Apostol's Solution to the Basel problem
The Basel problem asks for the solution to the infinite series of the reciprocal squares:
$$ \sum_{n=1}^\infty \frac 1 {n^2} = \, ? $$As many other things in mathematics, it was first solved by Euler, who found that the solution was, incredibly, the transcendental value:
$$ \frac{\pi^2}{6} $$If you read through the linked Wikipedia page, you’ll see many (quite beautiful) proofs, but they tend to rely on some decently heavy mathematical machinery.
I was going through some old maths notes I made, and stumbled across an elegant proof from Tom M. Apostol1 that I wanted to reproduce here.
Start by rewriting the sum so each term can be expressed as a product of two simple integrals:
$$ \begin{aligned} \sum_{n=1}^{\infty} \frac{1}{n^2} &= \sum_{n=0}^{\infty} \frac{1}{(n+1)^2} &&\text{(reindexing)} \\ &= \sum_{n=0}^{\infty} \left[\frac{x^{n+1}}{n+1}\right]_0^1\left[\frac{y^{n+1}}{n+1}\right]_0^1 \\ &= \sum_{n=0}^{\infty} \left(\int_0^1 x^n \, \mathrm{d}x\right)\left(\int_0^1 y^n \, \mathrm{d}y\right) \\ &= \sum_{n=0}^{\infty} \int_0^1 \int_0^1 (xy)^n \, \mathrm{d}x \, \mathrm{d}y && \text{(Fubini)} \\ &= \iint_{[0,1]^2} \sum_{n=0}^{\infty} (xy)^n \, \mathrm{d}A_{x,y} && \text{(Fubini)} \\ &= \iint_{[0,1]^2} \frac{1}{1-xy} \, \mathrm{d}A_{x,y} &&\text{(geometric sum)} \end{aligned} $$Where we interchange the series and the integrals by Fubini’s theorem, since $(xy)^n \ge 0$ on $[0,1]^2$.
Now use the linear change of variables $x=u-v$, $y=u+v$:
The Jacobian determinant has absolute value $2$, and the square maps to a diamond-shaped region $\Delta$.
$$ \begin{aligned} \sum_{n=1}^{\infty} \frac{1}{n^2} &= \iint_{\Delta} \frac{1}{1-(u-v)(u+v)}\left|\frac{\partial(x,y)}{\partial(u,v)}\right| \, \mathrm{d}u \, \mathrm{d}v \\ &= \iint_{\Delta} \frac{1}{1-u^2+v^2} \left|\begin{matrix}1 & -1 \\ 1 & 1\end{matrix}\right| \, \mathrm{d}u \, \mathrm{d}v \\ &= 2\iint_{\Delta} \frac{1}{1-u^2+v^2} \, \mathrm{d}u \, \mathrm{d}v \\ &= 2\int_0^{1/2} \int_{-u}^{u} \frac{1}{1-u^2+v^2} \, \mathrm{d}v \, \mathrm{d}u \\ &\quad + 2\int_{1/2}^{1} \int_{u-1}^{1-u} \frac{1}{1-u^2+v^2} \, \mathrm{d}v \, \mathrm{d}u. \end{aligned} $$By symmetry in $v$, fold each inner integral to the positive side:
$$ \begin{aligned} \sum_{n=1}^{\infty} \frac{1}{n^2} &= 4\int_0^{1/2} \int_0^{u} \frac{1}{1-u^2+v^2} \, \mathrm{d}v \, \mathrm{d}u \\ &\quad + 4\int_{1/2}^{1} \int_0^{1-u} \frac{1}{1-u^2+v^2} \, \mathrm{d}v \, \mathrm{d}u \\ &= 4\int_0^{1/2} \frac{1}{\sqrt{1-u^2}} \arctan\!\left(\frac{u}{\sqrt{1-u^2}}\right) \, \mathrm{d}u \\ &\quad + 4\int_{1/2}^{1} \frac{1}{\sqrt{1-u^2}} \arctan\!\left(\frac{1-u}{\sqrt{1-u^2}}\right) \, \mathrm{d}u. \end{aligned} $$We can evaluate these integrals with a trigonometric substitution:
$$ \begin{aligned} 4\int_0^{1/2} \frac{1}{\sqrt{1-u^2}} \arctan\!\left(\frac{u}{\sqrt{1-u^2}}\right) \, \mathrm{d}u &= 4\int_0^{\pi/6} \frac{1}{\cos\theta} \arctan\!\left(\frac{\sin\theta}{\cos\theta}\right) \cos\theta \, \mathrm{d}\theta \\ &\text{where } u=\sin\theta,\; \mathrm{d}u=\cos\theta\,\mathrm{d}\theta \\ &= 4\int_0^{\pi/6} \arctan(\tan\theta) \, \mathrm{d}\theta \\ &= 4\int_0^{\pi/6} \theta \, \mathrm{d}\theta. \end{aligned} $$and
$$ \begin{aligned} 4\int_{1/2}^{1} \frac{1}{\sqrt{1-u^2}} \arctan\!\left(\frac{1-u}{\sqrt{1-u^2}}\right) \, \mathrm{d}u &= 8\int_0^{\pi/6} \frac{1}{\sin(2\theta)} \arctan\!\left(\frac{1-\cos(2\theta)}{\sin(2\theta)}\right) \sin(2\theta) \, \mathrm{d}\theta \\ &\text{where } u=\cos(2\theta),\; \mathrm{d}u=-2\sin(2\theta)\,\mathrm{d}\theta \\ &= 8\int_0^{\pi/6} \arctan\!\left(\frac{2\sin^2\theta}{2\sin\theta\cos\theta}\right) \, \mathrm{d}\theta \\ &= 8\int_0^{\pi/6} \arctan(\tan\theta) \, \mathrm{d}\theta \\ &= 8\int_0^{\pi/6} \theta \, \mathrm{d}\theta. \end{aligned} $$Finally, putting both pieces together gives:
$$ \begin{aligned} \sum_{n=1}^{\infty} \frac{1}{n^2} &= 12\int_0^{\pi/6} \theta \, \mathrm{d}\theta = 12\left[\frac{\theta^2}{2}\right]_0^{\pi/6}\\ &= \boxed{\frac{\pi^2}{6}} \end{aligned} $$Note: I can’t actually access the original text (paywalled, alas), so I hope this is faithful to Apostol’s derivation. I’m not sure where I came across this argument initially, but I suspect it was in my Vector Calculus class.
Tom M. Apostol, “A Proof that Euler Missed: Evaluating $\zeta(2)$ the Easy Way”, The Mathematical Intelligencer (1983). DOI: https://doi.org/10.1007/BF03026576. ↩︎