Zeta(2)

Basel Problem 1644 asks to find the value of \displaystyle 1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\cdots

Euler (1735) showed that

\displaystyle \sum \frac{1}{n^2}=1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\cdots =\frac{\pi^2}{6}

The result, \displaystyle \frac{\pi^2}{6}, is a cornerstone of analysis, and there are diverse methods of proof from function theory, harmonic analysis, and geometry. All of them essentially uncover how integers \displaystyle \mathbb Z sit inside real numbers.

Euler’s Proof:

Euler’s original method relies on two representations of \displaystyle \sin(\pi x). The first is the Taylor series:

\displaystyle \sin (\pi x)=\pi x-\frac{(\pi x)^{3}}{3 !}+\frac{(\pi x)^{5}}{5 !}-\cdots

The second is the Weierstrass factorization, representing the function as an infinite product over its roots at the integers \displaystyle n \in \mathbb{Z}:

\displaystyle \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

\displaystyle =\pi x+\pi x^{3}\left(1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\cdots\right)+\pi x^{5}\left(\frac{1}{1 \cdot 4}+\frac{1}{1 \cdot 9}+\cdots+\frac{1}{4 \cdot 9}+\cdots\right)+\cdots

Comparing the coefficients of {x^3} we get,

\displaystyle \pi\left(1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\cdots\right)=\frac{\pi^{3}}{6}
\displaystyle 1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\cdots=\frac{\pi^{2}}{6}

Euler’s assertion that the sine function can be factored into a product over its roots was an act of profound intuition. While its rigorous proof was not available in his time, it is now understood as a consequence of the Weierstrass factorization theorem which guarantees that any entire function (a function analytic over the entire complex plane) of finite order can be represented by a product reflecting its zeroes. The sine function is a canonical example, and its factorization into the form Euler used is fully justified. For example, the function \displaystyle ( f(z) = e^z \sin(\pi z) ) has the exact same roots as \displaystyle ( \sin(\pi z) ) , but it is clearly a different function. What prevents such an arbitrary factor \displaystyle ( e^{g(z)} ) from appearing? Euler’s argument implicitly assumes that such a factor is not present, or is equal to 1. His intuition correctly sensed that the sine function was sufficiently “polynomial-like” for the analogy to hold.

Finite Products to Infinite Product: Euler’s product can be rigorously derived as the limiting case of an exact, finite product identity. Using De Moivre’s formula, one establishes the following identity for any odd integer n=2m+1 and complex number z:

\displaystyle \sin(z)=n\sin\left( \frac{z}{n} \right) \prod_{j=1}^{m} \left(1-\frac{\sin^2 (z/n)}{\sin^2 (j\pi/n)}\right)

The infinite product emerges upon analyzing this identity in the limit as n\to\infty. We examine the behavior of each component:

First, the leading factor converges to z: \displaystyle \lim_{n\to\infty} n\sin\left(\frac{z}{n}\right) = \lim_{n\to\infty} z \frac{\sin(z/n)}{z/n} = z

Second, for any fixed index j, the argument of the product has a well-defined limit as n\to\infty: \displaystyle \lim_{n\to\infty} \frac{\sin^2(z/n)}{\sin^2(j\pi/n)} = \frac{z^2}{j^2\pi^2}

As n\to\infty (and thus m\to\infty), the finite product converges to the infinite product. Combining these limits yields

\displaystyle \sin(z) = z \prod_{j=1}^{\infty} \left(1-\frac{z^2}{j^2\pi^2}\right)

Fourier Proof

This approach uses tools from harmonic analysis. Consider the function \displaystyle f(x) = x on \displaystyle [0, 1]. Its norm in \displaystyle L^2[0,1] is:

\displaystyle \langle f, f\rangle=\int_{0}^{1} x \bar{x} d x=\int_{0}^{1} x^{2} d x=\frac{1}{3}

The Fourier coefficients are \displaystyle \hat{f}(0) = 1/2 and \displaystyle \hat{f}(n) = \frac{i}{2\pi n} for \displaystyle n \neq 0.


\displaystyle \hat f (n) = \int_{0}^{1} x \exp (-2 \pi i n x) d x =-\frac{1}{2 \pi i n}
\displaystyle \hat f(0) =\int_{0}^{1} x=\frac{1}{2}

Parseval’s identity, \displaystyle \|f\|^2 = \sum_{n=-\infty}^{\infty} |\hat{f}(n)|^2, is an isometry between the function space \displaystyle L^2[0,1] and the sequence space of its coefficients \displaystyle l_2(\mathbb Z). Applying it:

\displaystyle \frac{1}{3}=\langle f, f\rangle=\sum |\hat f(n)|^2=\left(\frac{1}{2}\right)^{2}+\sum_{n }\left|-\frac{1}{2 \pi i n}\right|^{2}.

