Finding the polynomials (or power series) $p_n(x)$ such that $2\int x^ne^{x^2}dx = p_n(x) e^{x^2}$ By using integration by parts, one can obtain the following equation

$$2\int x^ne^{x^2}dx=x^{n-1}e^{x^2}-(n-1)\int x^{n-2}e^{x^2}dx .$$

A recursive formula for $\int x^ne^{x^2}dx$ that works if and only if $n$ is odd because:
$$2\int xe^{x^2}dx=e^{x^2}$$
Now, upon using this formula, I obtained the following polynomials:
$$2\int x^1e^{x^2}dx=(1)e^{x^2}$$
$$2\int x^3e^{x^2}dx=(x^2-1)e^{x^2}$$
$$2\int x^5e^{x^2}dx=(x^4-2x^2+2)e^{x^2}$$
$$2\int x^7e^{x^2}dx=(x^6-3x^4+6x^2-6)e^{x^2}$$
The solution is always in the form of a polynomial multiplied by $e^{x^2}$.  I was wondering if this polynomial could have an easy solution in terms of the exponent on the left that may be easily found:
$$p_n(x)=x^{n-1}-(n-1)x^{n-3}+?x^{n-5}+\dots$$
Any easy pattern?
Then, I would like to see if this can be made into a continuous polynomial that works for even $n$ so that I may evaluate $\int e^{x^2}dx$ as a polynomial multiplied by $e^{x^2}$, which I can imagine this polynomial will have an infinite amount of terms, a power series.
$$\int x^0e^{x^2}dx=x^{-1}e^{x^2}+x^{-3}e^{x^2}+?x^{-5}e^{x^2}+\dots$$
I know the solution can not be shown in a finite amount of elementary functions, as shown on other related questions, but couldn't this work?

Is there an easy form of my power series?  After that, apply this to try and evaluate $\int x^0e^{x^2}dx$.  Determine if the formula works for the result.

 A: Computing one more example is suggestive:
$$2 \int x^9 e^{x^2} dx = (x^8-4x^6+12x^4-24x^2+24)e^{x^2},$$ and we can see that the ratios between successive coefficients are $-4, -3, -2, 1, -1$, leading to the guess
$$\color{#bf0000}{\boxed{2 \int x^{2 k + 1} e^{x^2} dx = \left[ \sum_{i = 0}^k (-1)^{k - i} \frac{k!}{i!} x^{2k} \right] e^{x^2}}}$$ (for integers $k \geq 0$).
We can prove that this formula holds for general $k$ using a standard induction argument and the recurrence relation given in the first display equation of the question.
I do not know how to use this to derive the power series for the analogous even-exponent integrals, i.e., for $\int_0^x t^{2k} e^{t^2} dt$.
We can proceed as follows using another method to recover the power series in $x$: The power series for $$e^{-x^2} \int_0^x t^{2k} e^{t^2} dt$$ is
$$e^{-x^2} \int_0^x e^{t^2} dt \sim \sum_{i = 0}^{\infty} (-1)^i \frac{4^i i!}{(2i + 1)^!} x^{2 i + 1}$$
(which we can establish by writing $e^{t^2}$ as a power series, formally integrating, multiplying by the power series for $e^{-x^2}$, and proving by induction on degree that the series agree). Then, the series for a general even power is
$$\color{#bf0000}{\boxed{e^{-x^2} \int_0^x t^{2k} e^{t^2} dt \sim \sum_{i = 0}^{\infty} (-1)^i \frac{4^i i!}{(2i + 1)^!} x^{2 k + 2 i + 1} }}.$$
A: Assume $$\int_a^{x}t^n e^{t^2} dt = p(x) e^{x^2}$$ 
for some polynomial function $p(x)$ (you can prove this). You're looking for $p(x)$ - you can just differentiate both sides wrt x to get 
$$x^n e^{x^2}=p'(x)e^{x^2}+2xp(x)e^{x^2}$$
$e^{x^2}$ is never 0 so just divide through and you have a linear ordinary differential equation.
$$x^n=p'(x) + 2xp(x)$$
The solution to this ODE involves an incomplete gamma function:
$$p(x)=\frac{(-1)^{\frac{-(n+1)}{2}}}{2}\Gamma(\frac{1+n}{2},-x^2)$$
For odd n, i.e., $m=\frac{n+1}{2}\in\mathbb{Z}$, the incomplete gamma function has a simple power series: $$\Gamma(m, -x^2)=(m-1)!e^{x^2}\sum_{k=0}^{m-1}\frac{(-1)^kx^{2k}}{k!}$$
Hence
$$p_m(x)=\frac{(-1)^m}{2}(m-1)!\sum_{k=0}^{m-1}\frac{(-1)^kx^{2k}}{k!}$$
A: Allow me to replace $x^2$ by $-x^2$ (corresponding to multiplying $x$ by $i$).
The (physicist's) Hermite polynomials $H_n(x)$ satisfy $H_0(x) = 1$, $H_1(x) = 2x$, and (Rodrigues's formula)
$$ H_n(x) e^{-x^2} = (-1)^n \dfrac{d^n}{dx^n} e^{-x^2} = - \dfrac{d}{dx} \left(H_{n-1}(x) e^{-x^2}\right)$$
Thus for $n \ge 1$, $$\int H_n(x) e^{-x^2}\;dx = - H_{n-1}(x) e^{-x^2} + C$$ 
Now you can convert between the powers $x^n$ and the Hermite polynomials
$$ \eqalign{x^n &= \dfrac{n!}{2^n} \sum_{m=0}^{\lfloor n/2 \rfloor} \dfrac{H_{n-2m}(x)}{m!(n-2m)!}\cr
H_n(x) &= n! \sum_{m=0}^{\lfloor n/2\rfloor} \dfrac{(-1)^m}{m!(n-2m)!} (2x)^{n-2m}}$$
so that if $n$ is odd $\int x^n e^{-x^2}\; dx$ is expressed as an odd polynomial times $e^{-x^2}$, while if $n$ is even it is an even polynomial times $e^{-x^2}$ plus a multiple of $\int e^{-x^2}\; dx$.
A: Maple says
$$
2\,\int \!{{\rm e}^{{x}^{2}}}\,{\rm d}x=
\big({x}^{-1}+\frac{1}{2}\,{x}^{-3}+\frac{3}{4}\,{x
}^{-5}
+{\frac {15}{8}}x^{-7}+\frac{105}{16}x^{-9}+O(x^{-11})\big)e^{x^2}
$$
as $x \to \infty$.
