Values of the Riemann Zeta function for even parameters

Alexander Frolkin

3 May 2001

Our aim here is to derive value of ${\rm\zeta}(s) := \sum_{r=1}^{\infty} \frac{1}{r^s}$ for even $s$. From Fourier theory, we have:

$${\rm f}(x) = \frac{1}{2}a_0 + \sum_{r=1}^{\infty} a_r \cos rx + \sum_{r=1}^{\infty} b_r \sin rx$$

where ${\rm f}(x)$ is periodic in $[-\pi, \pi]$ with period $2 \pi$, and

$$a_r = \frac{1}{\pi} \int_{-\pi}^{\pi} {\rm f}(x) \cos rx {\rm d}x$$

and

$$b_r = \frac{1}{\pi} \int_{-\pi}^{\pi} {\rm f}(x) \sin rx {\rm d}x$$

We start by finding the Fourier series for ${\rm f}(x) = x^{2n}$.

$$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x^{2n} {\rm d}x = \frac{2 \pi^{2n}}{2n+1}$$

We need to find

$$b_r = \frac{1}{\pi} \int_{-\pi}^{\pi} x^{2n} \sin rx {\rm d}x$$

We can see that this vanishes, since

$$\begin{eqnarray*} \pi b_r & = & \int_{-\pi}^{0} x^{2n} \sin rx {\rm d}x + \int_... ...{\rm d}(-x) + \int_{0}^{\pi} x^{2n} \sin rx {\rm d}x\\ & = & 0 \end{eqnarray*}$$

Next, we need to find

$$\pi a_r = \int_{-\pi}^{\pi} x^{2n} \cos rx {\rm d}x$$

We will first deduce a formula for

$$I_n := \int x^{2n} \cos rx {\rm d}x$$

for $n \in \mathbb{N}$. Integrating by parts,

$$\begin{eqnarray*} I_n & = & \int x^{2n} {\rm d}\biggl(\frac{1}{r} \sin rx\biggr... ...+ \frac{2n}{r^2} x^{2n-1} \cos rx - \frac{2n(2n-1)}{r^2} I_{n-1} \end{eqnarray*}$$

We can easily find $I_0$:

$$I_0 = \int \cos rx {\rm d}x = \frac{1}{r} \sin rx$$

This gives

$$I_1 = \frac{1}{r} x^2 \sin rx + \frac{2}{r^2} x \cos rx - \frac{2}{r^3} \sin rx$$

and

$$I_2 = \frac{1}{r} x^4 \sin rx + \frac{4}{r^2} x^3 \cos rx - ... ...3\cdot 2}{r^4} x \cos rx + \frac{4\cdot 3\cdot 2}{r^5} \sin rx$$

We can now see the pattern emerging and after some educated guesswork, we can say:

$$\begin{eqnarray*} I_n & = & \sin rx \sum_{k=0}^n \frac{(-1)^k (2n)! x^{2n-2k}}{(... ...}{(2n - 2k - 1)! r^{2k + 2}}\\ & =: & S_n \sin rx + C_n \cos rx \end{eqnarray*}$$

We are now in a position to formally prove the stated identity, by induction. We first assume that the result holds true for $n$. We then have, from the recurrence relation:

$$\begin{eqnarray*} I_{n+1} & = & \frac{1}{r} x^{2(n + 1)} \sin rx + \frac{2(n+1)}... ...n+1)}{r^2} x^{2(n+1) - 1} - \frac{(2n+2)(2n+1)}{r^2} C_n \biggr) \end{eqnarray*}$$

Looking at the sine coefficient,

$$\begin{eqnarray*} & & \frac{1}{r} x^{2(n+1)} - \frac{(2n+2)(2n+1)}{r^2} \sum_{k=... ...)! x^{2(n+1) - 2k}}{(2(n+1) - 2k)! r^{2k + 1}}\\ & =: & S_{n+1} \end{eqnarray*}$$

Similarly, for the cosine coefficient,

$$\begin{eqnarray*} & & \frac{2(n+1)}{r^2} x^{2(n+1) - 1} - \frac{(2n + 2)(2n + 1)... ...2(n+1) - 2k - 1}}{(2(n+1)-2k - 1)! r^{2k + 2}}\\ & =: & C_{n+1} \end{eqnarray*}$$

Hence, we can conclude that our guess is correct, by induction, $\forall n \in \mathbb{N}$.

