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Define the function $$Li_s(z)=\sum_{k=1}^\infty \frac{z^k}{k^s}$$ for |z|<1. Let's focus on $s=2$. It can be extended to a holomorphic function on $\mathbb C \setminus [1,\infty)$ $$Li_2(z)= -\int_0^z \frac{du}u \, \log(1-u)$$ by the identity $$ \frac{d}{dz} Li_2(z) = -\frac 1z \log (1-z)$$ and by taking a branch for the logarithm.

How do we prove that its discontinuity through the branch cut is $2\pi i \log |z|$ ? if so, is the double discontinuity (i.e. the discontinuity of the discontinuity at the same branch cut) going to be zero?

EDIT 1: perhaps what I wanted to understand better is the following: define $$\phi (w, \bar{w})= \frac{1}{w - \bar{w}} \left[2 {\rm Li}_{2} \left(w \right)- 2 {\rm Li}_{2} \left(\bar{w} \right) + \ln (w \bar{w}) \ln \left( \frac{1 - w}{1 - \bar w} \right)\right]$$ Then its double discontinuity across the cuts $-\infty < w \le 0$ and $1\le \bar w<\infty$ should be $$ {\rm Disc}_{w} {\rm Disc}_{\bar w} \phi (w, \bar w) = {\pi^2 \over w - \bar w}$$

jj_p
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  • What is a double discontinuity? To what are you referring? The complex logarithm $\log(1-z)$ is discontinuous at the branch cut, however the cut is selected. – Mark Viola Apr 13 '17 at 16:24

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Using the definition, if $x>1$, $$ Li(x+i\epsilon)-Li(x-i\epsilon) = \int_{1+i\epsilon}^{x+i\epsilon} -\frac{1}{u} \log{(1-u)} \, du - \int_{1-i\epsilon}^{x-i\epsilon} -\frac{1}{u} \log{(1-u)} \, du + o(1), $$ where the $o(1)$ term comes from the integral from the integral over a small semicircle around the branch point $1$: the singularity is logarithmic, so the integral over this is $O(\epsilon\log{\epsilon})=o(1)$. We can change variables ($v=u-i\epsilon$ and $v=u+i\epsilon$) to write both remaining integrals under the same integral sign: $$ Li(x+i\epsilon)-Li(x-i\epsilon) = \int_{1}^{x} \left( -\frac{1}{v+i\epsilon} \log{(1-v-i\epsilon)} + \frac{1}{v-i\epsilon} \log{(1-v+i\epsilon)} \right) dv + o(1) $$ We make another approximation: since $v$ is not near zero, $$\frac{1}{v \pm i\epsilon} = \frac{1}{v} + O(\epsilon), $$ and since the rest of the integrand is bounded, we can absorb this into the $o(1)$ and $$ Li(x+i\epsilon)-Li(x-i\epsilon) = \int_{1}^{x} \frac{1}{v}\left( \log{(1-v+i\epsilon)} -\log{(1-v-i\epsilon)} \right) dv + o(1) $$ Here we see that the discontinuity is related to the discontinuity in the logarithm. $1-v<0$, and $$\log{(-r+i\epsilon)}-\log{(-r-i\epsilon)} = \int_{|z|=r} dz/z + O(\epsilon) = 2\pi i + O(\epsilon) $$ by the integral formula for the logarithm, so the remaining integral is $$ Li(x+i\epsilon)-Li(x-i\epsilon) = \int_{1}^{x} \frac{2\pi i}{v} \, dv + o(1) = 2\pi i \log{x} + o(1). $$

$\log{|z|}$ is only discontinuous at $z=0$. There's only one possible level of discontinuity, since the discontinuity is the difference of two continuous functions.

Chappers
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