Simplify $\int_{0}^{1}\ln(x-a)\ln x\,\mathrm{d}x$, for $a<0$ 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.
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$\ds{\int_{0}^{1}\ln\pars{x - a}\ln\pars{x}\,\dd x:\ {\large ?}\,,\qquad a < 0}$.

\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}

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