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

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Answer 1 · 2015-11-29T23:19:35

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

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

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

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

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