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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

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

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