The problem
I'm trying to derive a closed form expression for the values of the Riemann Zeta function for even natural numbers $k \geq 2$, meaning we want to find a closed form expression for $\zeta(k) := \sum_{n\geq 1} 1/n^k$.
Strategy
In order to solve this problem, I'll try to use myself of Fourier series expansions. My idea is as follows.
- Define $f_k(x) := x^k$ and try to find the Fourier series expansion for $f_k$ in $(-\pi,\pi)$ using the complex exponential as a basis. The Fourier coefficients will involve an integral of the form $\int f_k(x)e^{inx}dx$ which after partial integration $k$ times should give us something of the form $1/n^k$.
- Applying Parsevals formula will then help us to compute the Riemann Zeta function for even $k$'s since that'll involve some squaring.
Attempt
We want to expand $f_k(x)$ in a Fourier series with $\{e^{inx}\}_{n\in\mathbb{Z}}$ as our basis and for $x\in(-\pi,\pi)$. Meaning:
$$ f_k(x) = \sum_{n\in\mathbb{Z}} c_{n,k} e^{inx} $$
We want to find the coefficients $c_{n,k}$. This can be done utilizing the inner product as:
$$c_{n,k} = \frac{\langle f_k(x), e^{inx} \rangle }{2\pi} = \frac{1}{2\pi} \underset{:=I_{n,k}}{\underbrace{\int_{-\pi}^\pi x^k e^{inx}}}dx$$
Using integration by parts once, we land at the reccurence relation:
$$I_{n,k} + \frac{k}{in} I_{n,k-1} = \frac{(-1)^n \pi^k}{in} (1-(-1)^k)$$
Here's where my first problem comes. I want to solve this reccurence relation, but it's very messy for me and I've not studied recurrence relations alot. I'd be glad if anyone could propose a solution just regarding on how one can solve the recurrence relation to find a closed form expression for $I_{n,k}$. Thereafter, I can move on to solve my initial problem.
Also, if there's any other easier way to compute the integral I'd be glad if you presented it. I tried partial integration, but then I have to express the coefficients in a sum which will get messy when I then substitute it back in the Fourier expansion. That's why I neglected this idea.
Thanks in advance.