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If $B_k(x)$ are the Bernoulli polynomials, then (by definition, if you like) we get that

$$\sum_{k=0}^{\infty}B_k(x)\frac{t^k}{k!}=\frac{te^{tx}}{e^t-1}$$

My question is whether or not there is a known formula for

$$\mathcal{G}(x;t):=\sum_{k=0}^{\infty}B_{2k+1}(x)\frac{t^{2k+1}}{(2k+1)!}.$$

The motivation for this question is that while studying modular forms the formula

$$\sum_{k=1}^{\frac{m-1}{2}}k\mathcal{G}\left(\frac{k}{m},t\right)=\frac{mt}{8\sinh\left(\frac{t}{2}\right)\sinh\left(\frac{t}{2m}\right)}-\frac{t}{8\sinh^{2}\left(\frac{t}{2m}\right)}$$

appeared for every odd integer $m$. This seems to imply that there is something going on with $\mathcal{G}(x;t)$, at least for rational values of $x$.

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    $\begingroup$ Given the above gf of the Bernoully polynomials $F(x,t)$, why the odd part wrto t isn't just $$\mathcal G(x,t)=\frac12\Big( F(x,t)-F(x,-t)\Big)$$ ? $\endgroup$
    – Pietro Majer
    Commented Oct 3, 2021 at 21:10
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    $\begingroup$ The sums in the title and the text are different. I suspect the one in the title is the correct one, since the other one is just the odd part of the first display? $\endgroup$
    – Timothy Budd
    Commented Oct 3, 2021 at 21:11
  • $\begingroup$ The odd part of the series can be written as $$\frac{t\sinh(t(x-1/2))}{2\sinh(t/2)}.$$ Is this all that you are asking? $\endgroup$
    – Hjalmar Rosengren
    Commented Oct 4, 2021 at 4:58

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Using Pietro Majer's bisection formula we find by a straightforward computation (I did it with Maple, but I'm sure it could be done without too much difficulty by hand) that the OP's formula for $$\sum_{k=1}^{\frac{m-1}{2}}k\mathcal{G}\left(\frac{k}{m},t\right)$$ is indeed true.

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