# Three solutions to the Basel problem

The Basel problem was first posed in 1644 by Pietro Mengoli, author of the Novae quadraturae arithmeticae. It was to find the value of the sum

$\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \frac{1}{49} + \dots$

whose terms are the reciprocals of the positive square numbers.

Mengoli’s problem remained intractable for a lifetime (91 years) before it was famously solved by Leonhard Euler in 1735, who showed that

$\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi}{6}$

Later work by Bernhard Riemann defined the zeta function as

$\zeta(s) = \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^s}$

where $s \in \mathbb{C}$ is a complex number.

Thus the formula given by Euler is commonly written as $\zeta(2) = \frac{\pi}{6}$.

Proof 1:

We begin with Euler’s original proof of the theorem. Like all of Euler’s groundbreaking work it is a true jewel of the imagination.

Consider the MacLaurin expansion of $\sin(\pi x)$.

$\sin(\pi x) = \displaystyle\sum_{n=0}^{\infty} (-1)^{n+1} \frac{(\pi x)^{2n+1}}{(2n+1)!} = \pi x - \frac{(\pi x)^3}{3!} + \frac{(\pi x)^5}{5!} - \frac{(\pi x)^7}{7!} + \dots$

The solutions to $\sin(\pi x) = 0$ occur when $x \in \mathbb{Z}$. Thus we may factorise $\sin(\pi x)$ as

\begin{aligned} \displaystyle\sin(\pi x) & = \pi x \cdot \prod_{\substack{n \in \mathbb{Z} \\ n \neq 0}} \Big(1 - \frac{x}{n} \Big) \\ & = \displaystyle\pi x (1 - x)(1+x)\Big(1-\frac{x}{2}\Big)\Big(1+\frac{x}{2}\Big)\Big(1-\frac{x}{3}\Big)\Big(1+\frac{x}{3}\Big) \cdots \end{aligned}

The factor of $\pi$ follows from the MacLaurin expansion given above. Collecting pairs of factors as the difference of two squares we have

$\sin(\pi x) = \displaystyle \pi x (1 - x^2)\Big(1-\frac{x^2}{4}\Big)\Big( 1 - \frac{x^2}{9}\Big)\Big(1-\frac{x^2}{16} \Big)\Big(1-\frac{x^2}{25} \Big) \cdots$

The coeffecient of $x^3$ in the MacLaurin expansion of $\sin(\pi x)$ is $-\frac{\pi^3}{3!} = -\frac{\pi^3}{6}$. The coeffecient of $x^3$ in the product representation is

$- \pi \displaystyle\Big(1+\frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \dots \Big)$

as the contributions to the coefficient of $x^3$ come from the $\pi x$ factor multiplied with exactly one of the $x^2$ factors. Equating coefficients we have

$- \pi \displaystyle\Big(1+\frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \dots \Big) = -\frac{\pi^3}{6}$

and thus

$\displaystyle 1+\frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \dots = \frac{\pi^2}{6}$

as required.

Some technical issues are ignored in the proof above. Notably, how are we certain that the representation of the sine function as a product of linear factors is a permissible one? The necessary mathematical formalism to answer this question was provided by Weierstrass. See for example [4] which covers these details. Perhaps this treatment leaves some doubt as to the uniqueness of the series representation and the validity of equating coefficients in the final step of the proof. These are valid concerns which will not be dealt with here.

Proof 2:

The second proof is due to Apostol [1] and appeared in the in 1983 Mathematical Intelligencer. Apostol is a common noun to mathematical students everywhere owing to his seminal calculus textbook. Apostol’s proof has the advantage of being the work of a man rather than a magician. Apostol was lost to the world on May 8, 2016 at 92 years old. You may see for yourself (e.g in this lecture given in 2013) that he remained sharp until the end.

Consider the following integral

$I = \displaystyle\int_0^1 \int_0^1 \frac{1}{1-xy}~ \,dx \,dy$

We will show that $I = \zeta(2)$. Considering the domain of integration $(x,y) \in [0,1]^2$, we may write $\displaystyle\frac{1}{1-xy}$ as the geometric series

$\displaystyle\frac{1}{1-xy} = \sum_{n=0}^{\infty} (xy)^n$

and substituting this into the double integral we have

\begin{aligned} I & = \int_0^1 \int_0^1 \sum_{n=0}^{\infty} (xy)^n \,dx \,dy \\ & = \int_0^1 \sum_{n=0}^{\infty} y^n \Big[\frac{x^{n+1}}{n+1} \Big]_{x=0}^{x=1} \,dy \\ & = \int_0^1 \sum_{n=0}^{\infty} \frac{y^n}{n+1} \,dy \\ & = \sum_{n=0}^{\infty} \Big[ \frac{y^{n+1}}{(n+1)^2}\Big]_{y=0}^{y=1} \\ & = \sum_{n=0}^{\infty} \frac{1}{(n+1)^2} = \zeta(2) \end{aligned}

