Euler for another day

“Every sum is a trace” is a well-known folklore saying in certain automorphic circles (echoed by a no less convincing “Every sum is an expectation” around probabilists); in this spirit, let’s have a look at one of the most famous sums

$latex \zeta(2)=\sum_{n\geq 1}{\frac{1}{n^2}},$

which Euler was the first to evaluate.

It is possible to see it, and then compute it, as a trace, and I’m sure this has been done many times; here is (a variant of) the way I presented things for an exercise for my Spectral Theory class.

Consider the Hilbert space

$latex H=L^2([0,1])$

(for the Lebesgue measure), and the Volterra operator T:

$latex Tf(x)=\int_0^x{f(y)dy}$

which is a linear operator acting on H, in fact a Hilbert-Schmidt operator with kernel

$latex k(x,y)=\mathbf{1}_{\{y\leq x\}}$

which is bounded, and therefore certainly belongs to the space

$latex L^2([0,1]^2,dxdy).$

The operator T is therefore compact, and the operator S=T^*T is also compact, and in fact positive, so that the trace is well-defined as a non-negative real number, or infinity. The trace is well-known to be expressible in different ways: (i) as a sum of the series formed with the eigenvalues of S (with multiplicity); (ii) as the sum of the series

$latex \sum_{n}{\langle S(f_n),f_n\rangle}=\sum_{n}{||Tf_n||^2},$

for an arbitrary choice of orthonormal basis (f_n)_n of H; (iii) as the integral

$latex \int_0^1\int_0^1{|k(x,y)|^2dxdy}.$

This last integral is of course completely elementary: it is the area of the lower-half triangle in the square [0,1]² below the diagonal; in other words, it is 1/2.

For an alternate expression (hence an identity), we look at the series above for the easiest orthonormal basis available:

$latex f_n(t)=e^{2i\pi nt}\quad\quad\text{ for } n\in\mathbf{Z}.$

For the special case n=0, we have

$latex Tf_0(x)=x,$

hence

$latex \langle Sf_0,f_0\rangle =||Tf_0||^2=\int_0^1{x^2dx}=\frac{1}{3}.$

For non-zero n, we have

$latex Tf_n(x)=\frac{e^{2i\pi nx}-1}{2i\pi n},$

and therefore (Parseval, if you wish, or direct computation):

$latex \langle Sf_n,f_n\rangle = ||Tf_n||^2=\frac{1}{4\pi^2 n^2}\times (1+1)=\frac{1}{2\pi^2n^2}.$

Summing over all n and identifying the two expression for the trace, we get

$latex \frac{1}{2}=\frac{1}{3}+\sum_{n\not=0}{\frac{1}{2\pi^2n^2}}=\frac{1}{3}+\frac{1}{\pi^2}\zeta(2),$

and hence — unsurprisingly, I presume — we get

$latex \zeta(2)=\frac{\pi^2}{6}.$

(I said unsurprisingly, but I first managed to get confused enough about the computation — for a slightly different operator — that, for a while, I almost convinced myself that ╬Â(2)=¤Ç²/12).

As a proof (which, I repeat, is certainly not new, though it is not found in this collection), this is fairly close in flavor to the Fourier-expansion proofs, where one expands (typically) the function x-1/2 on [0,1] into Fourier series before applying the Parseval identity. (In fact, it seems this is “dual” in some simple way which could be made precise).

Like the Fourier-expansion argument, it has the nice feature of showing almost immediately that it will be possible to generalize the argument to

$latex \zeta(2k)=\sum_{n\geq 1}{\frac{1}{n^{2k}}}$

for k> 0 integer, using T^k instead of T; it also hints quite strongly that the result will be of the type

$latex \zeta(2k)=\alpha_k\pi^{2k},$

for some rational number ╬▒_k. But it is equally obvious that this will not work at all like this for zeta evaluated at odd positive integers, as it should…