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This question is related to my previous question here. While tinkering around for a solution I found that the integral there can be reduced to the problem of solving the following basic logarithmic integral:

Define the function $\mathcal{I}:[-1,\infty)^{2}\rightarrow\mathbb{R}$ via the integral representation, $$\mathcal{I}{\left(a,b\right)}:=\int_{0}^{1}\frac{\ln{\left(1+ax\right)}\ln{\left(1+bx\right)}}{x}\,\mathrm{d}x;~~~\small{\left(a,b\right)\in[-1,\infty)^{2}}.$$

If my calculations are correct, for $-1<a\neq0\land-1<b\neq0\land a\neq b$, the integral $\mathcal{I}$ can be evaluated in closed form as:

$$\begin{align} \mathcal{I}{\left(a,b\right)} &=\operatorname{Li}_{3}{\left(\frac{a}{1+a}\right)}+\operatorname{Li}_{3}{\left(-a\right)}+\operatorname{Li}_{3}{\left(\frac{b}{1+b}\right)}+\operatorname{Li}_{3}{\left(-b\right)}\\ &~~~~~-\operatorname{Li}_{3}{\left(\frac{b-a}{1+b}\right)}-\operatorname{Li}_{3}{\left(\frac{a-b}{1+a}\right)}+\color{red}{\operatorname{Li}_{3}{\left(\frac{b}{a}\right)}}-\color{red}{\operatorname{Li}_{3}{\left(\frac{(1+a)b}{a(1+b)}\right)}}\\ &~~~~~-\ln{\left(1+a\right)}\operatorname{Li}_{2}{\left(-a\right)}-\ln{\left(1+b\right)}\operatorname{Li}_{2}{\left(-b\right)}\\ &~~~~~+\ln{\left(\frac{1+a}{1+b}\right)}\left[\color{red}{\operatorname{Li}_{2}{\left(\frac{(1+a)b}{a(1+b)}\right)}}+\operatorname{Li}_{2}{\left(\frac{b-a}{1+b}\right)}\right]\\ &~~~~~+\frac12\ln^{2}{\left(\frac{1+a}{1+b}\right)}\color{red}{\ln{\left(\frac{a-b}{a(1+b)}\right)}}+\frac16\ln^{3}{\left(\frac{1+a}{1+b}\right)}\\ &~~~~~-\frac16\ln^{3}{\left(1+a\right)}-\frac16\ln^{3}{\left(1+b\right)}.\\ \end{align}$$

Now for my question. Except for the terms I highlighted in red, we would have a manifestly real expression for $\mathcal{I}$ in each of its individual terms. It seams feasible that some suitable combination of polylogarithmic identities could transform the red terms into ones such that all polylogarithmic arguments are less then unity and all logarithmic arguments are positive simultaneously, but none of the transformations I can think of seem to be doing the trick. Assuming it can be done, how might the sum of red terms be simplified to one in which all individual components are real?

Thank you for reading. -DH


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We have defined the function $\mathcal{I}:[-1,\infty)^{2}\rightarrow\mathbb{R}$ via the integral representation,

$$\mathcal{I}{\left(a,b\right)}:=\int_{0}^{1}\frac{\ln{\left(1+ax\right)}\ln{\left(1+bx\right)}}{x}\,\mathrm{d}x;~~~\small{\left(a,b\right)\in[-1,\infty)^{2}}.\tag{1}$$

The goal is to find a single closed-form expression for the integral $\mathcal{I}{\left(a,b\right)}$ in terms of elementary functions and polylogarithms that is valid over as large of the domain as possible. The specific definitions for the various polylogarithmic functions employed below have been collected in the appendix at the bottom.

It is obvious from definition $(1)$ that the function $\mathcal{I}{\left(a,b\right)}$ is symmetric under exchange of its arguments, i.e.,

$$\forall \left(a,b\right)\in[-1,\infty)^{2}:\mathcal{I}{\left(a,b\right)}=\mathcal{I}{\left(b,a\right)},\tag{2}$$

so we can assume without loss of generality that $a\le b$.

