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How can one derive the following identity, found here, relating the polylogarithm functions to Bernoulli polynomials?

$$\operatorname{Li}_n(z)+(-1)^n\operatorname{Li}_n(1/z)=-\frac{(2\pi i)^n}{n!}B_n\!\left(\frac12+\frac{\ln(-z)}{2\pi i}\right).$$

I'm curious because this formula turned out to be useful for me, but I have no idea where it comes from.

Edit: If there is a complicated proof of the above formula, but also a simpler procedure for deriving the formula for specific $n$, I would be interested in seeing both of these. I'm also only interested in the case where $z$ is real and negative.

Gary
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WillG
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    I would put $z = - {\rm e}^{2\pi {\rm i}w}$ with $ - \frac{1}{2} < w < \frac{1}{2}$, use the series expansion of $\operatorname{Li}_n$ and compare the left-hand side with $(24.8.3)$. The latter follows from $(24.2.3)$ via Cauchy's formula and the residue theorem. – Gary May 17 '23 at 04:03
  • @Gary Does (24.8.3) also hold for $x\in\mathbb C$? I would like $z$ (in the formula in my post) to be real (and negative), which will require using complex $x$ in (24.8.3). Specifically, on the right side we have $B_n(\frac12+ai)$, with $a\in\mathbb R$. – WillG May 17 '23 at 14:44
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    Both sides are analytic on the plane with a cut from $0$ to positive infinity. You can use analytic continuation to extended the identity to this larger domain. – Gary May 17 '23 at 22:29

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