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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?

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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 $-2$.

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