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

Question

6601 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-14T18:57:10

#int e^(-x^2) dx = sqrt(pi)/2 "erf"(x) + C#

#=sum_(n=0)^oo (-1)^n/((n!)(2n+1))x^(2n+1) + C#

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

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

So #int e^(-x^2) dx = sqrt(pi)/2 "erf"(x) + C#

Can we find out the value of the integral without using this special non-elementary function?

We can at least express it in terms of a power series:

#e^t = sum_(n=0)^oo t^n/(n!)#

Substituting #t = -x^2# we get:

#e^(-x^2) = sum_(n=0)^oo (-1)^n/(n!)x^(2n)#

So:

#int e^(-x^2) dx = int (sum_(n=0)^oo (-1)^n/(n!)x^(2n)) dx#

#=sum_(n=0)^oo (-1)^n/((n!)(2n+1))x^(2n+1) + C#

Hence the power series for #"erf"(x)# is simply given by:

#"erf"(x) = 2/sqrt(pi) sum_(n=0)^oo (-1)^n/((n!)(2n+1))x^(2n+1)#

since #"erf"(x)# is the integral from #0# to #x# and this sum is #0# when #x = 0#.