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Aymen Hafeez

The Riemann zeta function at even values

/ 7 min read

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I wanted to show the derivation of the expression for the Riemann zeta function at even values, i.e.

ζ(2k)=(−1)k+1(2π)2kβ2k2(2k)!\zeta(2k)=\frac{(-1)^{k+1}(2\pi)^{2k}\beta_{2k}}{2(2k)!}

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

sin⁡(πx)πx=∏n=1∞(1−x2n2)\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 sin⁡(πx)πx\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)=x2+7x+12f(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)f(x)=(x+4)(x+3)

However, it can also be expressed as

f(x)=(1+x4)(1+x3)f(x)=\left(1+\frac{x}{4}\right)\left(1+\frac{x}{3}\right)

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

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

sin⁡(πx)πx=(1−x)(1+x)(1−x2)(1+x2)(1−x3)(1+x3)⋯\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:

sin⁡(πx)πx=(1−x2)(1−x24)(1−x29)⋯\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 sin(πx)πx\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⁡sin⁡(πx)πx=log⁡(∏n=1∞(1−x2n2))\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⁡sin⁡(πx)πx=∑n=1∞log⁡(1−x2n2)\log{\frac{\sin(\pi x)}{\pi x}}=\sum_{n=1}^{\infty}\log{\left(1-\frac{x^2}{n^2}\right)}

log⁡(sin⁡(πx))=log⁡(πx)+∑n=1∞log⁡(1−x2n2)\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:

ddx[log⁡(sin⁡(πx))]=ddx[log⁡(πx)+∑n=1∞log⁡(1−x2n2)]\frac{\text{d}}{\text{d} x}\left[\log{(\sin(\pi x))}\right]=\frac{\text{d}}{\text{d} x}\left[\log{(\pi x)}+\sum_{n=1}^{\infty}\log{\left(1-\frac{x^2}{n^2}\right)}\right]

πcos⁡(πx)sin⁡(πx)=1x−∑n=1∞(2xn2)(11−x2n2)\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)

πxcot⁡(πx)=1−2∑n=1∞(x2n2)(11−x2n2)\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 11−x2n2\frac{1}{1-\frac{x^2}{n^2}} as its series expansion gives

πxcot⁡(πx)=1−2∑n=1∞∑k=0∞(x2n2)(x2n2)k\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

πxcot⁡(πx)=1−2∑n=1∞∑k=0∞(x2n2)k+1\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=1k=1 if we also change the exponent of the term being summed:

πxcot⁡(πx)=1−2∑n=1∞∑k=1∞x2kn2k\pi x \cot(\pi x)=1-2\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{x^{2k}}{n^{2k}}

πxcot⁡(πx)=1−2∑k=1∞ζ(2k)x2k\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 ζ(2k)\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

πxcot⁡(πx)=πxcos⁡(πx)sin⁡(πx)=πxeiπx+e−iπx22ieiπx−e−iπx=iπxeiπx+e−iπxeiπx−e−iπx \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 eiπxeiπx\frac{e^{i\pi x}}{e^{i\pi x}} (just because it seemed to work):

πxcot⁡(πx)=iπxeiπx+e−iπxeiπx−e−iπx⋅eiπxeiπx=iπxe2iπx+1e2iπx−1 \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

πxcot⁡(πx)=iπxe2iπx−1+2e2iπx−1=iπx(1+2e2iπx−1)=iπx+2iπxe2iπx−1 \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:

xex−1=∑k=0∞βkk!xk\frac{x}{e^x-1}=\sum_{k=0}^{\infty}\frac{\beta_k}{k!}x^k

Substituting this into our second cotangent expression gives

πxcot⁡(πx)=iπx+∑k=0∞βkk!(2iπx)k\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 x2kx^{2k} coefficients to be equated, which will give the expression for ζ(2k)\zeta({2k}).

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

πxcot⁡(πx)=iπx+β00!+β11!(2iπx)+∑k=2∞βkk!(2iπx)k\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

β0=1\beta_0=1 and β1=−12\beta_1=-\frac{1}{2}, and so we get

πxcot⁡(πx)=iπx+1−12(2iπx)+∑k=2∞βkk!(2iπx)k=1+∑k=2∞βkk!(2iπx)k \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 −2-2, just because we know what we’re working towards and the expression is getting close,

πxcot⁡(πx)=1−2∑k=2∞βkk!(2iπx)k−2\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:

πxcot⁡(πx)=1−2∑k=1∞β2k2k!(2iπx)2k−2x2k=1−2∑k=1∞i2k(2π)2kβ2k−2(2k)!x2k=1−2∑k=1∞(−1)k(2π)2kβ2k−2(2k)!x2k \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

πxcot⁡(πx)=1−2∑k=1∞(−1)k+1(2π)2kβ2k2(2k)!x2k\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:

πxcot⁡(πx)=1−2∑k=1∞ζ(2k)x2k\pi x \cot(\pi x)=1-2\sum_{k=1}^{\infty}\zeta(2k)x^{2k}

Equating the two cotangent expression, we get

ζ(2k)=(−1)k+1(2π)2kβ2k2(2k)!\zeta(2k)=\frac{(-1)^{k+1}(2\pi)^{2k}\beta_{2k}}{2(2k)!}