Derivation of a Formula for the Even Values of the Riemann Zeta Function

 

Lemma 1: Fourier Series of the Dirac Comb

A Dirac comb of period T is defined as \[{\mathrm{III}}_T(x)=\sum_{k=-\infty}^{\infty} \delta(x-kT)\] Where \(\delta(x)\) is the Dirac delta function. Since the Dirac comb is periodic with period T, we can expand it as a fourier series: \[\sum_{k=-\infty}^{\infty} \delta(x-kT)=\sum_{n=-\infty}^{\infty} A_n e^{i 2 \pi n x/T}\] We solve for the \(A_m\) in the usual way: \[ \int_{-T/2}^{T/2}\sum_{k=-\infty}^{\infty} \delta(x-kT)e^{-i 2 \pi m x/T} dx=1=\int_{-T/2}^{T/2}\sum_{n=-\infty}^{\infty} A_n e^{i 2 \pi (n-m) x/T} dx=T\cdot A_m \]\[ A_m=1/T \] Thus: \[\sum_{k=-\infty}^{\infty} \delta(x-kT)=\frac{1}{T}\sum_{n=-\infty}^{\infty} e^{i 2 \pi n x/T}\]

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Lemma 2: An Infinite Series

\[ \sum_{n=-\infty}^{\infty} \frac{1}{x+i n}=\frac{1}{x}+\sum_{n=1}^{\infty} \frac{1}{x+i n}+\frac{1}{x-i n}=\frac{1}{x}+2x\sum_{n=1}^{\infty} \frac{1}{x^2+n^2} \]\[ \sum_{n=-\infty}^{\infty} \frac{1}{x+i n}=\int_{0}^{\infty} \sum_{n=-\infty}^{\infty} e^{-y(x+i n)} dy \]\[ \sum_{n=-\infty}^{\infty} \frac{1}{x+i n}=\int_{0}^{\infty} e^{-yx} \sum_{n=-\infty}^{\infty} e^{-iyn} dy \]\[ \sum_{n=-\infty}^{\infty} \frac{1}{x+i n}=2\pi \int_{0}^{\infty} e^{-yx} \sum_{k=-\infty}^{\infty} \delta(x-2\pi k) dy \]\[ \sum_{n=-\infty}^{\infty} \frac{1}{x+i n}=2\pi \left (\frac{1}{2}+ \sum_{k=1}^{\infty} e^{-2\pi k x} \right ) \]\[ \sum_{n=-\infty}^{\infty} \frac{1}{x+i n}=2\pi \left (\frac{1}{2}+ \frac{e^{-2\pi x}}{1-e^{-2\pi x}} \right )= \pi \frac{e^{2\pi x}+1}{e^{2\pi x}-1} \] Therefore, combining the first and last expressions and rearranging, we find: \[ \sum_{n=1}^{\infty} \frac{1}{x^2+n^2}=\frac{\pi}{2x} \frac{e^{2\pi x}+1}{e^{2\pi x}-1}-\frac{1}{2x^2}=\frac{\pi}{2x} \coth(\pi x)-\frac{1}{2x^2} \] Additionally, by taking the limit as x approaches zero, we find: \[ \sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6} \]

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Theorem: Formula for the Even Values of the Riemann Zeta Function

Recall that, by definition: \[ \zeta(n)=\sum_{k=1}^{\infty}\frac{1}{k^n} \] Let us then analyze \[ f(x)=1-\frac{x}{2}+\sum_{n=2}^{\infty}\frac{x^{n}}{n!} A_{n} \] Where \[ A_n=-2 \cdot n! \cdot \cos(n\pi/2) \cdot 2^{-n}\pi^{-n} \zeta(n) \] Thus: \[ f(x)=1-\frac{x}{2}-2\sum_{n=1}^{\infty}\left (\frac{-x^2}{4\pi^2} \right )^n \zeta(2n) \]\[ f(x)=1-\frac{x}{2}-2\sum_{n=1}^{\infty}\left (\frac{-x^2}{4\pi^2} \right )^n \sum_{k=1}^{\infty}\frac{1}{k^{2n}} \]\[ f(x)=1-\frac{x}{2}-2\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\left (\frac{-x^2}{4\pi^2 k^2} \right )^n \]\[ f(x)=1-\frac{x}{2}-2\sum_{k=1}^{\infty} \frac{-x^2}{4\pi^2 k^2}\frac{1}{1+\frac{x^2}{4\pi^2 k^2}} \]\[ f(x)=1-\frac{x}{2}+\frac{x^2}{2\pi^2}\sum_{k=1}^{\infty} \frac{1}{k^2+\frac{x^2}{4\pi^2}} \]\[ f(x)=1-\frac{x}{2}+\frac{x^2}{2\pi^2} \left ( \frac{\pi^2}{x} \frac{e^x+1}{e^x-1} -\frac{2\pi^2}{x^2} \right ) \]\[ f(x)=\frac{x}{2} \left ( \frac{e^x+1}{e^x-1} -1 \right )=\frac{x}{e^x-1} \] Therefore, for n>1, \[ A_n=\lim_{x \rightarrow 0} \frac{\mathrm{d}^n }{\mathrm{d} x^n} \frac{x}{e^x-1} \] These numbers are called the Bernoulli Numbers, symbolized as \(B_n\) and they are easily found to be all rational. Thus, by rearranging, we find: \[ \zeta(2n)=\frac{\pi^{2n} 2^{2n-1} \left | B_{2n} \right |} {(2n)!} \] Thus, all the even values of the zeta function can be found by finding the appropriate Bernoulli number, which itself can be found by simple differentiation. Moreover, we see that all the values are rational multiples of the corresponding power of pi. Specifically, we find that: \[ \zeta(2)=\frac{\pi^2}{6} \]\[ \zeta(4)=\frac{\pi^4}{90} \]\[ \zeta(6)=\frac{\pi^6}{945} \]\[ \zeta(8)=\frac{\pi^8}{9450} \]\[ \zeta(10)=\frac{\pi^{10}}{93555} \]