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 $x$.



Closed form for odd part of Bernoulli Polynomial generating function, $\sum_{k=0}^{\infty}B_{2k+1}(x)\frac{t^{2k+1}}{(2k+1)!}$
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.


