What is the integral of #e^(2x^2)#?

1 Answer
Aug 3, 2015

Answer:

#int e^(2x^2) dx # cannot be expressed using elementary functions. You need the imaginary error function, erfi(x).

Explanation:

The imaginary error function is #2/sqrtpi int e^(x^2) dx#

#int e^(2x^2) dx # can be intergrated using substitution #u = sqrt2 x# so #du = sqrt2 dx# and we get:

#int e^(2x^2) dx = 1/sqrt2 int e^(u^2) du#

# = 1/sqrt2 sqrtpi/2 "erfi" u +C#

# = sqrt(2pi)/4 "erfi"(sqrt2x) +C#