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

Apr 27, 2015

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

(Other that to write: $\int {e}^{{x}^{2}} \mathrm{dx}$, 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 ${e}^{{x}^{2}} = 1 + {x}^{2} + {x}^{4} / 2 + {x}^{6} / 6 + \setminus \cdots$ (for all $x$).

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

$\setminus \int {e}^{{x}^{2}} \mathrm{dx} = \setminus \int \left(1 + {x}^{2} + {x}^{4} / 2 + {x}^{6} / 6 + \setminus \cdots\right) \mathrm{dx}$

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

Alternatively, you can also give the antiderivative a name. Wolfram Alpha writes the antiderivative whose graph goes through the origin as $\setminus \frac{\setminus \sqrt{\setminus \pi}}{2} \setminus m b \otimes \left\{e r f i\right\} \left(x\right)$, where $\setminus m b \otimes \left\{e r f i\right\} \left(x\right)$ is called the "imaginary error function".