Deriving an expression for the Riemann zeta function at even values

---

I wanted to show the derivation of the expression for the Riemann zeta function at even values, i.e. $\newcommand{\dd}{\mathrm{d}}$

\[\zeta(2k)=\frac{(-1)^{k+1}(2\pi)^{2k}\beta_{2k}}{2(2k)!}\]

We start by expressing the function $\frac{\sin(\pi x)}{\pi x}$ in its infinite product form:

\[\frac{\sin(\pi x)}{\pi x}=\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2}\right)\]

A way of understanding this representation is by realising that $\frac{\sin(\pi x)}{\pi x}$ can be expressed as a product of linear factors, as you would with any other polynomial. Let us consider a second degree polynomial as an example:

\[f(x)=x^2+7x+12\]

We know that it can be expressed as a product of its linear factors:

\[f(x)=(x+4)(x+3)\]

However, it can also be expressed as

\[f(x)=\left(1+\frac{x}{4}\right)\left(1+\frac{x}{3}\right)\]

This can be verified by seeing that $f(-4)$ and $f(-3)=0$. However, look into the Weierstrass factorisation theorem for a more rigorous explanation on expressing functions in this form.

$\frac{\sin(\pi x)}{\pi x}$ has roots at $x=n$ for $n \in \mathbb{Z}$ and $n \neq 0$, and so, expressing it as a product of its linear factors in a similar way gives

\[\frac{\sin(\pi x)}{\pi x}=\left(1-x\right)\left(1+x\right)\left(1-\frac{x}{2}\right)\left(1+\frac{x}{2}\right)\left(1-\frac{x}{3}\right)\left(1+\frac{x}{3}\right)\cdots\]

Making the observation that pairing terms gives the difference of two squares we get the infinite product form:

\[\frac{\sin(\pi x)}{\pi x}=\left(1-x^2\right)\left(1-\frac{x^2}{4}\right)\left(1-\frac{x^2}{9}\right)\cdots\]

From here we’re going to manipulate the product of $\frac{sin(\pi x)}{\pi x}$ to a point where we recognise something of the sort we are trying to derive. Firstly, we take the natural logarithm of both sides:

\[\log{\frac{\sin(\pi x)}{\pi x}}=\log\left(\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2}\right)\right)\]

Making use of the property that the logarithm of a product is equivalent to the sum of the individual logarithms gives

\[\log{\frac{\sin(\pi x)}{\pi x}}=\sum_{n=1}^{\infty}\log{\left(1-\frac{x^2}{n^2}\right)}\] \[\log{(\sin(\pi x))}=\log{(\pi x)}+\sum_{n=1}^{\infty}\log{\left(1-\frac{x^2}{n^2}\right)}\]

Taking the derivative of both sides:

\[\frac{\dd}{\dd x}\left[\log{(\sin(\pi x))}\right]=\frac{\dd}{\dd x}\left[\log{(\pi x)}+\sum_{n=1}^{\infty}\log{\left(1-\frac{x^2}{n^2}\right)}\right]\] \[\frac{\pi \cos(\pi x)}{\sin(\pi x)}=\frac{1}{x}-\sum_{n=1}^{\infty}\left(\frac{2x}{n^2}\right)\left(\frac{1}{1-\frac{x^2}{n^2}}\right)\] \[\pi x \cot(\pi x)=1-2\sum_{n=1}^{\infty}\left(\frac{x^2}{n^2}\right)\left(\frac{1}{1-\frac{x^2}{n^2}}\right)\]

Expressing $\frac{1}{1-\frac{x^2}{n^2}}$ as its series expansion gives

\[\pi x \cot(\pi x)=1-2\sum_{n=1}^{\infty}\sum_{k=0}^{\infty}\left(\frac{x^2}{n^2}\right)\left(\frac{x^2}{n^2}\right)^k\] \[\pi x \cot(\pi x)=1-2\sum_{n=1}^{\infty}\sum_{k=0}^{\infty}\left(\frac{x^2}{n^2}\right)^{k+1}\]

The lower limit of the inner summation can be changed to $k=1$ if we also change the exponent of the term being summed:

\[\pi x \cot(\pi x)=1-2\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{x^{2k}}{n^{2k}}\] \[\pi x \cot(\pi x)=1-2\sum_{k=1}^{\infty}\zeta(2k)x^{2k}\]

And so, we see that we have some sort of expression with $\zeta(2k)$ on the right-hand side. Let’s now take a closer look at the cotangent function on the left-hand side.