It is also obvious from definition $(1)$ that the function $\mathcal{I}{\left(a,b\right)}$ will trivially vanish in the special case that one or both of its arguments equals zero:

$$\forall \left(a,b\right)\in[-1,\infty)^{2}:\mathcal{I}{\left(a,0\right)}=\mathcal{I}{\left(0,b\right)}=0.\tag{3}$$

There are additional special cases with simple final expressions. One such special case is that of equal parameters. For $a\in[-1,\infty)$,

$$\begin{align} \mathcal{I}{\left(a,a\right)} &=\int_{0}^{1}\frac{\ln^{2}{\left(1+ax\right)}}{x}\,\mathrm{d}x\\ &=2\int_{0}^{1}\frac{\ln^{2}{\left(1+ax\right)}}{2x}\,\mathrm{d}x\\ &=2\,S_{1,2}{\left(-a\right)}.\tag{4}\\ \end{align}$$

Another special case occurs when either of the parameters equals negative unity. For $a\in[-1,\infty)$, we have

$$\begin{align} \mathcal{I}{\left(a,-1\right)} &=\int_{0}^{1}\frac{\ln{\left(1+ax\right)}\ln{\left(1-x\right)}}{x}\,\mathrm{d}x\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{x}\int_{0}^{1}\mathrm{d}y\,\frac{ax}{1+axy}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{a\ln{\left(1-x\right)}}{1+axy}\\ &=\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}x\,\frac{a\ln{\left(1-x\right)}}{1+ayx}\\ &=\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}x\,\frac{1}{1-x}\left[\frac{\ln{\left(1+ayx\right)}}{y}-\frac{\ln{\left(1+ay\right)}}{y}\right];~~~\small{I.B.P.s}\\ &=\int_{0}^{1}\frac{\mathrm{d}y}{y}\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+ayx\right)}-\ln{\left(1+ay\right)}}{1-x}\\ &=\int_{0}^{1}\frac{\mathrm{d}y}{y}\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1+ay-ayt\right)}-\ln{\left(1+ay\right)}}{t};~~~\small{\left[1-x=t\right]}\\ &=\int_{0}^{1}\frac{\mathrm{d}y}{y}\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-\frac{ay}{1+ay}t\right)}}{t}\\ &=\int_{0}^{1}\frac{\mathrm{d}y}{y}\left[-\operatorname{Li}_{2}{\left(\frac{ay}{1+ay}\right)}\right]\\ &=\int_{0}^{1}\frac{\operatorname{Li}_{2}{\left(-ay\right)}+\frac12\ln^{2}{\left(1+ay\right)}}{y}\,\mathrm{d}y;~~~\small{\text{Landen's Identity}}\\ &=\int_{0}^{1}\frac{\operatorname{Li}_{2}{\left(-ay\right)}}{y}\,\mathrm{d}y+\frac12\int_{0}^{1}\frac{\ln^{2}{\left(1+ay\right)}}{y}\,\mathrm{d}y\\ &=\operatorname{Li}_{3}{\left(-a\right)}+S_{1,2}{\left(-a\right)},\tag{5}\\ \end{align}$$

where the so-called dilogarithm identity of Landen mentioned in the above derivation refers to the identity,

$$\operatorname{Li}_{2}{\left(z\right)}+\operatorname{Li}_{2}{\left(\frac{z}{z-1}\right)}+\frac12\ln^{2}{\left(1-z\right)}=0;~~~\small{z\notin[1,\infty)}.\tag{6}$$


Main Result:

Suppose $-1<a<b\land a\neq0\land b\neq0$. Then, using the algebraic identity,

$$2xy=x^{2}+y^{2}-\left(x-y\right)^{2},\tag{7}$$

and the logarithmic identity,

$$\ln{\left(z\right)}-\ln{\left(w\right)}=\ln{\left(\frac{z}{w}\right)};~~~\small{z\ge0\land w\ge0},\tag{8}$$

