# Does the dilogarithm function (which is multi-valued) have a single-valued inverse?

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 $$\text{Re}(u_k), along a curve which is almost a vertical line, all having the same value of $$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?

• 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 $p$. – player3236 Dec 4 '20 at 3:38
• I meant the complex dilogarithm, not the real dilogarithm. – mr_e_man Dec 4 '20 at 4:04
• 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. – Sangchul Lee Dec 4 '20 at 9:14
• $Li_2(z) = \frac{-\log(1-z)}{z}, Li_2(1-z)' = \frac{\log(z)}{1-z}$ gives that $Li_2(1-e^{2i\pi} z)'=Li_2(1- z)'+\frac{2i\pi }{1-z}$ so $Li_2(1-e^{2i\pi} z)=Li_2(1- z)-2i\pi \log(1-z)$ – reuns Dec 7 '20 at 4:52
• @reuns - What's your point? It looks like you're showing that $\text{Li}_2$ is multi-valued. We already knew that. – mr_e_man Dec 7 '20 at 20:37

$$\text{Li}_2'(z)=\frac{-\log(1-z)}{z}$$ 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.

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

• 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... – mr_e_man Dec 8 '20 at 3:08