2) \log (t+3)}{t+1}\right\}$$

We see that the last term is exactly the opposite of the original integrand, so that the complicated part drops out (move it to the l.h.s. and get twice the original expression).

The remaining integrals are in fact easy to solve. Using Mathematica we get some terms in the complex domain which, however compensate when we use the relations

$$\operatorname{Li}_2\left(\frac{3}{2}\right)= -\operatorname{Li}_2\left(-\frac{1}{2}\right)+\frac{\pi ^2}{6}+\log (2) \log \left(\frac{3}{2}\right)-i \pi \log \left(\frac{3}{2}\right)$$

$$\operatorname{Li}_2(2)= \frac{\pi ^2}{6}+\frac{\pi ^2}{12}-i \pi \log (2)$$

The final result then becomes (I'm busy simplifying this)

$$i_{1} = -\operatorname{Li}_4(-2)+\operatorname{Li}_4\left(-\frac{1}{2}\right)-\frac{1}{2} \operatorname{Li}_2\left(\frac{1}{3}\right) \log ^2(2)-2 \operatorname{Li}_2\left(\frac{2}{3}\right) \log ^2(2)+\frac{1}{4} \operatorname{Li}_2\left(\frac{1}{4}\right) \log ^2(2)-\frac{1}{2} \operatorname{Li}_2\left(\frac{1}{3}\right) \log (3) \log (2)+\frac{1}{2} \operatorname{Li}_2\left(\frac{2}{3}\right) \log (27) \log (2)-\frac{1}{2} \operatorname{Li}_2\left(\frac{3}{4}\right) \log (8) \log (2)+\frac{1}{2} \operatorname{Li}_2\left(\frac{3}{4}\right) \log (9) \log (2)-\frac{1}{2} \operatorname{Li}_3(-2) \log (2)-\frac{1}{2} \text{Li}_3\left(\frac{1}{3}\right) \log (2)+\frac{3}{2} \operatorname{Li}_3\left(\frac{2}{3}\right) \log (2)+\frac{1}{4} \operatorname{Li}_3\left(\frac{1}{4}\right) \log (2)-\operatorname{Li}_3\left(\frac{3}{4}\right) \log (2)+\frac{1}{2} \operatorname{Li}_3(-2) \log (4)-\frac{1}{2} \zeta (3) \log (2)-\frac{3}{8} \zeta (3) \log (4)-\frac{1}{24} 35 \log ^4(2)+\frac{5}{2} \log (3) \log ^3(2)+\frac{1}{3} \log ^3(3) \log (2)+\frac{1}{2} \zeta(2) \log ^2(2)-2 \log ^2(3) \log ^2(2)+\frac{1}{6} \log (3) \log (64) \log ^2(2)$$

This is a linear combination of the components $\log(2)$, $\log(3)$, $\zeta(2)$, $\zeta(3)$, $\operatorname{Li}_{2,3,4}$. Notice that there is no $\pi$ or (Euler)-$\gamma$ appearing.

Answer 4

Not an answer but too long for a comment.

For the fun of it, I used Taylor expansion to $O(x^{n+1})$. Below are some numbers wich show a very slow convergence $$\left( \begin{array}{cc} n & \text{result} \\ 100 & \color{red} {0.295}088992683718 \\ 200 & \color{red} {0.2951}14319823039 \\ 300 & \color{red} {0.2951}19066043094 \\ 400 & \color{red} {0.29512}0734361895 \\ 500 & \color{red} {0.29512}1508301534 \\ 600 & \color{red} {0.29512}1929303681 \\ 700 & \color{red} {0.295122}183398346 \\ 800 & \color{red} {0.295122}348430736 \\ 900 & \color{red} {0.295122}461636524 \end{array} \right)$$