(This is a cleaned-up and somewhat expanded version of this Twitter thread.)

What follows is a silly little proof that

\[ \zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \]

where \( \zeta \) is the Riemann zeta function.

Consider the integral

\[ I := \int_0^1 \frac{\log(1 - x + x^2)}{x(x - 1)} \, dx. \]

We have, by using partial fractions and performing some other algebraic manipulations,

\[ \begin{align*} I &= -\int_0^1 \! \frac{\log(1 - x + x^2)}{x} \, dx - \int_0^1 \! \frac{\log(1 - x + x^2)}{1 - x} \, dx \\

&= -2\int_0^1 \! \frac{\log(1 - x + x^2)}{x} & (x \mapsto 1 - x ) \\

&= 2\left( \int_0^1 \! \frac{\log(1 + x)}{x} \, dx - \int_0^1 \! \frac{\log(1 + x^3)}{x} \, dx \right) \\

&= \frac{4}{3}\int_0^1 \frac{\log(1 + x)}{x} \, dx & (x \mapsto x^{1/3}). \end{align*} \]

To evaluate this integral, we take the Maclaurin series:

\[ \int_0^1 \! \frac{\log(1 + x)}{x} \, dx = \int_0^1 \! \sum_{n=1}^{\infty} \frac{(-1)^nx^{n-1}}{n} \, dx \]

Since for all positive integers \( N \),

\[ \left|\sum_{n=1}^N \frac{(-1)^nx^{n-1}}{n}\right| \le \sum_{n=1}^{N} \frac{x^{n-1}}{n} \le \sum_{n=1}^{\infty} \frac{x^{n-1}}{n} = -\frac{\log(1 - x)}{x} \]

on \( [0, 1) \) and

\[ -\int_0^1 \! \frac{\log(1 - x)}{x} \, dx < -\int_0^{1/2} \! \frac{\log(1 - x)}{x} \, dx - 2\int_{1/2}^1 \! \log(1-x) \, dx < \infty, \]

we can invoke the dominated convergence theorem to switch summation and limit processes. We then have

\[ \begin{align*} \int_0^1 \! \frac{\log(1 + x)}{x} \, dx &= \sum_{n=1}^{\infty} \int_0^1 \frac{(-x)^{n-1}}{n} \, dx \\

&= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} = \eta(2) \end{align*} \]

where \( \eta(s) \) is the Dirichlet eta function.

Now, we have \( \eta(s) = 2^{1-s}\zeta(s) \); for \(s > 1 \), this can be seen by separating and rearranging all even terms in the summation. Thus, we have

\[ I = \frac{4}{3}\eta(2) = \frac{2}{3}\zeta(2). \]

We look at another way to evaluate \( I \). Noticing that \( 1 - x + x^2 = 1 - x(1 - x) \), we can write the integrand as a power series in \( x (1 - x) \):

\begin{align*} I &= -\int_0^1 \! \frac{\log(1 - x(1 - x))}{x(1 - x)} \, dx \\

&= \int_0^1 \sum_{n=1}^{\infty} \frac{x^{n-1}(1 - x)^{n-1}}{n} \, dx \\

&= \sum_{n=1}^{\infty} \int_0^1 \! \frac{x^{n-1}(1 - x)^{n-1}}{n}

\end{align*}

where we justify the switch of integral and sum by the monotone convergence theorem (since the summand is nonnegative on \([0, 1]\)). Recall the Euler Beta function given by

\[ \mathrm{B}(m,n) = \int_0^1 \! x^{m-1}(1 - x)^{n-1} \, dx = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}. \]

In the case that \( m \) and \( n \) are integers, we get

\[ \mathrm{B}(m,n) = \frac{(m - 1)!(n - 1)!}{(m + n - 1)!} \]

and in particular

\[ \int_0^1 \! \frac{x^{n-1}(1 - x)^{n-1}}{n} \, dx = \frac{\mathrm{B}(n,n)}{n} = \frac{((n - 1)!)^2}{n(2n - 1)!} = \frac{2}{n^2\binom{2n}{n}}. \]

Thus,

\[ I = 2\sum_{n=1}^{\infty} \frac{1}{n^2\binom{2n}{n}}. \]

The next part may strike you as something I pulled out of nowhere. We invoke the identity

\[ \arcsin^2{x} = \frac{1}{2}\sum_{n=0}^{\infty} \frac{(2x)^{2n}}{n^2\binom{2n}{n}}, \qquad x \in [-1, 1] , \]

a nice proof of which can be found by following the links starting here. (Okay, I'll admit it's not a very commonly-taught series, and the only reason I recognized it is that I used to spend too much time on AoPS. As such, I feel bad about blackboxing it like this. But it's cute!) We deduce that

\[ I = 4\arcsin^2\left(\frac{1}{2}\right) = \frac{\pi^2}{9}. \]

Thus, \( (2/3)\zeta(2) = \pi^2/9 \), and so \( \zeta(2) = \pi^2/6 \). QED.

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