The cotangent can be expressed as the ratio of cosine to sine. Expressing the cosine and sine in their exponential function form gives

\[\begin{aligned} \pi x \cot(\pi x)&=\pi x\frac{\cos(\pi x)}{\sin(\pi x)} \\ &=\pi x \frac{e^{i\pi x}+e^{-i\pi x}}{2}\frac{2i}{e^{i\pi x}-e^{-i\pi x}} \\ &=i\pi x \frac{e^{i\pi x}+e^{-i\pi x}}{e^{i\pi x}-e^{-i\pi x}} \end{aligned}\]

Multiplying the right-hand side by $\frac{e^{i\pi x}}{e^{i\pi x}}$ (just because it seemed to work):

\[\begin{aligned} \pi x \cot(\pi x)&=i\pi x \frac{e^{i\pi x}+e^{-i\pi x}}{e^{i\pi x}-e^{-i\pi x}} \cdot \frac{e^{i\pi x}}{e^{i\pi x}} \\ &=i\pi x \frac{e^{2i\pi x}+1}{e^{2i\pi x}-1} \end{aligned}\]

And now, splitting the numerator, we get

\[\begin{aligned} \pi x \cot(\pi x)&=i\pi x \frac{e^{2i\pi x}-1+2}{e^{2i\pi x}-1} \\ &=i\pi x \left(1+\frac{2}{e^{2i\pi x}-1}\right) \\ &=i\pi x+\frac{2i\pi x}{e^{2i\pi x}-1} \\ \end{aligned}\]

Those last couple of steps may have seemed counterintuitive, however, it allows us to make use of the following relation:

\[\frac{x}{e^x-1}=\sum_{k=0}^{\infty}\frac{\beta_k}{k!}x^k\]

Substituting this into our second cotangent expression gives

\[\pi x \cot(\pi x)=i\pi x+\sum_{k=0}^{\infty}\frac{\beta_k}{k!}(2i\pi x)^k\]

You may notice that this looks quite similar in form to the first cotangent equation we derived. Manipulating the above equation into a similar form to the one we derived earlier will allow the $x^{2k}$ coefficients to be equated, which will give the expression for $\zeta({2k})$.

We begin by expanding out the first two terms of the summation:

\[\pi x \cot(\pi x)=i\pi x+\frac{\beta_0}{0!}+\frac{\beta_1}{1!}(2i \pi x)+\sum_{k=2}^{\infty}\frac{\beta_k}{k!}(2i\pi x)^k\]

$\beta_0=1$ and $\beta_1=-\frac{1}{2}$, and so we get

\[\begin{aligned} \pi x \cot(\pi x)&=i\pi x+1-\frac{1}{2}(2i \pi x)+\sum_{k=2}^{\infty}\frac{\beta_k}{k!}(2i\pi x)^k \\ &=1+\sum_{k=2}^{\infty}\frac{\beta_k}{k!}(2i\pi x)^k \end{aligned}\]

Taking out a factor of $-2$, just because we know what we’re working towards and the expression is getting close,

\[\pi x \cot(\pi x)=1-2\sum_{k=2}^{\infty}\frac{\beta_k}{k!}\frac{(2i\pi x)^k}{-2}\]

We can alter the lower summation limit again by altering the exponent:

\[\begin{aligned} \pi x \cot(\pi x)&=1-2\sum_{k=1}^{\infty}\frac{\beta_{2k}}{2k!}\frac{(2i\pi x)^{2k}}{-2}x^{2k} \\ &=1-2\sum_{k=1}^{\infty}\frac{i^{2k}(2\pi)^{2k}\beta_{2k}}{-2(2k)!}x^{2k} \\ &=1-2\sum_{k=1}^{\infty}\frac{(-1)^{k}(2\pi)^{2k}\beta_{2k}}{-2(2k)!}x^{2k} \end{aligned}\]

And so, we get

\[\pi x \cot(\pi x)=1-2\sum_{k=1}^{\infty}\frac{(-1)^{k+1}(2\pi)^{2k}\beta_{2k}}{2(2k)!}x^{2k}\]

Now recall the cotangent expression we got to earlier:

\[\pi x \cot(\pi x)=1-2\sum_{k=1}^{\infty}\zeta(2k)x^{2k}\]

Equating the two cotangent expression, we get

\[\zeta(2k)=\frac{(-1)^{k+1}(2\pi)^{2k}\beta_{2k}}{2(2k)!}\]