I want to express the Fourier Coefficients of the Bernoulli Polynomials in terms of sine or cosine, where
$$B_n: S^1 \rightarrow \mathbb{R}$$ with $S^1$ identified to $[0,1)$ (hence polynomials of period 1).
So far, I have that, if $n$ is odd, then
$$B_n(x) = \frac{-i(-1)^{\frac{n+1}{2}}n!}{(2\pi)^n} \sum_{m \neq 0} \frac{e^{2i\pi mx}}{m^n}$$
and if $n$ is even, then
$$B_n(x) = \frac{(-1)^{\frac{n}{2}+1}n!}{(2\pi)^n} \sum_{m \neq 0} \frac{e^{2i\pi mx}}{m^n}$$
However, I have also that $B_n(x)$ is even on if $n$ is even and odd otherwise.
Is there a way to derive the sine\cosine Fourier coefficients from those expressions?
I know that it's suppose to be $$B_n(x) = \frac{2(-1)^{\frac{n+1}{2}}n!}{(2\pi)^n} \sum_{m =1}^{\infty} \frac{\sin(2 \pi m x)}{m^n}$$
if n is odd, and
$$B_n(x) = \frac{2(-1)^{\frac{n}{2}+1}n!}{(2\pi)^n} \sum_{m =1}^{\infty} \frac{\cos(2 \pi m x)}{m^n}$$
if n is even, but I can't find a way to prove it.
Thanks in advance!