Algebraic rearrangement yields

\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.

An alternative, and perhaps more direct, Fourier-analytic proof avoids Parseval’s identity, instead leveraging the pointwise convergence of the series for f(x)=x^2 on the interval [-\pi, \pi]. Being an even function, its expansion is a pure cosine series, which can be calculated as:

\displaystyle x^2 = \frac{\pi^2}{3} + \sum_{n=1}^{\infty} (-1)^n \frac{4}{n^2} \cos(nx)

This identity holds for all x \in [-\pi, \pi]. The key is to evaluate it at a strategic point, x=\pi, where \cos(n\pi) = (-1)^n. The series simplifies dramatically:

\displaystyle \pi^2 = \frac{\pi^2}{3} + \sum_{n=1}^{\infty} (-1)^n \frac{4}{n^2} (-1)^n = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{1}{n^2}

Rearranging this simple algebraic equation directly yields the result. Here, the integers n are reinterpreted as defining the harmonic frequencies, and the sum emerges from evaluating the resulting vibration at its endpoint.

Connection: The link between the Euler and Fourier methods is the partial fraction expansion of the cotangent function. Taking the logarithmic derivative of Euler’s product formula yields:

\displaystyle \pi \cot(\pi x) = \frac{1}{x} + \sum_{n=1}^{\infty} \frac{2x}{x^2 - n^2}

This identity is essentially the Fourier series representation of a periodic function related to the sawtooth wave underpinning the second proof. Expanding the terms \displaystyle \frac{2x}{x^2 - n^2} = -2x \sum_{k=0}^{\infty} \frac{x^{2k}}{n^{2k+2}} and collecting powers of \displaystyle x provides a systematic way to find \displaystyle \zeta(2k), showing that Euler’s coefficient comparison was a specific instance of this more general harmonic analysis identity.

Apostol’s Proof:

This approach recasts the sum as a double integral, which is then evaluated using a geometric transformation.

\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \sum_{n=0}^{\infty} \frac{1}{(n+1)^2} = \int_0^1 \int_0^1 \sum_{n=0}^{\infty} (xy)^n dx dy = \int_0^1 \int_0^1 \frac{dx dy}{1-xy}

The coordinate change \displaystyle x=u-v, y=u+v with Jacobian 2 transforms the integral. Exploiting symmetry, the integral becomes:

\displaystyle \int_{0}^{1} \int_{0}^{1} \frac{1}{1-x y} d x d y =4 \int_{0}^{1 / 2} \int_{0}^{u} \frac{1}{1-u^{2}+v^{2}} d v d u

\displaystyle +4 \int_{1 / 2}^{1} \int_{0}^{1-u} \frac{1}{1-u^{2}+v^{2}} d v d u

Using {\displaystyle \int_{0}^{t} \frac{d t}{a^{2}+t^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right),} we get

\displaystyle \int_{0}^{1} \int_{0}^{1} \frac{1}{1-x y} d x d y =4 \int_{0}^{1 / 2} \tan ^{-1}\left(\frac{u}{\sqrt{1-u^{2}}}\right) \cdot \frac{d u}{\sqrt{1-u^{2}}}

\displaystyle +4 \int_{1 / 2}^{1} \tan ^{-1}\left(\frac{1-u}{\sqrt{1-u^{2}}}\right) \cdot \frac{d u}{\sqrt{1-u^{2}}}

The integral are computed below.

The first integral, via \displaystyle u=\cos(2\theta):

\displaystyle \begin{aligned} &4 \int_{1 / 2}^{1} \tan ^{-1}\left(\frac{1-u}{\sqrt{1-u^{2}}}\right) \cdot \frac{d u}{\sqrt{1-u^{2}}} \\ &=4 \int_{0}^{\pi / 6} \tan ^{-1}\left(\frac{1-\cos (2 \theta)}{\sqrt{1-\cos ^{2}(2 \theta)}}\right) \frac{2 \sin (2 \theta) d \theta}{\sqrt{1-\cos ^{2}(2 \theta)}} \\ &=8 \int_{0}^{\pi / 6} \tan ^{-1} \sqrt{\frac{1-\cos (2 \theta)}{1+\cos (2 \theta)}} d \theta \\ &=8 \int_{0}^{\pi / 6} \tan ^{-1} \sqrt{\frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta}} d \theta=\frac{\pi^{2}}{9} \end{aligned}

The second integral, via a substitution \displaystyle u=\sin\theta:

