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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:

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

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

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

and

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

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

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

We need to find

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

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

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

We will first deduce a formula for

\begin{displaymath}
I_n := \int x^{2n} \cos rx {\rm d}x
\end{displaymath}

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 $I_0$:

\begin{displaymath}
I_0 = \int \cos rx {\rm d}x = \frac{1}{r} \sin rx
\end{displaymath}

This gives

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

and

\begin{displaymath}
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
\end{displaymath}

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

\begin{displaymath}
{\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}}
\end{displaymath}

We now have

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

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

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

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,

\begin{displaymath}
\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}
\end{displaymath}

and so we have

\begin{displaymath}
4{\rm\zeta}(2) = \frac{2}{3} \pi^2
\end{displaymath}

and therefore,

\begin{displaymath}
{\rm\zeta}(2) = \frac{\pi^2}{6}
\end{displaymath}

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,

\begin{displaymath}
{\rm\zeta}(4) = \frac{\pi^4}{90}
\end{displaymath}

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,

\begin{displaymath}
{\rm\zeta}(6) = \frac{\pi^6}{945}
\end{displaymath}

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*}






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Alexander Frolkin 2001-05-03