we have

$$\begin{align} \mathcal{I}{\left(a,b\right)} &=\int_{0}^{1}\frac{\ln{\left(1+ax\right)}\ln{\left(1+bx\right)}}{x}\,\mathrm{d}x\\ &=\int_{0}^{1}\frac{2\ln{\left(1+ax\right)}\ln{\left(1+bx\right)}}{2x}\,\mathrm{d}x\\ &=\frac12\int_{0}^{1}\frac{\ln^{2}{\left(1+ax\right)}+\ln^{2}{\left(1+bx\right)}-\left[\ln{\left(1+ax\right)}-\ln{\left(1+bx\right)}\right]^{2}}{x}\,\mathrm{d}x\\ &=\frac12\int_{0}^{1}\frac{\ln^{2}{\left(1+ax\right)}}{x}\,\mathrm{d}x+\frac12\int_{0}^{1}\frac{\ln^{2}{\left(1+bx\right)}}{x}\,\mathrm{d}x\\ &~~~~~-\frac12\int_{0}^{1}\frac{\ln^{2}{\left(\frac{1+ax}{1+bx}\right)}}{x}\,\mathrm{d}x\\ &=S_{1,2}{\left(-a\right)}+S_{1,2}{\left(-b\right)}-\frac12\int_{0}^{1}\frac{\ln^{2}{\left(\frac{1+ax}{1+bx}\right)}}{x}\,\mathrm{d}x\\ &=S_{1,2}{\left(-a\right)}+S_{1,2}{\left(-b\right)}-\frac12\int_{1}^{\frac{1+a}{1+b}}\frac{\ln^{2}{\left(y\right)}}{\left(\frac{1-y}{by-a}\right)}\cdot\frac{a-b}{\left(by-a\right)^{2}}\,\mathrm{d}y;~~~\small{\left[\frac{1+ax}{1+bx}=y\right]}\\ &=S_{1,2}{\left(-a\right)}+S_{1,2}{\left(-b\right)}-\frac12\int_{\frac{1+a}{1+b}}^{1}\frac{\left(b-a\right)\ln^{2}{\left(y\right)}}{\left(1-y\right)\left(by-a\right)}\,\mathrm{d}y\\ &=S_{1,2}{\left(-a\right)}+S_{1,2}{\left(-b\right)}-\frac12\int_{\frac{1+a}{1+b}}^{1}\frac{\ln^{2}{\left(y\right)}}{1-y}\,\mathrm{d}y-\frac12\int_{\frac{1+a}{1+b}}^{1}\frac{b\ln^{2}{\left(y\right)}}{by-a}\,\mathrm{d}y\\ &=S_{1,2}{\left(-a\right)}+S_{1,2}{\left(-b\right)}-\frac12\int_{0}^{\frac{b-a}{1+b}}\frac{\ln^{2}{\left(1-t\right)}}{t}\,\mathrm{d}t;~~~\small{\left[1-y=t\right]}\\ &~~~~~-\frac12\int_{\frac{1+a}{1+b}}^{1}\frac{b\ln^{2}{\left(y\right)}}{by-a}\,\mathrm{d}y\\ &=S_{1,2}{\left(-a\right)}+S_{1,2}{\left(-b\right)}-S_{1,2}{\left(\frac{b-a}{1+b}\right)}-\frac12\int_{\frac{1+a}{1+b}}^{1}\frac{b\ln^{2}{\left(y\right)}}{by-a}\,\mathrm{d}y.\tag{9}\\ \end{align}$$

Next, note that $\ln{\left(y\right)}=\ln{\left(\left|y\right|\right)}$ for $y>0$, and $0<\frac{1+a}{1+b}\le y$ for $\left(a,b\right)\in(-1,\infty)^{2}$. Thus,

