What is the integral of #e^(2x^2)#?
1 Answer
Aug 3, 2015
Explanation:
The imaginary error function is
# = 1/sqrt2 sqrtpi/2 "erfi" u +C#
# = sqrt(2pi)/4 "erfi"(sqrt2x) +C#
The imaginary error function is
# = 1/sqrt2 sqrtpi/2 "erfi" u +C#
# = sqrt(2pi)/4 "erfi"(sqrt2x) +C#