To find $\pi a_r$, we note that for $r \in \mathbb{Z}$, $\sin r\pi = 0$ and $\cos r\pi = (-1)^r$. Hence, the $S_n \sin rx$ part vanishes and we are left with

$${\rm A}_n(r) := \int_{-\pi}^{\pi} x^{2n} \cos rx {\rm d}x =... ...rac{(-1)^k (2n)! \pi^{2n - 2k - 1}}{(2n - 2k - 1)! r^{2k + 2}}$$

We now have

$$x^{2n} = \frac{\pi^{2n}}{2n + 1} + \frac{1}{\pi} \sum_{r=1}^{\infty} {\rm A}_n(r) \cos rx$$

Now put $n = 1$ and $x = \pi$ to obtain

$$\pi^2 = \frac{\pi^2}{3} + \frac{1}{\pi} \sum_{r=1}^{\infty}(-1)^r {\rm A}_1(r)$$

We now look at the summand:

$$\begin{eqnarray*} & & (-1)^r {\rm A}_1(r)\\ & = & 2(-1)^{2r} \sum_{k=0}^{0} \fr... ...i^{2 - 2k - 1}}{(2-2k-1)! r^{2k+2}}\\ & = & 2 \frac{2 \pi}{r^2} \end{eqnarray*}$$

Substituting back,

$$\pi^2 = \frac{\pi^2}{3} + \frac{2}{\pi} \sum_{r=1}^{\infty} ... ...}{r^2} = \frac{\pi^2}{3} + 4 \sum_{r=1}^{\infty} \frac{1}{r^2}$$

and so we have

$$4{\rm\zeta}(2) = \frac{2}{3} \pi^2$$

and therefore,

$${\rm\zeta}(2) = \frac{\pi^2}{6}$$

Similarly, we can deduce the value of ${\rm\zeta}(4)$ by putting $n=2$.

$$\begin{eqnarray*} {\rm A}_2(r) & = & 2(-1)^r \sum_{k=0}^1 \frac{(-1)^k 4! \pi^{4... ...= & 8(-1)^r\pi \Biggl[ \frac{\pi^2}{r^2} - \frac{6}{r^4} \Biggr] \end{eqnarray*}$$

Now substitute into the Fourier series,

$$\begin{eqnarray*} \pi^4 & = & \frac{\pi^4}{5} + 8 \sum_{r=1}^{\infty} (-1)^{2r} ... ... & = & \frac{\pi^4}{5} + \frac{8}{6} \pi^4 - 48 {\rm\zeta}(4)\\ \end{eqnarray*}$$

Rearranging,

$$\begin{eqnarray*} 48 {\rm\zeta}(4) & = & \pi^4 \biggl( \frac{1}{5} + \frac{4}{3} - 1 \biggr)\\ & = & \frac{8}{15} \pi^4 \end{eqnarray*}$$

Hence,

$${\rm\zeta}(4) = \frac{\pi^4}{90}$$

To find ${\rm\zeta}(6)$, put $n=3$.

$$\begin{eqnarray*} {\rm A}_3(r) & = & 2(-1)^r \sum_{k=0}^2 \frac{(-1)^k 6! \pi^{5... ...rac{\pi^4}{r^2} - \frac{20 \pi^2}{r^4} + \frac{120}{r^6} \Biggr] \end{eqnarray*}$$

Plugging back into the Fourier series,

$$\begin{eqnarray*} \pi^6 & = & \frac{\pi^6}{7} + 12 \sum_{r=1}^{\infty} \Biggl[ \... ...+ \frac{12}{6} \pi^6 - \frac{240}{90} \pi^6 + 1440 {\rm\zeta}(6) \end{eqnarray*}$$

Rearrange,

$$\begin{eqnarray*} 1440{\rm\zeta}(6) & = & \pi^6 \biggl( 1 - \frac{1}{7} - 2 + \frac{24}{9} \biggr)\\ & = & \frac{32}{21}\pi^6 \end{eqnarray*}$$

Therefore,

$${\rm\zeta}(6) = \frac{\pi^6}{945}$$

Given enough patience and time, we can proceed as far as we like in the same way to find ${\rm\zeta}(2n)$, having already found ${\rm\zeta}(2m)$, $m = 1, 2, \ldots, n - 1$.

$$\begin{eqnarray*} {\rm\zeta}(2) & = & \frac{\pi^2}{6}\\ {\rm\zeta}(4) & = & \fr... ...57546764463363635252374414183254365234375}\pi^{50}\\ & \vdots & \end{eqnarray*}$$