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The $p$-logarithm is defined for $|z|<1$ by

$$\text{Li}_p(z)=\sum_{n=1}^\infty\frac{z^n}{n^p}$$

and defined elsewhere in $\mathbb C$ by analytic continuation, though it may be multi-valued, depending on the path of continuation.

For an integer $p<0$, the $p$-logarithm is a rational function, whose inverse is an algebraic function, which is multi-valued. For example,

$$\text{Li}_{-1}(z)=\frac{z}{(1-z)^2},\qquad\text{Li}_{-1}^{-1}(z)=\frac{2z+1\pm\sqrt{4z+1}}{2z}.$$

For $p=0$ we have

$$\text{Li}_0(z)=\frac{z}{1-z},\qquad\text{Li}_0^{-1}(z)=\frac{z}{1+z}=-\text{Li}_0(-z),$$

and for $p=1$ we have a form of the ordinary logarithm, whose inverse is an entire function:

$$\text{Li}_1(z)=-\ln(1-z),\qquad\text{Li}_1^{-1}(z)=1-e^{-z}.$$

So, for $p<0$ the $p$-logarithm is single-valued while its inverse is multi-valued, for $p=0$ the $p$-logarithm and its inverse are both single-valued, and for some $p>0$ the $p$-logarithm is multi-valued while its inverse is single-valued.

This isn't much reason to expect the "pattern" to hold for all integers $p>0$. Nevertheless, I'd like to know if it does hold. There's a lot of information on polylogarithms on Wikipedia, but I didn't see any obvious answer to this simple question: Is $\text{Li}_p^{-1}$ single-valued? Equivalently, is $\text{Li}_p$ injective?

("Injective" is usually defined for single-valued functions. Here it means, if $z_1\neq z_2$, then the sets $\text{Li}_p(z_1)$ and $\text{Li}_p(z_2)$ don't intersect.)

We may focus on the case $p=2$.


Let's consider the values of $\text{Li}_2(z)$ near the branch point $z=1$, or equivalently of $\text{Li}_2(1-z)$ near $z=0$, noting that $\text{Li}_2(1)=\pi^2/6$:

$$\frac{\pi^2}{6}-\text{Li}_2(1-z)=\text{Li}_2(z)+\ln(z)\ln(1-z)$$

$$=\left(z+\frac{z^2}{4}+\frac{z^3}{9}+\cdots\right)-\ln(z)\left(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots\right)$$

(throw away the higher powers of $z$ since $z\approx0$)

$$\approx z-z\ln(z)=z\big(1-\ln(z)\big)$$

(use $\ln(z)\approx-\infty$)

$$\approx z\big({-\ln(z)}\big)=-z\ln(z).$$

Now let $z=e^u$ where $\text{Re}(u)\approx-\infty$, so that $\ln(z)=u$ (as one of the possible values), and

$$\frac{\pi^2}{6}-\text{Li}_2(1-e^u)\approx-ue^u.$$

This latter function is certainly not injective. For any $a\in\mathbb R$, there is a sequence of points $u_k\in\mathbb C$ with $\text{Re}(u_k)<a$, along a curve which is almost a vertical line, all having the same value of $u_ke^{u_k}$, and all having different values of $z_k=e^{u_k}$. I can see this by graphing and analyzing the level curves of $|ue^u|$ and $\arg(ue^u)$: if $u=x+yi$, these curves have the form $y=\pm\sqrt{C^2e^{-2x}-x^2}$ and $x=y\cot(D-y)$, respectively.

Can we use this to prove that $\text{Li}_2$ is not injective? Or is something lost in these approximations? What other methods can we use?

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    $\begingroup$ Related: math.stackexchange.com/questions/73515/…. Alternatively one may see that $Li_2(x)$ is strictly increasing, and hence injective, by considering its derivative. This extends easily to all positive integers $p$. $\endgroup$ – player3236 Dec 4 '20 at 3:38
  • $\begingroup$ I meant the complex dilogarithm, not the real dilogarithm. $\endgroup$ – mr_e_man Dec 4 '20 at 4:04
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    $\begingroup$ Numerical computation suggests that $$\operatorname{Li}_2(z)=-\int_{0}^{z}\frac{\log(1-\xi)}{\xi}\,\mathrm{d}\xi=\int_{1}^{\infty}\left(\frac{1}{x-z}-\frac{1}{x}\right)\log x\,\mathrm{d}x$$ defined for $z\in\mathbb{C}\setminus(1,\infty)$ is indeed injective. However, its inverse function seems not extend to an entire function, since the images of some other branches of $\operatorname{Li}_2$ overlap. $\endgroup$ – Sangchul Lee Dec 4 '20 at 9:14
  • $\begingroup$ $Li_2(z) = \frac{-\log(1-z)}{z}, Li_2(1-z)' = \frac{\log(z)}{1-z}$ gives that $Li_2(1-e^{2i\pi} z)'=Li_2(1- z)'+\frac{2i\pi }{1-z}$ so $Li_2(1-e^{2i\pi} z)=Li_2(1- z)-2i\pi \log(1-z)$ $\endgroup$ – reuns Dec 7 '20 at 4:52
  • $\begingroup$ @reuns - What's your point? It looks like you're showing that $\text{Li}_2$ is multi-valued. We already knew that. $\endgroup$ – mr_e_man Dec 7 '20 at 20:37
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$\text{Li}_2'(z)=\frac{-\log(1-z)}{z}$ doesn't vanish so any curve $\gamma:0\to ?$ along which (the continuation of) $\text{Li}_2(z)$ is analytic gives a curve $\text{Li}_2^{-1}(\gamma):0\to ?$ along which $\text{Li}_2^{-1}$ is analytic.

Next $\text{Li}_2'(z)=\frac{-\log(1-z)}{z}$ shows that $$\text{Li}_2(1-z)=\frac{\pi^2}{6}-\text{Li}_2(z)-\log(z)\log(1-z)$$ for $z\in (0,1)$ and it is meant the principal branch of each term.

Continuing analytically by starting at $1/2$ and rotating one time around $z=0$ we get a new branch

$$\text{Li}_2(1-z)^{new\ branch}=\frac{\pi^2}{6}-\text{Li}_2(z)-(\log(z)+2i\pi)\log(1-z)$$ analytic for $\Im(z)>0,|z|<1$.

This branch has a zero near $0.08+0.18 i$, and since $\text{Li}_2(0)=0$ too it means that $\text{Li}_2^{-1}$ is multivalued, ie. there are some curves $\Gamma:0\to 0$ along which $\text{Li}_2^{-1}$ is analytic but not the same at its departure and arrival.

enter image description here

(it is the argument principle which proves that there is a zero on this plot)

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  • $\begingroup$ The argument principle requires a continuous curve, not a sequence of hundreds of points/pixels. :) Of course we can get around this by appealing to continuity... $\endgroup$ – mr_e_man Dec 8 '20 at 3:08

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