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

Answer 1

Samantha Adkison

#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#

#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)#

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Answer 2

Christopher Allers

The integral of (e^{-x^2}) from negative infinity to positive infinity is equal to the square root of pi ((\sqrt{\pi})). This result is a fundamental part of the Gaussian integral, which has various applications in probability theory, statistics, and physics.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.