Basel's Problem - New proof and some calculus

Sorry for the long hold up on another post. This new post is a result on my investigations on a proof of \(\zeta(2) = \frac{\pi^2}{6}\) by Daniele Ritelli. Yes, it's another post on something Euler solved first. But I did things different this time around: I discussed the few (advanced) Calculus theorems that played crucial roles in the proof.

Theorem. \(\displaystyle \qquad \zeta(2) = \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} \)

Proof: The proof relies on evaluating the integral \(\displaystyle \int_0^\infty \int_0^\infty \frac{1}{(1+y)(1+x^2 y)} \, dA \) in two different ways, first with \(dA = dx\, dy\) and second with \(dA = dy\, dx\). The first is easiest since \(\dfrac{1}{1+y}\) can be treated as a constant and factored out the inner integral to give:
\[ \begin{aligned} \int_0^\infty \int_0^\infty \frac{1}{(1+y)(1+x^2 y)} \, dx \, dy &= \int_0^\infty \frac{1}{1+y} \int_0^\infty \frac{1}{1+x^2 y} \, dx \, dy \\ &= \int_0^\infty \frac{1}{1+y} \int_0^\infty \frac{1}{1+x^2 y} \, dx \, dy \\ &= \int_0^\infty \frac{1}{1+y} \left. \left[\frac{\tan^{-1}\left(x\sqrt{y}\right)}{\sqrt{y}} \right] \right|_0^\infty \, dy \\ &= \int_0^\infty \frac{\pi}{(1+y) \cdot 2\sqrt{y}}\, dy \\ &= \pi \left. \tan^{-1}\left(\sqrt{y}\right)\right|_0^\infty \\ &= \frac{\pi^2}{2} \end{aligned}\]

To evaluate the integral using second method, partial fractions come to the rescue and treat \(x\) as a constant: \[ \frac{1}{(1+y)(1+x^2 y)} = \frac{ ^1/_{1-x^2}}{1+y} + \frac{^{x^2}/_{x^2-1}}{1+x^2 y} \qquad (x \neq 1) \]

Therefore,
\[ \begin{aligned} \int_0^\infty \int_0^\infty \frac{1}{(1+y)(1+x^2 y)} \, dy \, dx &= \int_0^\infty \frac{1}{1-x^2} \int_0^\infty \frac{1}{1+y} - \frac{x^2}{1+x^2y} \, dy \, dx \\ &= \int_0^\infty \frac{1}{1-x^2} \left. \left[ \ln\left| \frac{1+y}{1+x^2 y} \right| \right] \right|_0^\infty \, dx \\ &= \int_0^\infty \frac{-2\ln x}{1-x^2} \, dx \\ &= \int_0^1 \frac{-2\ln x}{1-x^2} \, dx + \int_1^\infty \frac{-2\ln x}{1-x^2} \, dx \\ &= \int_0^1 \frac{-2\ln x}{1-x^2} \, dx + \int_0^1 \frac{-2\ln u}{1-u^2} \, du \quad \text{substitution } u=x^{-1} \\ &= \int_0^1 \frac{-4\ln x}{1-x^2} \, dx \end{aligned} \]

Recalling the first half of the proof, we can now set: $$ \int_0^1 \frac{-\ln x}{1-x^2} \, dx = \frac{\pi^2}{8} $$

Using the geometric expansion \(\displaystyle \frac{1}{1-x^2} = \sum_{n=0}^\infty x^{2n}\) defined for \(x \in (-1,1)\), then:

\[ \begin{aligned} \frac{\pi^2}{8} &= \int_0^1 \frac{-\ln x}{1-x^2} \, dx \\ &= \int_0^1 \sum_{n=0}^\infty -x^{2n} \ln x \, dx \\ &= \sum_{n=0}^\infty \int_0^1-x^{2n} \ln x \, dx \\ &= \sum_{n=0}^\infty \left( \lim_{a\rightarrow 0^+} \left.\left[ \frac{-x^{2n+1} \ln x}{2n+1} \right] \right|_a^1 + \int_0^1\frac{x^{2n}}{2n+1} \, dx \right) \\ &= \sum_{n=0}^\infty \left. \left[ \frac{x^{2n+1}}{(2n+1)^2} \right] \right|_0^1 \\ &= \sum_{n=0}^\infty \frac{1}{(2n+1)^2} \\ &= \zeta(2) - \frac{1}{4}\zeta(2) \\ &= \frac{3}{4}\zeta(2) \\ \zeta(2) &= \frac{\pi^2}{6} \end{aligned} \] \(\Box\)

The rest of the post consists of various important steps and their justifications. This was a great opportunity to see (1) the subtlety involved in infinite processes and (2) the care that needs to be taken when writing proofs.

