Integration of exponential function 4

Question

How can I find the integral $\int e^{x^2}dx$ in terms of finite terms?

I know that this integration can be solved by using power series technique.

I am stuck on it. Could anyone help me? Thanks.

Answer 1

If expressing your integral in terms of a special function constitutes writing it in finite terms, then a primitive for your integral can be written in terms of the so-called imaginary error function $\text{erfi} (x)$.

Defined as $$\text{erfi} (x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{t^2} \, dt,$$ we see that $$\int e^{x^2} \, dx = \frac{\sqrt{\pi}}{2} \text{erfi} (x) + C.$$

Note that despite the name imaginary error function, $\text{erfi}⁡ (x)$ is real when $x$ is real.

Answer 2

Not sure what finite terms means, but here is the power series technique you are talking about. Recall that $$ e^x = \sum_{k=0}^\infty \frac{x^k}{k!}, $$ thus $$ e^{x^2} = \sum_{k=0}^\infty \frac{\left(x^2\right)^k}{k!} = \sum_{k=0}^\infty \frac{x^{2k}}{k!}, $$ so you have $$ \int e^{x^2} dx = \int \sum_{k=0}^\infty \frac{x^{2k}}{k!} dx = \sum_{k=0}^\infty \int \frac{x^{2k}}{k!} dx = \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1) \cdot k!} + C $$ but I don't think there an analytic representation of this series.