The evaluation of Riemann’s zeta function

at even integers is well-known, going back to the venerable Leonhard Euler and his answer to the Basel problem, namely

The Basel problem stumped even the great Bernoulli’s, and Euler’s solution required the ingenuity and insight found throughout his work. Nearly three-hundred years later, is this easier to answer? I claim that not only is it easy, but that it is entirely routine to both guess the answer and to prove it, provided that you allow some basic Fourier analysis. Let’s see how.

Given a suitably nice function $f$ on $[0, 1]$, we define its Fourier transform by

These are the coefficients of the Fourier series

There is a deep theory about what functions are suitably nice and what properties the Fourier transform satisfies, but for now we are only interested in one fact:

The family $\{e^{2\pi i n x}\}_n = \{e(n)\}_n$ is an orthonormal basis for the square-integrable functions on $[0, 1]$ equipped with the usual integral inner product

\[(f, g) = \int_0^1 f \overline{g}.\]

In particular,

\[||f||_2 = \sum_k |(f, e(n))|^2 = \sum_k |\hat{f}(k)|^2.\]

In one direction, this tells us that sums whose terms are Fourier coefficients can be evaluated as an integral. This is exactly what we will do.

First, we need a humanly-proved lemma (though in principle a computer could likely figure this out).

Lemma. The Fourier transform of $f_n(x) = x^n$ at integers satisfies $\hat{f_n}(0) = (n + 1)^{-1}$, and $\hat{f}_n(k)$ is a polynomial in $(2\pi i k)^{-1}$ of degree $n$ for all nonzero integers $k$.

Proof. Apply integration by parts to prove that the sequence $\hat{f}_n(k)$ satisfies

The claim follows immediately by induction once we note that $\hat{f}_0(k) = 0$. $\blacksquare$

At this point we are in possession of a very powerful ansatz. The Fourier transform of polynomials at integers gives us reciprocals of powers of integers! There must be a connection with the zeta function. In particular, we want to find a polynomial

such that $\hat{f}(k)$ is some multiple of $k^{-n}$ times a factor independent of $k$. Given the lemma above, we know that $\hat{f}(k)$, for nonzero $k$, is a polynomial in $(2\pi i)^{-1}$, meaning that we can look at their coefficients and require ones smaller than $(2\pi i)^{-n}$ to vanish. We can stipulate that $a_n = -1$ and that $\hat{f}(0) = 0$, which gives us a total of $n + 1$ linear equations in $n + 1$ unknowns, which we can (probably) solve!

So, in light of this, it suffices to write a program that will equate these coefficients and solve the resulting linear equations. This will then, a posteriori, provide evaluations of $\zeta(n)$ for positive, even integers $n$. (Only the evens since, of course, we must take squares.)

The program

Here is one such program that will do the job.

import sympy as sp
from sympy.abc import x, t
from sympy import I, pi

def zeta_ansatz(n):
    xs = sp.symbols("a:{}".format(n + 1))
    f = sum(xs[k] * x**k for k in range(n + 1))

    k = sp.symbols("k", integer=True, zero=False)
    ft = sp.integrate(f * sp.exp(-2 * pi * I * k * x), (x, 0, 1))
    ft = ft.simplify().expand()
    # Replace 2 pi I with a dummy variable 1 / t to grab coefficients.
    ft = ft.subs(k, 1).subs(I, 1).subs(pi, 1 / t)

    coeffs = ft.collect(t, evaluate=False)
    zero_eqns = [coeff for key, coeff in coeffs.items() if key != t**n]
    eqns = zero_eqns + [xs[-1] + 1, sum(x / (k + 1) for k, x in enumerate(xs))]
    soln = sp.solve(eqns, xs)

    f = sum(soln[xs[k]] * x**k for k in range(n + 1))

    coeff = sp.factorial(n) / (2 * pi)**n

    return f, sp.integrate(abs(f)**2, (x, 0, 1)) / coeff**2 / 2

An example:

In [1]: time [zeta_ansatz(k) for k in range(1, 5)]
CPU times: user 1.85 s, sys: 19 µs, total: 1.85 s
Wall time: 1.85 s
Out[1]: 
⎡⎛          2⎞  ⎛               4⎞  ⎛          2        6⎞  ⎛                         8 ⎞⎤
⎢⎜         π ⎟  ⎜   2       1  π ⎟  ⎜   3   3⋅x    x   π ⎟  ⎜   4      3    2   1    π  ⎟⎥
⎢⎜1/2 - x, ──⎟, ⎜- x  + x - ─, ──⎟, ⎜- x  + ──── - ─, ───⎟, ⎜- x  + 2⋅x  - x  + ──, ────⎟⎥
⎣⎝         6 ⎠  ⎝           6  90⎠  ⎝        2     2  945⎠  ⎝                   30  9450⎠⎦

So there, we have answered the Basel problem plus evaluated the next three terms of the sequence $\zeta(2n)$, all in under two seconds and without any foreknowledge of the answer. Not too shabby, eh, Euler?

Connections to known results

These results are obviously not new. There are known closed-form evaluations of $\zeta(2n)$ for all positive integers $n$. A cursory glance suggests that the polynomials we get are exactly the negatives of the Bernoulli polynomials $B_n(x)$, and mentioned in that article is that the Fourier transform of $B_n(x)$ is

which matches exactly what we have said here.

Does this idea uniquely define the Bernoulli polynomials? The Fourier transform is invertible, so saying “let $B_n(x)$ be the preimage of such and such function under the Fourier transform on $[0, 1]$” is a fine definition. It is surprising that such a thing is a polynomial, but nevertheless true, and lucky for us that it was. Conversely, I am pretty sure that polynomials will only ever give you linear combinations of the $\zeta$ function evaluated at even integers, so we have completely exhausted the usefulness of polynomials coupled with Fourier transforms.