One of the counter-examples that was proposed by Cauchy to LaGrange’s functional idea was . This function, when integrated only has a series solution that cannot be represented in some closed form. You can however, evaluate it as a definite integral from to .
Let
\begin{equation} I = \int_{-\infty}^{\infty} e^{-x^2} \end{equation}
It might be easier to evaluate its square, so lets do that instead.
\begin{equation} I^2 = \int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy = \end{equation}
\begin{equation}
\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2}e^{-y^2}dx dy =
\end{equation}
\begin{equation} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)}dxdy \end{equation}
Well notice that is really and if we switch to polar,
\begin{equation} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)}dxdy = \end{equation}
\begin{equation} \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2}rdrd\theta \end{equation}
Now we can do a substition. Let and .
\begin{equation} \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2}rdrd\theta \end{equation}
\begin{equation} - \frac{1}{2} \int_{0}^{2\pi} \int_{0}^{-\infty} e^u du d\theta = \end{equation}
\begin{equation} - 2\pi \frac{1}{2} \int_{0}^{-\infty} e^u du = \end{equation}
\begin{equation} -\pi e^{-r^2} \bigg|_{0}^{\infty} \end{equation}
\begin{equation} -\pi(e^{- \infty} - e^0) = \pi \end{equation}
Since , then . To summarize:
\begin{equation} \int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi} \end{equation}
This is like the coolest thing I’ve ever seen for a few reasons. It looks maliciously easy, especially to calc I students. Integrals and derivatives of 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 to derive an approximation of Schrodinger’s equation.
This post was also really just a testing ground to see if I could get mathjax working. It looks sucessful.