Here on page $6$ and $7$, Bruce Berndt and Armin Straub show that
$$ (- \beta)^{-m}\,\sum_{n = 1}^{\infty}\dfrac{\coth(\beta n)}{n^{2m + 1}} - \alpha^{-m}\,\sum_{n = 1}^{\infty}\dfrac{\coth(\alpha n)}{n^{2m + 1}}$$
$$ = 2^{2m}\,\sum_{k = 0}^{m + 1}\,(-1)^{k}\,\dfrac{B_n\,B_{2m + 2 - 2k}}{(2k)!\,(2m + 2 - 2k)!}\,\,\alpha^{m + 1 - k}\beta^k \tag{1}\label{1}$$
where $B_n$ is $n$-th Bernoulli number. Substituting $m = 2m + 1$ and $\alpha = \beta = \pi$ in $\eqref{1}$ we get
$$\sum_{n = 1}^{\infty}\dfrac{\coth(\pi n)}{n^{4m + 3}} = 2^{4m + 2}\pi^{4m+3}\sum_{k = 0}^{2m + 2}\,(-1)^{k+ 1}\dfrac{B_{2k}}{(2k)!}\dfrac{B_{4m + 4 - 2k}}{(4m + 4 - 2k)!} \tag{3}$$
However what if we substitute $m = 2m$ and $\alpha = \beta = \pi$ in $\eqref{1}$?
Do we get a an equation similar to $(3)$? I'm struggling with the substitution for quite some time. I'm only interested in the expression we get after substituting $m = 2m$ and $\alpha = \beta = \pi$ in $\eqref{1}$.
Thanks.
