By the general of integration can we find the integration of #e^(x^2)# ?

1 Answer
Apr 5, 2018

# int_0^x \ e^(t^2) \ dt = sqrt(pi)/2 \ erfi(x)#

Explanation:

In theory any integral can be evaluated (either numerically, or using a powers series), However not every integral has an associated anti-derivative defined in terms of the well known elementary functions that all scientists know and love.

The integral:

# int \ e^(x^2) \ dx #

Falls into this category.

That is, there is no elementary function that we can differentiate to get the function #e^(x^2)#. As such the solution to the problem is to define such a new function, thus we define the error function :

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

Along with the Imaginary error function:

# erfi(x) := i \ erf(ix) = 2/sqrt(pi) \ int_0^x \ e^(t^2) \ dt #

And so the solution the the given integral is:

# int_0^x \ e^(t^2) \ dt = sqrt(pi)/2 \ erfi(x)#