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Let $a<0$. The following integral:

$$\int_{0}^{1}\ln(x-a)\ln x\,\mathrm{d}x$$

can be computed to yield the result:

$$2+\frac{\pi^{2}}{6}a-\ln\left(-a\right)-\left(1-a\right)\ln\left(1-\frac{1}{a}\right)-\frac{1}{2}a\left[\ln^{2}\left(1-a\right)-\ln^{2}\left(-a\right)\right]-a\mathrm{Li}_{2}\left(\frac{a}{a-1}\right)$$

I believe this can be simplified. See for example, this answer, where a similar integral yields a shorter expression. I've been struggling with this but I am not conversant enough with the properties of the dilogarithm function.

Community
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a06e
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2 Answers2

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1}\ln\pars{x - a}\ln\pars{x}\,\dd x:\ {\large ?}\,,\qquad a < 0}$.

\begin{align}&\color{#c00000}{\int_{0}^{1}\ln\pars{x - a}\ln\pars{x}\,\dd x} =\left.\bracks{x\ln\pars{x} - x}\ln\pars{x - a} \vphantom{\Large A}\right\vert_{x\ =\ 0}^{x\ =\ 1} -\int_{0}^{1}\bracks{x\ln\pars{x} - x}\,{1 \over x - a}\,\dd x \\[3mm]&=-\ln\pars{1 - a} +\int_{0}^{1}\bracks{{x \over x - a} - \ln\pars{x} - a\,{\ln\pars{x} \over x - a}}\,\dd x \\[3mm]&=-\ln\pars{1 - a} + \bracks{1 + a\ln\pars{1 - a} - a\ln\pars{-a} -\pars{-1}} - a\color{#00f}{\int_{0}^{1}{\ln\pars{x} \over x - a}\,\dd x} \qquad\qquad\pars{1} \end{align}

\begin{align}&\color{#00f}{\int_{0}^{1}{\ln\pars{x} \over x - a}\,\dd x} =-\int_{0}^{1/a}{\ln\pars{ax} \over 1 - x}\,\dd x =\left.\ln\pars{1 - x}\ln\pars{ax}\vphantom{\Large A} \right\vert_{x\ =\ 0}^{x\ =\ 1/a} - \int_{0}^{1/a}{\ln\pars{1 - x} \over x}\,\dd x \\[3mm]&=\int_{0}^{1/a}{{\rm Li}_{1}\pars{x} \over x}\,\dd x =\int_{0}^{1/a}\totald{{\rm Li}_{2}\pars{x}}{x}\,\dd x =\color{#00f}{{\rm Li}_{2}\pars{1 \over a}} \end{align} where $\ds{{\rm Li_{s}}\pars{z}}$ are PolyLogarithm Functions and we used well known properties of them as explained in the above link.

With expression $\pars{1}$: \begin{align}&\color{#66f}{\large\int_{0}^{1}\ln\pars{x - a}\ln\pars{x}\,\dd x} \\[3mm]&=\color{#66f}{\large 2 -\pars{1 - a}\ln\pars{1 - a} - a\ln\pars{-a} -a\ {\rm Li}_{2}\pars{1 \over a}}\,,\qquad a < 0 \end{align}

Felix Marin
  • 89,464
2

Note that:

$$\text{Li}_2(z)=-\text{Li}_2(\frac{z}{z-1})-\frac{1}{2}\ln^2(1-z)$$ By identity #$9$ here: http://functions.wolfram.com/ZetaFunctionsan...

Now multiplying both sides by $z$ and adding the right most expression to the left side gives:

$$\frac{z}{2}\ln^2(1-z)+z\text{Li}_2(z)=-z\text{Li}_2(\frac{z}{z-1})$$

Now setting $z\rightarrow a$ and substituting this expression in for the dilogarithm appearing in your equality gives:

$$\int_{0}^1\ln(x-a)\ln(x) dx$$ $$=2+\frac{\pi^{2}}{6}a-\ln\left(-a\right)-\left(1-a\right)\ln\left(1-\frac{1}{a}\right)-\frac{1}{2}a\left[\ln^{2}\left(1-a\right)-\ln^{2}\left(-a\right)\right]+\left(\frac{a}{2}\ln^2(1-a)+a\text{Li}_2(a)\right)$$

Now if we multiply out the factor of $-\frac{1}{2}a$ into $[\ln^{2}\left(1-a\right)-\ln^{2}\left(-a\right)]$ we can rewrite the result:

$$2+\frac{\pi^{2}}{6}a-\ln\left(-a\right)-\left(1-a\right)\ln\left(1-\frac{1}{a}\right)-\frac{a}{2}\ln^{2}(1-a)+\frac{a}{2}\ln^{2}(-a)+\left(\frac{a}{2}\ln^2(1-a)+a\text{Li}_2(a)\right)$$ And canceling like terms gives: $$2+\frac{\pi^{2}}{6}a-\ln\left(-a\right)-\left(1-a\right)\ln\left(1-\frac{1}{a}\right)+\frac{a}{2}\ln^2(-a)+a\text{Li}_2(a)$$

Now if we consider the expression: $$-\left(1-a\right)\ln\left(1-\frac{1}{a}\right)=(a-1)\ln\left(1-\frac{1}{a}\right)=a\ln\left(1-\frac{1}{a}\right)-\ln\left(1-\frac{1}{a}\right)$$ $$=a\ln\left(1-\frac{1}{a}\right)-\ln\left(\frac{a-1}{a}\right)=a\ln\left(1-\frac{1}{a}\right)-\ln\left(\frac{1-a}{-a}\right)$$ $$=a\ln\left(1-\frac{1}{a}\right)-\big{(}\ln(1-a)-\ln(-a)\big{)}=a\ln\left(1-\frac{1}{a}\right)-\ln(1-a)+\ln(-a)$$

And substitute that in for where the expression occurred in our original formula, after canceling out a common factor of $\ln(-a)$ we get that your integral is equal to:

$$2+\frac{\pi^{2}}{6}a-\ln(1-a)+a\ln\left(1-\frac{1}{a}\right)+\frac{a}{2}\ln^2(-a)+a\text{Li}_2(a)$$ $$=2-\ln(1-a)+a\Big{(}\frac{\pi^2}{6}+\ln(1-\frac{1}{a})+\frac{1}{2}\ln^2(-a)+\text{Li}_2(a)\Big{)}$$ $$=2-\ln(1-a)+a\Big{(}\ln(1-\frac{1}{a})-\text{Li}_2(\frac{1}{a})\big{)}$$ Where the last identity follows from #$3$ here: http://functions.wolfram.com/ZetaFunctionsan...

Thus we can rewrite: $$\int_{0}^1\ln(x-a)\ln(a) dx=2-\ln(1-a)+a\ln(1-\frac{1}{a})-a\text{Li}_2(\frac{1}{a})$$

Ethan Splaver
  • 10,613