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I just noticed that $B_{2n}(\frac12+ix)\in\mathbb R$, where:

$x\in\mathbb R$, $n\in\mathbb N$, and $B_n(x)$ is the $n$th Bernoulli Polynomial.

Why? Is there a simple, slick proof? Does it follow from one of the common identities for the Bernoulli polynomials?

I noticed this because one identity for the polylogarithm functions involves $B_n(\frac12+ix)$ in a context where $x\in\mathbb R$ must yield a real number:

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

Letting $z=-e^{-2\pi x}$ and taking $n$ even shows that $B_{2n}(\frac12+ix)\in\mathbb R$. So one approach would be to prove the above identity, but this seems convoluted, requires proving things about the polylogarithm functions, and doesn't seem to offer any intuition for what's going on.

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  • $\begingroup$ Have you tried proving the expression you have is equal to its complex conjugate? $\endgroup$
    – Integral fan
    Commented Jun 30, 2022 at 19:49
  • $\begingroup$ Expand the generating function for the Bernoulli Polynomials. $\endgroup$
    – DecarbonatedOdes
    Commented Jun 30, 2022 at 19:52
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    $\begingroup$ If the identity holds true for complex numbers, let $f(t)=te^{(1/2+ix)t}/(e^t-1)$. You can show that $f(t)+f(-t)$ is real? $\endgroup$
    – mich95
    Commented Jun 30, 2022 at 19:52
  • $\begingroup$ @Integralfan The first identity (in the Bernoulli polynomial wiki) under "Symmetries" implies that $B_{2n}(1/2+ix)=B_{2n}(1/2-ix)$. But I don't think this does the trick? $\endgroup$
    – WillG
    Commented Jun 30, 2022 at 19:59
  • $\begingroup$ If you let $f(t)=te^{(1/2+ix)t}/(e^t-1)$. You can verify that $f(t)+f(-t)=te^{t/2}\cos(tx)/(e^t-1)$. Now, add up the power series of$ f(t) $and$ f(-t).$ You're left with the coefficients of the even powers, which is what you want (assuming that t is real) $\endgroup$
    – mich95
    Commented Jun 30, 2022 at 20:22

1 Answer 1

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For any polynomial $p(z)$, we have $p(z^*)=(p(z))^*$. For the Bernoulli polynomials, $B_n(x)=(-1)^nB_n(1-x).$ Thus, for even $n,$ we have $B_n\left(\frac12+ix\right)=B_n\left(\frac12-ix\right)$. With $x\in\mathbb R$, letting $z=\frac12+ix$, we have $B_n(z)=B_n(z^*).$ Thus $B_n(z)=B_n(z)^*$ by the first point, which implies $B_n(z)\in\mathbb R$.

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