Limit of dilogarithm function

Let $\zeta$ denote the Riemann zeta function and $\mathrm{Li}_s$ the polylogarithm of order $s$ . Let $s \geq 2$ . Evaluate the limit:

$\ell = \lim_{x \rightarrow 1^-} \left ( \zeta(s) - \mathrm{Li}_s(x) \right ) \ln (1-x)$

Solution

We distinguish cases.

For $s>2$ we have:
$\begin{align*} \ell &= \lim_{x \rightarrow 1^-} \left ( \zeta(s) - \mathrm{Li}_s(x) \right ) \ln (1-x) \\ &=\lim_{x\rightarrow 1^-} \frac{\zeta(s)- \mathrm{Li}_s(x)}{1-x} \cdot \left ( 1-x \right ) \ln (1-x) \\ &= \left ( \lim_{x\rightarrow 1^-} \frac{\zeta(s)- \mathrm{Li}_s(x)}{1-x} \right ) \cdot \left ( \lim_{x\rightarrow 1^-} (1-x) \ln (1-x) \right ) \\ &=\mathrm{Li}'_s(1) \cdot 0 \\ &=0 \end{align*}$
The dilogarithm function satisfies the following identity
$\mathrm{Li}_2(x) + \mathrm{Li}_2\left ( 1-x \right ) = \zeta(2) - \ln x \ln (1-x) \quad , \quad x \in (0, 1)$

Hence for $s=2$ we have:

$\begin{align*} \ell &= \lim_{x \rightarrow 1^-} \left ( \zeta(2) - \mathrm{Li}_2(x) \right ) \ln (1-x) \\ &=\lim_{x\rightarrow 1^-} \left ( \frac{\pi^2}{6} - \mathrm{Li}_2(x) \right ) \ln (1-x) \\ &=\lim_{x\rightarrow 1^-} \left ( \mathrm{Li}_2(1-x) + \ln x \ln (1-x) \right ) \ln (1-x) \\ &= \lim_{x\rightarrow 0^+} \left ( \mathrm{Li}_2(x) + \ln x \ln (1-x) \right ) \ln x =0 \end{align*}$

since

$\lim_{x\rightarrow 0^+} \mathrm{Li}_2(x) \ln x = \lim_{x \rightarrow 0^+} \frac{\mathrm{Li}_2(x)}{x} \cdot x \ln x = 1 \cdot 0 = 0$