What is the integral of e x raised to the power 2?

I had this problem when I was studying advanced mathematics. After searching through search engines, I found that it was difficult to find the correct answer to this question. Most of the respondents calculated it as definite integrals, while others used the integral method by parts and then uploaded an incorrect answer.
First of all, it is impossible to calculate the result using the integral method of parts . The result is as follows: < a i=4> ∫ e x 2 d x = x e x − ∫ x d e x 2 = x e x − ∫ e x 2 ⋅ 2 x d x \int e^{x^2}\,\mathrm{d}x =xe^x-\int x \,\mathrm{d}{e^{x^2}} =xe^x-\int e^{x^2}\cdot 2x\,\mathrm{d}x
$$\int {It\; is}^{x^2}\mathrm{d}x=x{e}^x-\int x\mathrm{d}{e}^{x^2}=x{e}^x-\int {It\; is}^{x^2}\cdot 2xdx$$
After this, there is no way to go on. The wrong person is probably going to $2x$extracted.

So does this function not have an original function?
According to the existence theorem of original function, it is not difficult to find ${It\; is}^{x^2}$ is a continuous function in the entire interval, so its original function must exist.

However, having a primitive function does not mean that it can be expressed in the form of basic elementary functions.

So we can consider using Taylor's formula to ${It\; is}^{x^{Expand\; 2}}$, calculate its convergence domain, and calculate the indefinite integral on its convergence domain.
① will ${It\; is}^{x^2}$ expands into a power series.
$${It\; is}^{x^2}=1+x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+\cdots +\frac{x^{2n}}{n!}+\cdots =\sum _{n=0}^{\infty }\frac{x^{2n}}{n!}$$
② Find the method based on the convergence domain of the power series:
If${\lim }_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|=\rho$ , 则 ${\sum }_{n=0}^{\infty }{a}_n{x}^n$ Target radius $R$ Definitely
$$R=\{\begin{array}{ll}\frac{1}{r},& r\ne 0\\ +\infty ,& r=0\\ 0,& r=+\infty \end{array}$$
Find the convergence radius R of the power series obtained in ①:
$$r=\underset{n\to \infty }{\lim }\left|\frac{\frac{1}{(n+1)!}}{\frac{1}{n!}}\right|=\underset{n\to \infty }{\lim }\frac{1}{n+1}=0\Rightarrow R=+\infty$$
则①中幂级继继浶嶺为$I=(-\infty ,+\infty )$。

③According to the properties of the power series summation function:
Power series${\sum }_{n=0}^{\infty }{a}_n{x}^n$ sum function $S\left(x\right)$ Where it is located < /span>$I$ 上可积，且有逐项积分公式
$${\int }_0^{x}S(t)\mathrm{d}t={\int }_0^x\sum _{n=0}^{\infty }{a}_n{t}^{n}dt=\sum _{n=0}^{\infty }{a}_n{\int }_0^x{t}^{n}dt=\sum _{n=0}^{\infty }\frac{a_n}{n+1}{x}^{n+1}(x\in I)$$
New parity number${\sum }_{n=0}^{\infty }\frac{a_n}{n+1}{x}^{n+1}$Area radius given source class number homology.

Inko${It\; is}^{x^2}$的不定积分终于可以计算了：
$$\begin{array}{rl}\int {It\; is}^{x^2}\mathrm{d}x& ={\int }_a^x{It\; is}^{t^2}\mathrm{d}t\\ & ={\int }_a^0{It\; is}^{t^2}dt+{\int }_0^x{It\; is}^{t^2}dt\\ & ={\int }_a^0{It\; is}^{t^2}dt+{\int }_0^x\sum _{n=0}^{\infty }\frac{t^{2n}}{n!}dt\\ & =\sum _{n=0}^{\infty }\frac{x^{2n+1}}{(2n+1)n!}+C,x\in (-\infty ,+\infty )\end{array}$$
Hope it helps readers