$$\begin{align} \mathcal{I}{\left(a,b\right)} &=S_{1,2}{\left(-a\right)}+S_{1,2}{\left(-b\right)}-S_{1,2}{\left(\frac{b-a}{1+b}\right)}-\frac12\int_{\frac{1+a}{1+b}}^{1}\frac{b\ln^{2}{\left(y\right)}}{by-a}\,\mathrm{d}y\\ &=S_{1,2}{\left(-a\right)}+S_{1,2}{\left(-b\right)}-S_{1,2}{\left(\frac{b-a}{1+b}\right)}-\frac12\int_{\frac{1+a}{1+b}}^{1}\frac{b\ln^{2}{\left(\left|y\right|\right)}}{by-a}\,\mathrm{d}y\\ &=S_{1,2}{\left(-a\right)}+S_{1,2}{\left(-b\right)}-S_{1,2}{\left(\frac{b-a}{1+b}\right)}\\ &~~~~~-\frac12\int_{-\frac{\left(1+a\right)b}{a\left(1+b\right)}}^{-\frac{b}{a}}\frac{\ln^{2}{\left(\left|\frac{at}{b}\right|\right)}}{1+t}\,\mathrm{d}t;~~~\small{\left[y=-\frac{at}{b}\right]}\\ &=S_{1,2}{\left(-a\right)}+S_{1,2}{\left(-b\right)}-S_{1,2}{\left(\frac{b-a}{1+b}\right)}\\ &~~~~~-\frac12\int_{-\frac{\left(1+a\right)b}{a\left(1+b\right)}}^{-\frac{b}{a}}\frac{\left[\ln{\left(\left|t\right|\right)}+\ln{\left(\left|\frac{a}{b}\right|\right)}\right]^{2}}{1+t}\,\mathrm{d}t\\ &=S_{1,2}{\left(-a\right)}+S_{1,2}{\left(-b\right)}-S_{1,2}{\left(\frac{b-a}{1+b}\right)}\\ &~~~~~-\frac12\int_{-\frac{\left(1+a\right)b}{a\left(1+b\right)}}^{-\frac{b}{a}}\frac{\ln^{2}{\left(\left|t\right|\right)}+2\ln{\left(\left|\frac{a}{b}\right|\right)}\ln{\left(\left|t\right|\right)}+\ln^{2}{\left(\left|\frac{a}{b}\right|\right)}}{1+t}\,\mathrm{d}t\\ &=S_{1,2}{\left(-a\right)}+S_{1,2}{\left(-b\right)}-S_{1,2}{\left(\frac{b-a}{1+b}\right)}-\frac12\int_{-\frac{\left(1+a\right)b}{a\left(1+b\right)}}^{-\frac{b}{a}}\frac{\ln^{2}{\left(\left|t\right|\right)}}{1+t}\,\mathrm{d}t\\ &~~~~~-\ln{\left(\left|\frac{a}{b}\right|\right)}\int_{-\frac{\left(1+a\right)b}{a\left(1+b\right)}}^{-\frac{b}{a}}\frac{\ln{\left(\left|t\right|\right)}}{1+t}\,\mathrm{d}t-\frac12\ln^{2}{\left(\left|\frac{a}{b}\right|\right)}\int_{-\frac{\left(1+a\right)b}{a\left(1+b\right)}}^{-\frac{b}{a}}\frac{\mathrm{d}t}{1+t}\\ &=S_{1,2}{\left(-a\right)}+S_{1,2}{\left(-b\right)}-S_{1,2}{\left(\frac{b-a}{1+b}\right)}-\frac12\int_{0}^{-\frac{b}{a}}\frac{\ln^{2}{\left(\left|t\right|\right)}}{1+t}\,\mathrm{d}t\\ &~~~~~+\frac12\int_{0}^{-\frac{\left(1+a\right)b}{a\left(1+b\right)}}\frac{\ln^{2}{\left(\left|t\right|\right)}}{1+t}\,\mathrm{d}t-\ln{\left(\left|\frac{a}{b}\right|\right)}\int_{0}^{-\frac{b}{a}}\frac{\ln{\left(\left|t\right|\right)}}{1+t}\,\mathrm{d}t\\ &~~~~~+\ln{\left(\left|\frac{a}{b}\right|\right)}\int_{0}^{-\frac{\left(1+a\right)b}{a\left(1+b\right)}}\frac{\ln{\left(\left|t\right|\right)}}{1+t}\,\mathrm{d}t-\frac12\ln^{2}{\left(\left|\frac{a}{b}\right|\right)}\int_{-\frac{\left(b-a\right)}{a\left(1+b\right)}}^{-\frac{\left(b-a\right)}{a}}\frac{\mathrm{d}u}{u};~~~\small{\left[1+t=u\right]}\\ &=S_{1,2}{\left(-a\right)}+S_{1,2}{\left(-b\right)}-S_{1,2}{\left(\frac{b-a}{1+b}\right)}-\frac12\,\Lambda_{3}{\left(-\frac{b}{a}\right)}+\frac12\,\Lambda_{3}{\left(-\frac{\left(1+a\right)b}{a\left(1+b\right)}\right)}\\ &~~~~~-\ln{\left(\left|\frac{a}{b}\right|\right)}\,\Lambda_{2}{\left(-\frac{b}{a}\right)}+\ln{\left(\left|\frac{a}{b}\right|\right)}\,\Lambda_{2}{\left(-\frac{\left(1+a\right)b}{a\left(1+b\right)}\right)}\\ &~~~~~-\frac12\ln^{2}{\left(\left|\frac{a}{b}\right|\right)}\int_{\frac{1}{1+b}}^{1}\frac{\mathrm{d}v}{v};~~~\small{\left[-\frac{\left(b-a\right)u}{a}=v\right]}\\ &=S_{1,2}{\left(-a\right)}+S_{1,2}{\left(-b\right)}-S_{1,2}{\left(\frac{b-a}{1+b}\right)}-\frac12\left[\Lambda_{3}{\left(-\frac{b}{a}\right)}-\Lambda_{3}{\left(-\frac{\left(1+a\right)b}{a\left(1+b\right)}\right)}\right]\\ &~~~~~-\ln{\left(\left|\frac{a}{b}\right|\right)}\left[\Lambda_{2}{\left(-\frac{b}{a}\right)}-\Lambda_{2}{\left(-\frac{\left(1+a\right)b}{a\left(1+b\right)}\right)}\right]\\ &~~~~~-\frac12\ln^{2}{\left(\left|\frac{a}{b}\right|\right)}\int_{1}^{1+b}\frac{\mathrm{d}w}{w};~~~\small{\left[v=w^{-1}\right]}\\ &=S_{1,2}{\left(-a\right)}+S_{1,2}{\left(-b\right)}-S_{1,2}{\left(\frac{b-a}{1+b}\right)}-\frac12\left[\Lambda_{3}{\left(-\frac{b}{a}\right)}-\Lambda_{3}{\left(-\frac{\left(1+a\right)b}{a\left(1+b\right)}\right)}\right]\\ &~~~~~-\ln{\left(\left|\frac{a}{b}\right|\right)}\left[\Lambda_{2}{\left(-\frac{b}{a}\right)}-\Lambda_{2}{\left(-\frac{\left(1+a\right)b}{a\left(1+b\right)}\right)}\right]-\frac12\ln^{2}{\left(\left|\frac{a}{b}\right|\right)}\ln{\left(1+b\right)}.\tag{10}\\ \end{align}$$

