Polylogarithm

hbghlyj

Li的定义
$\def\Li{{\rm Li}} \Li_{-n}(z) = \sum_{k=0}^n k! \;  S(n+1,k+1) \left( \dfrac z{1-z} \right)^{k+1}$
where n is a non-negative integer, and S(n,k) denote the Stirling's number of the second kind.
证明:对n使用数学归纳法.当n=0时$S(1,1)=1,\dfrac z{1-z}=\Li_0(z)$所以成立.
假设公式对n成立,利用第二类Striling数的递推公式$S(n+2,k+1)=(k+1)\;S(n+1,k+1)+S(n+1,k)$,
$\sum_{k=0}^{n+1} k! \;  S(n+2,k+1) \left( \dfrac z{1-z} \right)^{k+1}$
$=\sum_{k=0}^{n+1} (k+1)! \;  S(n+1,k+1) \left( \dfrac z{1-z} \right)^{k+1}+\sum_{k=0}^{n+1} k!\;  S(n+1,k) \left( \dfrac z{1-z} \right)^{k+1}$
(注意到S(n+1,0)=S(n+1,n+2)=0)
$=\sum_{k=0}^{n} (k+1)! \;  S(n+1,k+1) \left( \dfrac z{1-z} \right)^{k+1}+\sum_{k=0}^{n}(k+1)!\;  S(n+1,k+1) \left( \dfrac z{1-z} \right)^{k+2}$
$=\sum_{k=0}^n (k+1)! \;  S(n+1,k+1) \left( \dfrac z{1-z} \right)^{k}\dfrac{z}{(1-z)^2}$
$=z \; \dfrac {\partial}{\partial z}\Li_{-n}(z)$
$=\Li_{-n-1}(z)$.
命题对n+1也成立,由归纳原理得证.

hbghlyj

$$\Gamma(s)\Li_s(z)= \int_0^\infty\frac{x^{s-1}}{\frac{e^x}{z}-1} dx\tag1\label1$$
证明$$\int_0^\infty\frac{x^{s-1}}{\frac{e^x}{z}-1} dx=\int_0^\infty x^{s-1}\sum_{l=1}^{\infty}(ze^{-x})^ldx=\sum_{l=1}^{\infty}\frac{z^l}{l^s}\;\int_0^\infty (lx)^{s-1}e^{-lx}d(lx)=\Li_s(z)\Gamma(s)$$

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在\eqref{1}中取$z=1$就得$\Gamma(s) \zeta(s) =\int_0^\infty\dfrac{x^{s-1}}{e^x-1} dx.$

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将\eqref{1}对$z$求导,得\begin{align*}
\frac{\mathrm d}{\mathrm dz}\Gamma(s)\Li_s(z)&=\frac {\mathrm d}{\mathrm dz}\int_0^{\infty}t^{s-1}\frac{z}{e^t-z}\mathrm dt\\
\Gamma(s)\frac{\Li_{s-1}(z)}{z}&=\int_0^{\infty}t^{s-1}\frac{(e^t-z)+z}{(e^t-z)^2}\mathrm dt\\
\Gamma(s)\frac{\Li_{s-1}(z)}{z}&=\int_0^{\infty}\frac{t^{s-1}}{e^t-z}\mathrm dt+z\int_0^{\infty}\frac{t^{s-1}}{(e^t-z)^2}\mathrm dt\\
\Gamma(s)\frac{\Li_{s-1}(z)}{z}&=\Gamma(s)\frac{\Li_{s}(z)}{z}+z\int_0^{\infty}\frac{t^{s-1}}{(e^t-z)^2}\mathrm dt\\
\therefore~\int_0^{\infty}\frac{t^{s-1}}{(e^t-z)^2}\mathrm dt&=\Gamma(s)\left[\frac{\Li_{s-1}(z)}{z^2}-\frac{\Li_{s}(z)}{z^2}\right]
\end{align*}