# Fourier series of Bernoulli polynomials

We know that Bernoulli polynomials $$\phi_n(x)$$ can be defined as below:$$\phi_0(x)=1,\dot\phi_n(x)=n\phi_{n-1}(x),\int^1_0{\phi_n(x)=0}$$ and I want to use this to prove:$$\phi_k(x)=(-1)^{1+\frac{k}{2}}\frac{2(k!)}{(2\pi)^k}\sum_{n=1}^{\infty}{\frac{cos2\pi nx}{n^k}}$$where k is even and$$\phi_k(x)=(-1)^{\frac{k+1}{2}}\frac{2(k!)}{(2\pi)^k}\sum_{n=1}^{\infty}{\frac{sin2\pi nx}{n^k}}$$ where k is odd , but I have no idea about it , can anyone help me?

I'll use the more familiar notation (for me) $$B_n(x)$$ for the Bernoulli polynomials. The defining characteristics should make clear for which $$n$$ they apply, namely, \begin{align} B_0(x) &= 1 \\ \frac{d}{dx} B_n(x) = n B_{n-1}(x) \quad&\text{and} \quad \int_0^1 B_n(x) dx = 0 \quad\text{when } n \geqslant 1 \end{align} It follows $$B_1(x) = x - \frac{1}{2}$$ and $$B_n(1) = B_n(0)$$ when $$n \geqslant 2$$ because \begin{align} 0 &= \int_0^1 B_{n-1}(x) dx \\ &= \Big[ \frac{1}{n} B_{n}(x) \Big]_0^1 \\ &=B_n(1)-B_n(0) \end{align} Therefore, each $$B_n$$ is continuously differentiable in $$[0,1]$$, being a polynomial, and its end point values are the same (except for $$B_1$$) so its Fourier series will converge pointwise.
Then, if $$B_n(x)$$ has Fourier series $$\displaystyle \sum_{k=-\infty}^{\infty}a_{n,k}e^{2\pi i k x}$$, we have \begin{align} a_{n,k} = \int_0^1 B_n(x) e^{-2\pi i k x} dx \end{align} By direct calculation, when $$n=0$$, $$a_{0,k} = \left\{ \array{0&\text{when }k \neq 0 \\ 1 &\text{when }k = 0}\right.$$ and when $$n\neq 0, k=0$$ $$a_{n,0} = 0$$.
Otherwise, for $$n \geqslant 2$$, \begin{align} a_{n,k} &= \int_0^1 B_n(x) e^{-2\pi i k x} dx \\ &= -\frac{1}{2\pi i k}\Big[ B_n(x) e^{-2\pi i k x}\Big]_0^1 + \frac{1}{2\pi i k }\int_0^1 nB_{n-1}(x) e^{-2\pi i k x} dx. \end{align} If $$n=1$$ and $$k \neq 0$$ this becomes, \begin{align} a_{1,k} = -\frac{1}{2\pi i k} \end{align} and otherwise, we derive, $$n \geqslant 2, k\neq 0$$ \begin{align} a_{n,k}=\frac{n}{2\pi i k}a_{n-1,k}. \end{align} Applied repeatedly and using the special cases derived above, we can see that \begin{align} a_{n,k} = -(-i)^n \frac{n!}{(2\pi k)^n} =\left\{ \array{ (-1)^{n/2+1} \frac{n!}{(2\pi k)^n}&\text{when } n \geqslant 1 \text{ and even} \\ -(-1)^{(n+1)/2} \frac{n!}{(2\pi k)^n} \cdot i &\text{when } n \geqslant 1 \text{ and odd} \\ 1 & \text{when } n = 0 \text{ and } k = 0 \\ 0 &\text{otherwise.} } \right. \end{align} These are expressed in terms of Fourier terms in $$e^{2\pi i k x}$$. To convert to cosine and sine series simply pair the $$k$$ and $$-k$$ coefficients. Even numbered pairs give the cosine terms with a factor of $$2$$ and odd numbered pairs the sine terms with a factor of $$-2$$.