The integral of the Gaussian distribution
This one is just beautiful, links \(\pi\) and \(e\) together even though they really don’t seem related.
\[\begin{align*} \int{e^{-x^2} dx} &= \sqrt{\int{e^{-x^2} dx} \int{e^{-x^2} dx}} \\ \end{align*}\]Switching to a dummy variable:
\[\begin{align*} &= \sqrt{\int{e^{-x^2} dx} \int{e^{-y^2} dy}} \\ &= \sqrt{\int\int{e^{-(x^2 + y^2)} dx\, dy}} \\ \end{align*}\\\]We can now swap to polar coordinates by using well-known identities.
\[\begin{align*} x &= r \cos(\theta) \\ y &= r \sin(\theta) \\ dx\, dy &= r dr\, d\theta\\ \end{align*}\]Subbing everything in:
\[\begin{align*} \int\int{e^{-(x^2 + y^2)} dx\, dy} &= \int\int{e^{-(r \cos(\theta))^2 - (r \sin(\theta))^2} r dr\, d\theta } \\ &= \int\int{e^{-r^2 (\cos(\theta)^2+\sin(\theta)^2 )} r dr\, d\theta} \\ &= \int\int{e^{-r^2} r dr\, d\theta} \\ &= \int_0^{2\pi} d\theta \int_0^{\infty} e^{-r^2} r dr \\ &= 2\pi \int_0^{\infty} e^{-r^2} r dr \\ &= 2\pi \int_0^{\infty} e^{-u} \frac{1}{2} du \\ &= \pi\\ \text{And so:}\\ \int{e^{-x^2} dx} &= \sqrt{\pi} \end{align*}\]