The very first line of the proof already requires a rather advanced theorem of Calculus. The fact that $$ \iint_R f(x,y) \, dA = \iint_R f(x,y) \, dx \, dy = \iint_R f(x,y) \, dy \, dx $$
should not be taken for granted. It is a result of our \(f(x,y)\) being positive and continuous on the region \(R = [0,\infty) \times [0, \infty)\). My Calculus textbook gives without proof the analogous theorem applied to bounded regions and refers to advanced calculus texts for a proof. A counterexample I saw for the bounded region version:

\[ \int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2 + y^2)^2} \, dx \, dy \neq \int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2 + y^2)^2} \, dy \, dx \] You can evaluate the two integrals to verify for yourself. These integrals fail to behave as I expected because \(f(x,y)\) is not continuous or even defined at the origin. Fortunately, if you let \(0 < b <1\) then: \[ \int_b^1 \int_b^1 \frac{x^2-y^2}{(x^2 + y^2)^2} \, dx \, dy = \int_b^1 \int_b^1 \frac{x^2-y^2}{(x^2 + y^2)^2} \, dy \, dx = 0 \]

In my Probability and Stats class, we change the order of integration however we want and I had assumed this was given by the definition of double integrals. It's actually justified because the probability densities are nonnegative and continuous on the plane.

The second step I will justify is the partial fraction. As you may have noticed, the partial fraction as written requires \(x \neq 1\) but the partial fraction in the integral should be defined for \([0, \infty)\). The proper way to fix this would have been to split the original as I did later on: \[ \int_0^\infty \int_0^\infty f(x,y) \, dy \, dx = \int_0^1 \int_0^\infty f(x,y) \, dy \, dx + \int_1^\infty \int_0^\infty f(x,y) \, dy \, dx \]

But since the partial fraction consisted of two terms, I chose to save space and postpone the split for later. This was mostly an aesthetic decision.

I'll save the next big step for last and deal with a rather minor one: Is it justified that I claimed: \[ \sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \sum_{n=1}^\infty \frac{1}{n^2} - \frac{1}{4} \sum_{n=1}^\infty \frac{1}{n^2} \]

Well it would definitely be true if I knew that the summations on the right-hand side converged. For this, one need only use the integral test. Since \(\int_1^\infty x^{-2} \, dx = 1\) then \(\zeta(2)\) exists. In fact, it would be nice to check this first before starting the proof. Know a limit exists before attempting to find it.

Now the most crucial of steps, if you have not guessed it by now is the simple-looking: \[ \int_0^1 \frac{-\ln x}{1-x^2}\, dx = \sum_{n=0}^\infty \int_0^1-x^{2n} \ln x \, dx \]

I suspected the justification lay somewhere in the fact that series was uniformly convergent and somehow uniform convergence allows one to “extract” summation signs from integrals. In the referenced article, the author claims the step follows from the “monotone convergence theorem”. Thinking he was referring to the elementary version I learnt in class, I struggled for weeks trying to justify it with this in mind. Eventually, I found out that the theorem he mentioned was another advanced theorem involving Lebesgue integrals. Without the drive to try and learn measure theory from scratch I returned to the uniform convergence idea. What follows is my attempt to justify the step without recourse to Lebesgue integrals. I would not be surprise if there is a gap in my proof.

I did a lot of research on what uniform convergence really means; read the theorems and proofs that explain how uniform convergent function sequences/series behave as we'd expect. Eventually, I realized the functions I had did not uniformly converge on \([0,1]\); they were not even defined at the origin. Fixing this problem simply let \(\displaystyle \frac{-\ln x}{1- x^2} = -\ln x + \frac{-x^2 \ln x}{1-x^2}\).

Unfortunately, it is still not uniformly convergent since the series converges to the discontinuous function: $$ \sum_{n=1}^\infty -x^{2n} \ln x = \begin{cases} 0 & x=0\\ {~} \\ \dfrac{-x^2\ln x}{1-x^2} & 0<x<1 \\ {~} \\ 0 & x = 1 \end{cases} $$

However, it is indeed uniformly convergent on any interval \([0,k]\) when \(0<k<1\). So for any such \(k\): $$ \int_0^k \frac{-x^2\ln x}{1-x^2} \, dx = \sum_{n=1}^\infty \int_0^k -x^{2n} \ln x \, dx $$

By the boundedness theorem: $$ \left| \frac{-x^2 \ln x}{1-x^2 } \right| < M \qquad \text{ for some } M $$

And since each term of the series is positive, for all \(m \in \mathbb{N}\) and all \(x \in [0,1]\): $$ 0 \le \sum_{n=1}^m -x^{2n} \ln x \le \frac{-x^2 \ln x}{1-x^2} < M $$
So: $$ 0 \le \sum_{n=1}^m \int_k^1 -x^{2n} \ln x , dx \le \int_k^1 \frac{-x^2\ln x}{1-x^2} , dx < M(1-k) $$

Since we can make the difference between the two integrals arbitrarily small by choosing an appropriate \(k \in (0,1)\), then it must be that: $$ \int_0^1 \frac{-x^2\ln x}{1-x^2} \, dx = \sum_{n=1}^\infty \int_0^1 -x^{2n} \ln x \, dx $$

This concludes the commentary. In my readings on these calculus topics, I've come to learn so much about the limitations of what one can do with Riemann Integrals. I get why there was a need for Lebesgue integrals and hopefully I will find some time to get acquainted with it. Lastly, the two big theorems I used to justify steps in this post are so rich they deserve standalone posts. I can guarantee this is not the last you'll see on uniform convergence on this blog.

Bibliography
Ritelli, Daniele. "Another Proof of \(\zeta(2)=\frac{\pi^2}{6}\) Using Double Integrals." Amer. Math. Monthly 120.7 (2013): 642-645.