I am trying to prove the following property of polylogarithm. $$\lim\limits_{Re(s)\rightarrow -\infty} Li_s(e^{w}) = \Gamma(1-s)(-w)^{s-1} \text{ for } -\pi < \Im(w) < \pi.$$ As per Wikipedia's "Polylogarithm" entry, the above limit follows from the relationship of polylogarithm with Hurwitz-Zeta function as given below. $$\displaystyle \operatorname {Li}_{s}(z)={\Gamma (1-s) \over (2\pi )^{1-s}}\left[i^{1-s}~\zeta \left(1-s,~{\frac {1}{2}}+{\ln(-z) \over {2\pi i}}\right)+i^{s-1}~\zeta \left(1-s,~{\frac {1}{2}}-{\ln(-z) \over {2\pi i}}\right)\right]$$ How do I observe this? Also is the above limit uniform for all $w$ by any chance?
$\begingroup$
$\endgroup$
4
-
1$\begingroup$ $\lim_{s\to +\infty} a^{s} \zeta(s,a) = 1$ and your first line is an equivalent $\sim$ not a limit, and did you mean $Li_s(e^{-w})$ instead of $Li_s(e^w)$ $\endgroup$– reunsCommented Jun 6, 2020 at 10:15
-
$\begingroup$ Yes, I too understand it as the behaviour of $Li_s(e^w)$ for large enough $-Re(s)$ value. The wikipedia entry states the formula (1) as I had quoted here. It says $Li_s(e^w)$. I will try with your hint. $\endgroup$– user166305Commented Jun 6, 2020 at 10:46
-
$\begingroup$ By uniform limit, what I wanted to know is: Suppose both $s$ and $w$ are variables, then in the asymptotic $Li_s(e^w) \sim \Gamma(1-s)(-w)^{s-1}$ for $-\Re(s) \geq \sigma_0$, is the choice of $\sigma_0$ independent of $w$? $\endgroup$– user166305Commented Jun 6, 2020 at 11:17
-
$\begingroup$ @reuns: could you give your comment on my last query? $\endgroup$– user166305Commented Jun 7, 2020 at 16:00
Add a comment
|