Result $(10)$ provides us with a compact formula for $\mathcal{I}{\left(a,b\right)}$ in terms of logarithms, Kummer polylogarithms, and Nielsen generalized polylogarithms under the parameter conditions, $-1<a<b\land a\neq0\land b\neq0$.


Appendix:

The natural logarithm can be defined for positive real argument via the integral representation,

$$\ln{\left(z\right)}:=\int_{1}^{z}\frac{\mathrm{d}t}{t};~~~\small{z\in\mathbb{R}^{+}}.$$

For positive integer order $n\in\mathbb{N}$, the polylogarithm $\operatorname{Li}_{n}{\left(z\right)}$ may be defined for real argument $z$ recursively via,

$$\begin{cases} &\operatorname{Li}_{1}{\left(z\right)}=-\ln{\left(1-z\right)};~~~\small{z<1},\\ &\operatorname{Li}_{n}{\left(z\right)}=\int_{0}^{z}\frac{\operatorname{Li}_{n-1}{\left(t\right)}}{t}\,\mathrm{d}t;~~~\small{n\ge2\land z\le1}.\\ \end{cases}$$

An auxiliary function related to the polylogarithm is Kummer's polylogarithm. For positive integer $n\in\mathbb{N}$ and real argument $z\in\mathbb{R}$, we define

$$\Lambda_{n+1}{\left(z\right)}:=\int_{0}^{z}\frac{\ln^{n}{\left(\left|t\right|\right)}}{1+t}\,\mathrm{d}t$$

For positive integer indices $\left(n,p\right)\in\mathbb{N}^{2}$ and real argument $z\in(-\infty,1]$, define the Nielsen generalized polylogarithm via the integral representation,

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

In particular,

$$S_{1,p}{\left(z\right)}:=\frac{\left(-1\right)^{p}}{p!}\int_{0}^{1}\frac{\ln^{p}{\left(1-zt\right)}}{t}\,\mathrm{d}t=\frac{\left(-1\right)^{p}}{p!}\int_{0}^{z}\frac{\ln^{p}{\left(1-x\right)}}{x}\,\mathrm{d}x.$$


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  • $\begingroup$ This is very interesting. I wonder is it known-- for generic values of p -- how to reduce the Nielsen generalized and the Kummer's polylogarithms to polylogarithms and logarithms only? $\endgroup$ – Przemo Aug 17 '16 at 17:02
  • $\begingroup$ @Przemo Sorry for only now getting back you. Yes, the Kummer functions of any order can be decomposed into a linear combination of polylogs. The formula is pretty easy to work out, and can also be found on the function's Wikipedia page. Same goes for the Nielsen polylog for arbitrary $p$, ASSUMING that $n=1$. If however $n>1$, things become more dicey. $\endgroup$ – David H Sep 5 '16 at 11:33

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