We now ‘double count’ by evaluating the integral in a different way. We rotate the coordinate axes by 45 degrees, so that the point $(x,y)$ maps to $(u,v)$ where

\begin{aligned} \begin{bmatrix} x \\ y \end{bmatrix} & = \begin{bmatrix} \cos(\frac{\pi}{4}) & -\sin(\frac{\pi}{4})\\ \sin(\frac{\pi}{4}) & \cos(\frac{\pi}{4}) \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} \\ & = \displaystyle \frac{1}{\sqrt{2}}\begin{bmatrix} u-v \\ u+v \end{bmatrix} \end{aligned}

so that $\displaystyle 1-xy$ becomes $\displaystyle 1- \frac{u^2 - v^2}{2} = \frac{2 - u^2 + v^2}{2}$.

The region of integration becomes the square with vertices $(0,0), (\pm 1/ \sqrt{2},1/ \sqrt{2}), (0,\sqrt{2})$ in the $xy$-plane.

Symmetry of the integral about the diagonal $u = v$ allows us to write

\begin{aligned} \displaystyle I & = 2 \int_0^{1/ \sqrt{2}} \Big( \int_0^u \frac{2}{2 - u^2 + v^2} \,dv \Big) \,du \\ & + 2 \int_{1/ \sqrt{2}}^{\sqrt{2}} \Big( \int_0^{\sqrt{2}-u} \frac{2}{2 - u^2 + v^2} \,dv \Big) \,du \end{aligned}

The reason for making this transformation is that is allows us to write the unknown integral in terms of a known formula for the inverse tangent function. Namely the identity

$\displaystyle \int_0^x \frac{\,dt}{a^2 + x^2} = \frac{1}{a} \arctan \frac{x}{a}$

Thus

$\displaystyle \int_0^u \frac{1}{2 - u^2 + v^2} = \frac{1}{\sqrt{2-u^2}} ~ \arctan \frac{u}{\sqrt{2-u^2}}$

and

$\displaystyle \int_0^{\sqrt{2}-u} \frac{1}{2 - u^2 + v^2} = \frac{1}{\sqrt{2-u^2}} ~ \arctan \frac{\sqrt{2}-u}{\sqrt{2-u^2}}$

so that

\begin{aligned} \displaystyle I & = 4 \int_0^{1 / \sqrt{2}} \frac{1}{\sqrt{2-u^2}} \arctan \frac{u}{\sqrt{2-u^2}} \,du ~~ + ~~ 4 \int_{1/ \sqrt{2}}^{\sqrt{2}} \frac{1}{\sqrt{2-u^2}} \arctan \frac{\sqrt{2}-u}{\sqrt{2-u^2}} \,du \\ \end{aligned}

Let these two integrals be $I_1$ and $I_2$ respectively. Making the substitution $u = \sqrt{2} \sin \theta$ in $I_1$ so that $\,du = \sqrt{2} \cos \theta$ it follows that $\frac{u^2}{2} = \sin ^2 \theta$ and by the Pythagorean identity $\frac{u^2}{2} + \cos^2 \theta = 1$  Hence $\,du = \sqrt{2} \cos \theta = \sqrt{2 - u^2}$ and therefore $\tan \theta = u / \sqrt{2-u^2}$. Substituting this information into $I_1$ gives

\begin{aligned} I_1 & = \displaystyle 4 \int_0^{\pi/6} \theta \,d\theta \\ & = 4 \Big[ \frac{\theta^2}{2}\Big]_{\theta = 0}^{\theta = \pi /6} \\ & = 2 \Big( \frac{\pi}{6} \Big)^2 \end{aligned}

The second integral $I_2$ may be handled using a similar process. We make the change of coordinates $u = \sqrt{2} \cos 2\theta$ so that

\begin{aligned} \,du & = - 2 \sqrt{2} \sin 2 \theta \,d\theta \\ & = -2 \sqrt{2} \sqrt{1 - \cos^2 2\theta} \,d\theta \\ & = -2 \sqrt{2 - u^2} \,d\theta \end{aligned}

Finally

\begin{aligned} \frac{\sqrt{2}-u}{\sqrt{2-u^2}} & = \frac{\sqrt{2}-\sqrt{2}\cos 2\theta}{\sqrt{2-2\cos^2 2\theta}} \\ & = \sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta }} \\ & = \sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}} = \tan \theta \end{aligned}

Substituting the information above into the integral $I_2$, the calculation is reduced to

$I_2 = \displaystyle 8 \int_0^{\pi/6} \theta \,d\theta = 4 \Big( \frac{\pi}{6}\Big)^2$

And therefore

$\zeta(2) = \displaystyle I_1 + I_2 = 6 \Big( \frac{\pi}{6} \Big)^2 = \frac{\pi^2}{6}$

as expected.

