What is the integral of the error function?

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Answer 1 · 2016-02-20T01:39:40

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

We will use the definition of the error function:

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

Let #u = "erf"(x)# and #dv = dt#

#du = 2/sqrt(pi)e^(-x^2)# and #v = x#

By the integration by parts formula #intudv = uv - intvdu#

#int"erf"(x)dx = x"erf"(x) - int2/sqrt(pi)xe^(-x^2)dx#

To evaluate the remaining integral, let #u = -x^2#

Then #du = -2xdx# and so

#-int2/sqrt(pi)xe^(-x^2)dx = 1/sqrt(pi)inte^udu#

#=1/sqrt(pi)e^u + C#

#=e^(-x^2)/sqrt(pi)+C#

Putting it all together, we get our final result:

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