Question

What is `int_-oo^oo e^(-x^2)dx`?

Answer 1

`int_(-oo)^oo e^(-x^2) dx = sqrt(pi)`

Explanation:

The error function `"erf"(x)` is defined as follows:

`"erf"(x) = 2/sqrt(pi) int_0^x e^(-t^2) dt`

This non-elementary function is suitably scaled so that:

`lim_(x->-oo) "erf"(x) = -1`

`lim_(x->+oo) "erf"(x) = 1`

So

`int_(-oo)^oo e^(-x^2) dx = sqrt(pi)/2(lim_(x->+oo) "erf"(x) - lim_(x->-oo) "erf"(x))`

`=sqrt(pi)/2(1-(-1)) = sqrt(pi)`