Maybe I was wrong to say Apostol was not a magician…

Proof 3

The final proof comes from a math.stackexchange post (and possibly elsewhere) where several other methods for showing $\zeta(2) = \pi / 6$ are also given. The proof relies on the expansion of the function $f(x) = x^2$ on the domain $-\pi \leq x \leq \pi$ into its Fourier series.

$\displaystyle \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx)$

where

$a_n = \displaystyle \frac{1}{\pi} \int_{-\pi}^{~ \pi} f(x) \cos (nx) \,dx$

for $n = 0,1,2,3, \dots$ and

$b_n = \displaystyle \frac{1}{\pi} \int_{-\pi}^{~ \pi} f(x)\sin (nx) \,dx$

for $n = 1,2,3, \dots.$ We note immediately that $b_n = 0$ for $n \in \mathbb{N}$ as $f(x)$ is an even function, $\sin(nx)$ is an odd function, and the product of an odd and even function is an odd function.

Calculating the initial coefficient

$a_0 = \displaystyle \frac{1}{\pi}\int_{-\pi}^{~ \pi} x^2 \,dx = \frac{2}{\pi} \int_0^{\pi} x^2 \,dx = \frac{2\pi^2}{3}$

For $n = 1,2,3,\dots$

\begin{aligned} a_n & = \displaystyle \frac{1}{\pi} \int_{-\pi}^{~ \pi} x^2 \cos (nx) \,dx \\ & = \frac{2}{\pi} \int_{0}^{\pi} x^2 \cos (nx) \,dx \end{aligned}

Integrating by parts twice gives (up to an additive constant)

\begin{aligned} \int x^2 \cos(nx) \,dx & = \frac{x^2 \sin(nx)}{n} - \int \frac{2x \sin(nx)}{n} \,dx \\ & = \frac{x^2 \sin(nx)}{n} + \frac{2x\cos(nx)}{n^2} - \int \frac{2 \cos (nx)}{n^2}\,dx \\ & = \frac{x^2 \sin(nx)}{n} + \frac{2x\cos(nx)}{n^2} - \frac{2 \sin (nx)}{n^3} \end{aligned}

As $\sin(2\pi n) = 0$ and $\cos(2\pi n) = (-1)^n$ for integer values of $n$ it follows that

\begin{aligned} a_n & = \displaystyle \frac{2}{\pi} \Big[\frac{2x \cos(nx)}{n^2} \Big]_0^{\pi} \\ & = \frac{4(-1)^n}{n^2} \end{aligned}

Thus we have the Fourier series expansion

$\displaystyle f(x) = \frac{\pi^2}{3} + \sum_{n=1}^{\infty} \Big( \frac{4(-1)^n}{n^2} \cos(nx) \Big)$

As $f(\pi) = \pi^2$

\begin{aligned} \displaystyle \pi^2 & = \frac{\pi^2}{3} + \sum_{n=1}^{\infty} \Big( \frac{4(-1)^n(-1)^n}{n^2} \Big) \\ & = \frac{\pi^2}{3} + 4 \sum_{n=1}^{\infty} \frac{1}{n^2} \end{aligned}

Thus $\displaystyle \zeta(2) = \frac{\pi^2}{4} - \frac{\pi^2}{12} = \frac{\pi^2}{6}$.

It is interesting that the evaluation occurs on the boundary of the domain. The final evaluation at $x = \pi$ seems like a remarkable confluence. This method may also be used to calculate the values of $\zeta(2n)$ for $n \in \mathbb{N}$. I still wonder what intuition guides these proofs. For now it will have to be enough that they are beautiful.

References

[1] Apostol, T. M. “A Proof That Euler Missed: Evaluating $\zeta(2)$ the Easy Way.” Math. Intel. 5, 59-60, 1983.

[2] F. Beukers,  “A note on the irrationality of $\zeta(2)$ and $\zeta(3)$, Bull. Lon. Math. Soc. 11, 268-272, 1979.

[3] O’Connor, J.J. and Robertson, E.F. “Pietro Mengoli” http://www-history.mcs.st-and.ac.uk/Biographies/Mengoli.html

[4] Sullivan, B.W. “Numerous Proofs of $\zeta(2) = \frac{\pi}{6}$http://math.cmu.edu/~bwsulliv/MathGradTalkZeta2.pdf