I don’t know how much you’ll enjoy this post, but calculating values of the zeta function is one of those things that struck high-school me as being incredibly difficult and requiring mastery over the arcane arts of mathematics, so it’s pretty cool to know that I can do this now ¹. Without further ado, let’s find a formula for $\zeta(2k)\dots$ ²

Let’s start with Parseval. Recall that we have a nice theory of Fourier analysis on the circle $S^1$ for the space $C^2(S^1)$ of twice differentiable (with continuous second derivative) functions. Note furthermore that viewing circle functions as functions $f:[0,1]\to\C$ with $f(0)=f(1)$ gives an embedding $C^2(S^1)\into C^2([0,1])$. Note that the set $\bracks{e_n}_{n\in\Z}$ where $e_n(x)=e^{-2\pi inx}$ is orthonormal with respect to the following inner product on $C^2([0,1])$:

\[\angles{f,g}=\int_0^1f(x)\conj{g(x)}\dx.\]

We claim (without proof) that the (linear) span of $\bracks{e_n}$ is dense in $C^2([0,1])$, letting us make a limit argument to show that for any $f\in C^2([0,1])$, one has

\[\sum_{n=-\infty}^\infty\abs{\angles{f,e_n}}^2=\|f\|^2:=\int_0^1\abs{f(x)}^2\dx.\]

The comment about embedding $C^2(S^1)$ into $C^2([0,1])$ is meant to help you recognize that $\angles{f,e_n}$ is exactly the $n$th Fourier coefficient of $f$. Our strategy for calculating $\zeta(2k)$ will be to find some function $f\in C^2([0,1])$ whose Fourier series has coefficients (roughly) of the form $c_n=\frac1{n^k}$ and then apply Parseval’s above identity.

The functions we’ll use are the (Jacob) Bernoulli polynomials $B_k(x)$ defined by the following identity ³

\[\int_x^{x+1}B_k(u)\d u=x^k.\]

In particular, $\int_0^1B_k(x)\dx=0^k$ (Here, $0^0=1$). Note that one can show that $B_k’(x)=kB_{k-1}(x)$ ⁴, and for good measure, one can calculate

\[\begin{align*} B_0(x) &= 1\\ B_1(x) &= x - \frac12\\ B_2(x) &= x^2 - x + \frac16, \end{align*}\]

and so on. Note that this derivative relation gives

\[B_k(1)-B_k(0)=\int_0^1B_k'(x)\dx=k\int_0^1B_{k-1}(x)\dx=0\]

for $k>1$ whereas $B_1(1)-B_1(0)=1$. Let the $k$th Bernoulli number $B_k$ be $B_k(0)$ (some authors use $B_k(1)$ instead). With this, we’ve done enough set up, so let’s get to the zeta stuff. ⁵

Define $c_k(n):=\int_0^1B_k(x)e^{-2\pi inx}\dx$, the $n$th Fourier coefficient of $B_k(x)$, and calculate

\[c_k(n)=\int_0^1B_k(x)e^{-2\pi inx}\dx=\frac{-1}{2\pi in}\sqbracks{\left.B_k(x)e^{-2\pi inx}\right|_0^1-\int_0^1 B_k'(x)e^{-2\pi inx}\dx}=\frac{k c_{k-1}(n)}{2\pi in}\]

for $k>1$ (since the product vanishes) while $c_1(n)=-1/(2\pi in)$ (since the integral vanishes). This gives

\[c_k(n) = \frac{k c_{k-1}(n)}{2\pi i n}=\frac{k(k-1)c_{k-2}(n)}{(2\pi in)^2}=\cdots=\frac{k!c_1(n)}{(2\pi in)^{k-1}}=-\frac{k!}{(2\pi in)^k}\]

where the $\cdots$ indicates that an induction argument is taking place behind the scenes. Now, the careful reader should be up in arms with the above equalities because they only hold for $n\neq0$ (this is needed for the integration by parts to work as claimed). By definition of the Bernoulli polynomials, we have $c_k(0)=0$. At this point, Parseval says

\[\sum_{\substack{n=-\infty\\n\neq0}}^\infty \abs{c_k(n)}^2=\int_0^1B_k(x)^2\dx.\]

The left hand side simplifies as

\[\sum_{\substack{n=-\infty\\n\neq0}}^\infty \abs{c_k(n)}^2=2\sum_{n=1}^\infty\frac{(k!)^2}{(2\pi n)^{2k}}=\frac{2(k!)^2}{(2\pi)^{2k}}\zeta(2k),\]

so our fabled $\zeta$ finally appears. To calculate the right hand side, integrate by parts (and induct) some more:

\[\begin{align*} \int_0^1B_k(x)^2\dx &= \frac1{k+1}\sqbracks{\left.B_k(x)B_{k+1}(x)\right|_0^1-k\int_0^1B_{k-1}(x)B_{k+1}(x)\dx}\\ &= -\frac k{k+1}\int_0^1B_{k-1}(x)B_{k+1}(x)\dx\\ &= -\frac k{(k+1)(k+2)}\sqbracks{\left.B_{k-1}(x)B_{k+2}(x)\right|_0^1-(k-1)\int_0^1B_{k-2}(x)B_{k+2}(x)\dx}\\ &= \frac{k(k-1)}{(k+1)(k+2)}\int_0^1B_{k-2}(x)B_{k+2}(x)\dx\\ &=\cdots\\ &= (-1)^{k-1}\frac{(k!)^2}{(2k-1)!}\int_0^1B_1(x)B_{2k-1}(x)\dx\\ &= (-1)^{k-1}\frac{(k!)^2}{(2k)!}\sqbracks{\left.B_1(x)B_{2k-1}(x)\right|_0^1-\int_0^1B_{2k}(x)\dx}\\ &= (-1)^{k-1}B_{2k}\frac{(k!)^2}{(2k)!}. \end{align*}\]

Putting these two sides together gives

\[\begin{align*} \sum_{n=-\infty}^\infty\abs{\angles{B_k(x),e^{-2\pi inx}}}^2 &= \int_0^1B_k(x)^2\dx\\ \frac{2(k!)^2}{(2\pi)^{2k}}\zeta(2k) &= (-1)^{k-1}\frac{B_{2k}(k!)^2}{(2k)!}\\ \zeta(2k) &= (-1)^{k-1}B_{2k}\frac{(2\pi)^{2k}}{2(2k)!} \end{align*}\]

In particular, plugging in $k=1$ recovers the classic

\[\zeta(2)=(-1)^{1-1}B_2\frac{(2\pi)^2}{2(2!)}=\frac{\pi^2}6\]

which is a good sign ⁶. Furthermore, I haven’t actually checked this myself yet, but I’m pretty sure that using the functional equation for the $\zeta$ function along with this calculation let’s you show that $\zeta(1-2k)\in\Q$ for all $k$ (and so $\zeta(1-k)\in\Q$ for all $k$ since it’s 0 when $k$ is odd) which is quite surprising. To end, I’ll really indulge a high-school fantasy by “proving” everyone’s favorite “theorem.”

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