What is #int_-oo^oo e^(-x^2)dx#?
1 Answer
Nov 29, 2015
Explanation:
The error function
#"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)#