# Polylogarithms: How to prove the asympotic expression $z \le \mathrm{Li}_{s}(z) \le z(1+2z 2^{-s}), \;z<-1, \;s \gg \log_2|z|$

For $|z| < 1, s > 0$ the polylogarithm has the power series $$\mathrm{Li}_{s}(z) = \sum_{k=1}^\infty {z^k \over k^s} = z + {z^2 \over 2^s} + {z^3 \over 3^s} + \cdots = z\left(1+ {z \over 2^s} + {z^2 \over 3^s} + \cdots\right) \quad(*)$$

I seek an asymptotic expression for $\mathrm{Li}_{s}(z)$ for real $z< -1$ and large real $s$.

Wikipedia gives the limit $$\lim_{\mathrm{Re}(s) \rightarrow \infty} \mathrm{Li}_s(z) = z$$ It is easy to show that for $z < 0$ we have $\mathrm{Li}_s(z) \ge z$, this follows from the integral representation (https://dlmf.nist.gov/25.12.E11)

$$\mathrm{Li}_{s}(z)=\frac{z}{\Gamma\left(s\right)}\int_{0}^{\infty} \frac{x^{s-1}}{e^{x}-z}dx \ge \frac{z}{\Gamma\left(s\right)}\int_{0}^{\infty} \frac{x^{s-1}}{e^{x}}dx = z$$ For fixed $z$ the series $(*)$ suggest the asymptotics

$$z \le \mathrm{Li}_{s}(z) \le z(1+2z 2^{-s}), \quad z<-1, \;s \gg \log_2|z|$$

Numerical experiments with Maple confirm this conjecture, but how can it be proved? Or if it is wrong, what is a correct tight upper bound?

I happen to recently have wondered the same kind of thing. Although I don't have the answer to your exact question, I do have asymptotics for the polylogarithm (I only lack some uniformity in $$z$$, but we probably can extract explicit bounds from the proof). The idea is to use Laplace's method to the integral representation of the polylogarithm (plus of course Stirling's formula for the Gamma function, but this is actually the same proof).

To simplify the notations, let us set $$s' = s-1$$.

We want an asymptotic for the integral $$\int_0^{+\infty} \frac{x^{s'}}{e^x-z}dx$$. We write it as $$\int_0^{+\infty}\frac{x^{s'}}{e^x-z}dx = \int_0^{+\infty} \frac{e^{s'\ln(x) - x}}{1-ze^{-x}}dx$$ and to force the appearance of a phase of the form $$s'\phi(y)$$ we make the change of variables $$x = s' y$$: $$\int_0^{+\infty}\frac{x^{s'}}{e^x-z}dx = \int_0^{+\infty} \frac{e^{s'\ln(s')+s'(\ln(y) - y)}}{1-ze^{-s'y}}s'dy$$ so that with $$\phi(y) = \ln(y)-y$$ and $$u_{s,z}(y) = 1/(1-ze^{-s'y})$$, we have $$\int_0^{+\infty}\frac{x^{s'}}{e^x-z}dx = s'^{s'+1}\int_0^{+\infty} e^{s'\phi(y)}u_{s,z}(y)dy.$$

Now, for a function $$\phi$$ with a unique maximum at $$x_0$$ between $$a$$ and $$b$$, Laplace's method gives us the following asymptotics : $$\int_a^b e^{\lambda \phi(x)}u(x)dx = e^{\lambda\phi(x_0)}\sqrt{\frac{2\pi}{\lambda|\phi''(x_0)|}}u(x_0)(1+R(s))$$ with $$|R(s)|< \frac Cs |u|_{L^\infty}$$ where the $$L^\infty$$ norm is taken in a small complex neighborhood of $$[a,b]$$.*

We apply this to our $$\phi$$, with its maximum at $$x_0=1$$, and with $$\phi(x_0) = -1$$, $$\phi''(x_0) = -1$$, and we get $$\int_0^{+\infty}\frac{x^{s'}}{e^x-z}dx = s'^{s'+1}e^{-s'}\sqrt{\frac{2\pi}{s'}}\left(\frac{1}{1-ze^{-s'}} + \mathcal O(\frac1{s'})\right)$$ that is $$\int_0^{+\infty}\frac{x^{s'}}{e^x-z}dx = \sqrt{2\pi s'}\left(\frac{s'}e\right)^{s'}\left(1 + \mathcal O(\frac1{s'})\right).$$

Combining this with Stirling's approximation (which is this formula with $$z=0$$), and the integral formula for the polylogarithm, we have the asymptotics $$\text{Li}_s(z) = z\left(1+\mathcal O(\frac 1s)\right).$$

Now, regarding your conjecture that the error is at most $$2z^2/2^s$$, I don't exactly know, but we can get the second term of the asymptotics. For this, we study the incomplete polylogarithm (I don't know if the terminology is standard), defined by $$\text{Li}_{s,N}(z) = \sum_{k> N} \frac{z^k}{k^s}$$, and we have the integral formula $$\text{Li}_{s,N}(z) = \frac{z^{N+1}}{\Gamma(s)}\int_0^{+\infty}\frac{x^{s-1}e^{-Nx}}{e^x-z}dx$$ and the same proof as for the (complete) polylogarithm gives us $$\text{Li}_{s,N}(z) = \frac{z^{N+1}}{(N+1)^s}\left(1+\mathcal O(\frac1s)\right).$$

Since $$\text{Li}_s(z) = z + \text{Li}_{s,1}(z)$$, this proves that the second term is $$z^2/2^s$$.

All of this looks like something that should have been done quite a while ago in the literature, but I failed to find it (I blame my searching skills). The advantage of the proof I gave is that it works for any $$z$$ in the Riemann surface of the polylograithm, which was what I wanted when I started to look at this. And sorry, but I leave it to you to see what explicit bounds we can get from this proof, the mere idea of doing it is discouraging to me.

*Here I take advantage of the analycity of $$u$$ here to simplify the notations, but we could get something if $$u$$ was only $$C^\infty$$, for instance.