How do you find the integral of #e^(x^2)#?

2 Answers
Apr 27, 2015

Use some kind of approximation method. There is no nice, finitely expressible antiderivative.

(Other that to write: #int e^(x^2) dx#, of course.)

Apr 29, 2015

One symbolic way to do it is to use infinite series. Since #e^{x}=1+x+x^{2}/{2!}+x^{3}/{3!}+\cdots=1+x+x^{2}/2+x^{3}/6+\cdots# (for all #x#), it follows that #e^{x^{2}}=1+x^{2}+x^{4}/2+x^{6}/6+\cdots# (for all #x#).

It is valid in this example to now integrate term-by-term (the result is true for all #x#):

#\int e^{x^{2}} dx=\int (1+x^{2}+x^{4}/2+x^{6}/6+\cdots) dx#

#=C+x+x^{3}/3+x^{5}/10+x^{7}/42+\cdots#.

Alternatively, you can also give the antiderivative a name. Wolfram Alpha writes the antiderivative whose graph goes through the origin as #\frac{\sqrt{\pi}}{2}\mbox{erfi}(x)#, where #\mbox{erfi}(x)# is called the "imaginary error function".