Pace’s derivation of Euler’s sum of reciprocals of squares

One of my favourite identities in mathematics is the sum of the reciprocal of the squares
$$1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots = \frac{\pi^2}{6}.$$ This summation, first derived by Euler, is known as the Basel problem. It is perhaps the most natural sum to consider after the harmonic sum $1 + 1/2 + 1/3 + \cdots$, which diverges. Another way to write the sum of reciprocal of squares is with the Greek letter $\zeta$, as in
$$\zeta(2) = \sum_{k=1}^\infty\frac{1}{k^2}$$ because the sum of reciprocal of squares is in fact the value of the famous Riemann zeta function at $2$.

The sum of reciprocal of squares does not diverge, since we can bound the partial sums as
$$\sum_{k=1}^n \frac{1}{k^2} \leq \int_1^n \frac{1}{x^2}{\rm d}x + 1.$$
This integral inequality shows that
$$1\leq \zeta(2)\leq 3.$$ This is a pretty crude bound as $\pi^2/6 = 1.6449…$. So how do we go about showing that? One interesting proof was shown by Luigi Pace in [1], published in the American Mathematical Monthly, which we now explain.

We first recast the problem in terms of random variables. All of the random variables will be positive continuous random variables. That is, random variables that can be specified by a probability density function on the interval $(0,\infty)$. Let $X_1$ and $X_2$ be two such random variables, specified by probability density functions $p_{1}$ and $p_2$ respectively. A standard derivation in the theory of probability, which follows directly from Fubini’s theorem, is that the density function of the random variable $Y = X_1/X_2$ if $X_1$ and $X_2$ are independent, is given by
$$p_Y(y) = \int_0^\infty xp_1(xy)p_2(x){\rm d}x.$$ We are going to apply this formula to a specific distribution, the so-called half-Cauchy distribution. By definition, $X$ has such a distribution if its probability density function is given by
$$p(x) = \frac{2}{\pi(1 + x^2)}.$$ That this is a valid probability density function can be verified by using that the derivative of $\tan^{-1}(x)$ is $(1 + x^2)^{-1}$. I’ll leave that as an exercise.

So, now we can use our previous formula and consider the distribution for $Y = X_1/X_2$ where $X_1$ and $X_2$ are independent and identically distributed according to the half-Cauchy distribhution. This states that the density function of $Y$ is given by
$$p_Y(y) = \int_0^\infty \frac{4x}{\pi^2(1 + x^2y^2)(1 + x^2)}{\rm d}x.$$ This integral looks a little tricky. One way to solve it is to try a bunch of primitives involving logarithms. Looking at the denominator, we see we’ll need factors like $(1 + x^2y^2)$ and $(1 + x^2)$ in there so let’s try:
$$\log\left(\frac{1 + x^2y^2}{1 + x^2}\right).$$ Its derivative is easily verified to be
$$\frac{2x(y^2 – 1)}{(1 + x^2y^2)(1 + x^2)}.$$ That is pretty lucky, and of course it looks easier once I don’t include a lot of random scribbles, but from my experience this technique works decently. Therefore, we come back to our original integral expression for $p_Y(y)$, and we determine that
$$p_Y(y) = \frac{-4\log(y)}{\pi^2(1 – y^2)}.$$
Now, let us consider how we defined $Y$: it is the ratio of two half-Cauchy variables $X_1$ and $X_2$. That is, $Y = X_1/X2$. Therefore, $P(Y \leq 1) = 1/2$. Therefore, we obtain the formula
$$\int_0^1 \frac{-\log(y)}{1 – y^2}{\rm d}y = \frac{\pi^2}{8}.$$
We can transform the left-hand side of this equation by noting that $1/(1 – y^2)$ can be replaced by $1 + y^2 + y^4 + \cdots$, as we are on the interval $(0,1)$. Also, we can integrate term by term, again by Fubini’s theorem. Integrating term by term is a simple integration by parts, and this gives us:
$$\sum_{k=0}^\infty\frac{1}{(2k + 1)^2} = \frac{\pi^2}{8}.$$ This is very close to the original Basel problem of determining $\zeta(2)$. We can get back to the original Basel problem by noticing that
$$\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \cdots = \frac{1}{4}\zeta(2).$$ Therefore, we have
$$\frac{3}{4}\zeta(2) = \frac{\pi^2}{8},$$ or equivalently,
$$\zeta(2) = \frac{\pi^2}{6}.$$

Admittedly, the key stroke of this problem was to consider the half-Cauchy distribution, and that is certainly not an obvious step. Still, there is some sense to it: even though some density functions such as the one we derived for $Y$ are complicated, integrating it on some intervals might be straightforward, and this was enough in this case to give us the information that we needed. I believe something more systematic can be made of this method, and if anyone has any pointers to literature, please share them in the comments.

[1] Pace, Luigi. Probabilistically proving that $\zeta(2)=\pi^2/6$. Amer. Math. Monthly 118 (2011), no. 7, 641–643.

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