Why ignoring the function $f(x)=1/(1+x^2)$? I heard that the function $$f(x)=e^{-x^2}$$
Is extremely important in probability and statistics, because it looks like the normal distribution (or something like that). But i noticed that the graph of this function is similar to the function $$g(x)=\frac{1}{1+x^2}$$
Furthermore, the area under this curve is pretty easy to calculate, since
$$\int_{0}^z \frac{\,dx}{1+x^2}=\arctan(z).$$
So why the first function is more important than the second one, although they look pretty similar?
 A: I don't think this  function is being ignored.  It was studied by an Italian mathematician Agnesi, and before that Fermat, Newton and Grandi.  It has been called the "Witch of Agnesi". With no apparent connection to the circle, the area under it is $\pi$.  Try searching.  Apparently it is also useful in probability and statistics.
A: That function is also important in statistics. It is (a scaled version) of the $t$-distribution with one degree of freedom. See this Wikipedia article.
As the number of degrees of freedom approaches $\infty$, the $t$-distribution approaches the Normal distribution, which is a scaled version of $e^{-x^2}$. So the two functions are at different ends of a certain spectrum of functions.
A: They are not very similar.
First let us note that $\displaystyle\int_{-\infty}^{+\infty} e^{-x^2/2} \, dx = \sqrt{2\pi}$ and $\displaystyle\int_{-\infty}^{+\infty} \frac{dx}{1+x^2} = \pi. $
But $\displaystyle\int_0^{+\infty} x^n \cdot e^{-x^2/2} \, dx<+\infty,$ no matter how big $n>0$ is, whereas $\displaystyle\int_0^{+\infty} x\cdot\frac{dx}{1+x^2} = +\infty,$ and a fortiori $\displaystyle \int_0^{+\infty} x^2 \cdot\frac{dx}{1+x^2} = \infty$
The reason for dividing by $2,$ so that $x^2/2$ appears in the exponent rather than just $x^2,$ is that that makes the variance, and hence the standard deviation, equal to $1.$
A large number of theorems characterize the standard normal distribution $\dfrac 1 {\sqrt{2\pi}} e^{-z^2/2}\,dz. $ Here are two of them:

*

*Suppose $X_1,X_2,X_3,\ldots$ are independent and identically distributed random variables with expected value $\mu$ and variance $\sigma^2<+\infty.$ Then the distribution of $$ \frac{(X_1+\cdots+X_n)/n-\mu}{\sigma/\sqrt n} $$ approaches the standard normal distribution as $n\to\infty.$ In particular, the standard deviation of this sample mean is $\sigma/\sqrt n$ rather than $\sigma.$ Thus the dispersion of the sample mean goes to $0$ as the sample size grows.By contrast, if $X_1,X_2,X_3,\ldots$ are independent and are distributed as $\dfrac{dx}{\pi(1+x^2)},$ the Cauchy distribution then $(X_1+\cdots+X_n)/n$ also has that same Cauchy distribution. Its dispersion does not get smaller.


*Suppose the vector $(X_1,\ldots,X_n)$ has a probability density that depends on the $n$ arguments only through the sum of their squares, i.e. the distribution is spherically symmetric and centered at the origin. Suppose further that these $n$ scalar components are mutually independent. Then each of these random variables separately is normally distributed.
