$\int e^ {x^2}dx$ : Integration of e to the power of x^2

Question

$$\int e^{x^2}dx$$

How does one integrate the expression above?

The upper limit of the integral is $3$ and the lower limit $3y$.It is actually a question on double integrals but I am only interested in antiderivative of the expression above as I have never seen it come up in my calculus studies. Thanks in advance.

I realize if it is does not have a simple antiderivative, how would one solve it if it was given as

$$\int^{2}_{0} e^{x^2}dx$$

Answer 1

If you're looking for an exact solution, you're probably out of luck. If you're looking for a numerical solution, you're going to want to use the equation $e^x = \sum_{n=0}^\infty\frac{x^n}{n!}$. So, for instance, for the example you gave, $\int_0^2e^{x^2}dx$, you would do

$$\int_0^2 e^{x^2}dx = \int_0^2\sum_{n=0}^\infty\frac{x^{2n}}{n!}dx = \sum_{n=0}^\infty\frac{2^{2n+1}}{(2n+1)n!} = 2 + \frac{8}{3} + \frac{16}{5} + \frac{64}{21} + \frac{64}{27} + \cdots \approx 16.5$$

I got the final answer from Wolfram, but if you really wanted to work it out mathematically, working out eight or more terms, doubling the final would give you an upper bound; in general, for an arbitrary positive limit of integration x, you would have to write out enough terms that for the last $\frac{(n+1)(2n+3)}{(2n+1)} > 2x^2$. (Since it's an even function, you don't really need to consider negative bounds.)

Answer 2

If you have the integration bounds, assume it is $x\in[a,b]$ then you basically now calculate $\int_a^{b}e^{x^2}dx$.

One simple way is using coordinate transformation,such as cylinder coordinate where you would have $x=rcos\theta$ ,$y=rsin \theta$.

Now denote $I = \int_a^{b}e^{x^2}dx$, then you would know $I^2=\int_a^{b} \int_a^{b}e^{x^2+y^2}dxdy$, then you consider coordinate transformation and you would get $I$.

It is worth noting that above method only apply the extreme cases, such as $[a,b]=[0,\infty]$, or $[a,b]=[-\infty,\infty]$, the reason for this is that for those extreme cases, the polar angle $\theta$ from coordinate transformation has a integrable region. But for regular cases, there is no integrable region for $\theta$.