One of the counterexamples that was proposed by Cauchy to LaGranges functional idea was $e^{-x^2}$. This function, when integrated only has a series solution that cannot be represented by aanother function. You can however, evaluate it as a definite integral from $-\infty$ to $\infty$

Let

$$I = \int_{-\infty}^{\infty} e^{-x^2}$$

It might be easier to evaluate its square, so lets do that instead.

$$I^2 = \int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy =$$

\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2}e^{-y^2}dx dy =

$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)}dxdy$$

Well notice that $dxdy$ is really $dA$ and if we switch to polar, $-(x^2 + y^2) = -r^2$

$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)}dxdy =$$

$$\int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2}rdrd\theta$$

Now we can do a $u$ substition. Let $u = -r^2$ and $du = -2r dr$.

$$\int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2}rdrd\theta$$

$$- \frac{1}{2} \int_{0}^{2\pi} \int_{0}^{-\infty} e^u du d\theta =$$

$$- 2\pi \frac{1}{2} \int_{0}^{-\infty} e^u du =$$

$$-\pi e^{-r^2} \bigg|_{0}^{\infty}$$

$$-\pi(e^{- \infty} - e^0) = \pi$$

Since $I^2 = \pi$, then $I = \sqrt{\pi}$. To summarize:

$$\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}$$

This is like the coolest thing Ive ever seen for a few reasons. It looks maliciously easy, especially to calc I students. Integrals and derivatives of $e$ are supposed to be easy. This also has a lot of applications in statistics with the gaussian. I haven’t taken a course in probability yet so I don’t really know. Theres also a neat trick where Feynman uses $ae^{-x^2}$ to derive an approximation of Schrodingers equation.

This post was also really just a testing ground to see if I could get mathjax working. It looks sucessful.