### Welcome

Hi,

Welcome to my little math blog.

This is where I will dump thoughts on recreational mathematics, fun puzzles, and other stuff every so often.

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### A silly little derivation of $$\zeta(2)$$

(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