\displaystyle \begin{aligned} &4 \int_{0}^{1 / 2} \tan ^{-1}\left(\frac{u}{\sqrt{1-u^{2}}}\right) \cdot \frac{d u}{\sqrt{1-u^{2}}} \\ &=4 \int_{0}^{\pi / 6} \tan ^{-1}\left(\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}\right) \frac{\cos \theta d \theta}{\sqrt{1-\sin ^{2} \theta}} \\ &=4 \int_{0}^{\pi / 6} \tan ^{-1}(\tan \theta) d \theta \\ &=4 \int_{0}^{\pi / 6} \theta d \theta=4 \cdot \frac{1}{2}\left(\frac{\pi}{6}\right)^{2}=\frac{\pi^{2}}{18} \end{aligned}

Hence summing the results gives the final answer:

\displaystyle \int_{0}^{1} \int_{0}^{1} \frac{1}{1-x y} d x d y = \sum_{n \geq 1} \frac{1}{n^{2}} =\frac{\pi^{2}}{18}+\frac{\pi^{2}}{9} =\frac{\pi^{2}}{6}

The Calabi Substitution:

A remarkably elegant proof, related to the double integral method, tackles a different sum first. Consider the sum over the odd integers, which can be represented as an integral over the unit square \displaystyle S=(0,1) \times (0,1):

\displaystyle \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \int_0^1 \int_0^1 \frac{dx\,dy}{1-x^2y^2}

Here is a clever change of variables proposed by Eugenio Calabi to evaluate the integral:

\displaystyle x = \frac{\sin u}{\cos v}, \quad y = \frac{\sin v}{\cos u}

This transformation maps the open triangle \displaystyle T = \{u, v > 0, u + v < \pi/2\} bijectively to the interior of the unit square \displaystyle S. The magic lies in its Jacobian determinant:

\displaystyle |J| = \det \begin{pmatrix} \frac{\cos u}{\cos v} & \frac{\sin u \sin v}{\cos^2 v} \\ \frac{\sin v \sin u}{\cos^2 u} & \frac{\cos v}{\cos u} \end{pmatrix} = 1 - \frac{\sin^2 u \sin^2 v}{\cos^2 u \cos^2 v} = 1 - x^2 y^2

The Jacobian is constructed to be precisely the denominator of the integrand. The integral therefore transforms with astonishing simplicity:

\displaystyle \iint_S \frac{dx\,dy}{1-x^2y^2} = \iint_T \frac{1}{1-x(u,v)^2 y(u,v)^2} |J|\,du\,dv = \iint_T 1 \,du\,dv

The problem is thus reduced to finding the area of the triangle \displaystyle T, which has vertices at \displaystyle (0,0), (\pi/2, 0), and \displaystyle (0, \pi/2). The area is:

\displaystyle \text{Area}(T) = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^2}{8}

The full Basel sum is now easily recovered by separating it into even and odd parts:

\displaystyle \sum_{n=1}^\infty \frac{1}{n^2} = \sum_{k=1}^\infty \frac{1}{(2k)^2} + \sum_{k=0}^\infty \frac{1}{(2k+1)^2} = \frac{1}{4}\sum_{k=1}^\infty \frac{1}{k^2} + \frac{\pi^2}{8}

Letting \displaystyle S = \sum_{n=1}^\infty \frac{1}{n^2}, we have \displaystyle S = \frac{1}{4}S + \frac{\pi^2}{8}, which gives \displaystyle \frac{3}{4}S = \frac{\pi^2}{8}, and thus \displaystyle S = \frac{\pi^2}{6}.

A Deeper Perspective: Polylogarithms and “Motives”

The question of why the Basel problem, \displaystyle \zeta(2), has so many elegant proofs while the value of \displaystyle \zeta(3) (Apéry’s constant) remains so mysterious can be understood through the modern lens of algebraic geometry and the theory of “motives”. This hierarchy of complexity is mirrored in the behavior of the special functions that generalize the zeta function, known as polylogarithms, defined by

\displaystyle \text{Li}_s(z)=\sum_{k=1}^{\infty} \frac{z^k}{k^s}.

Note that \displaystyle \zeta(s)=\text{Li}_s(1).

  • The Dilogarithm (\text{Li}_2): This function, related to \displaystyle \zeta(2), is governed by beautiful and relatively simple functional equations, like the “five-term relation.” These symmetries are the analytic reflection of the simple geometry of its underlying motive. Apostol’s proof for \displaystyle \zeta(2) works because the change of variables is a geometric trick that unlocks one of these simple dilogarithm identities, relating the integral to \displaystyle \pi^2.
  • The Trilogarithm (\text{Li}_3) and Beyond: The trilogarithm, related to \displaystyle \zeta(3), and higher polylogarithms have much more complicated and less understood functional equations. They lack the simple symmetries of the dilogarithm. This complexity is a direct consequence of the more intricate geometry of the “motives” they represent.

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