On the fourth page in deriving (5), it says that it can be easily obtained from
$I(k,m)=-\frac{2k(2k-1)}{m^2\pi^2}I(k-1,m)$
that
$I(k,m)=\frac{(-1)^{k-1}(2k)!}{m^{2k}\pi^{2k}}$ for m even, and 0 otherwise.
Here $I(k,m)=\int_0^1B_{2k}(t)cos(m\pi t)dt$ where $B_k(t)$ is the Bernoulli Polynomial.
https://arxiv.org/pdf/1209.5030.pdf
I must be missing something but I don't see the connection