How do we calculate this exponential integral if we change limit from $\infty$ to $x_1$ We know the solutions of this integral (Bronstein-Semendijajev mathematics manual [page474]): 
\begin{align}
\int\limits_{0}^{\infty}x^n \cdot e^{-ax^2}dx = \frac{1\cdot3\dots(2k-1)\,\,\sqrt{\pi}}{2^{k+1}a^{k+1/2}}\longleftarrow\substack{\text{$n$ is the exponent over $x$}\\\text{while $k=n/2$}}
\end{align}
But how do I calculate the integral if I change the upper limit from $\infty$ to $x_1$ which is constant - for example how do I calculate this integral:
\begin{align}
\int\limits_{0}^{x_1}x^2 \cdot e^{-ax^2}dx
\end{align}

$\int_0^{x_1}x^2e^{-ax^2}~dx$
$=\int_0^{x_1}x^2\sum\limits_{n=0}^\infty\dfrac{(-1)^na^nx^{2n}}{n!}dx$
$=\int_0^{x_1}\sum\limits_{n=0}^\infty\dfrac{(-1)^na^nx^{2n+2}}{n!}dx$
$=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^na^nx^{2n+3}}{n!(2n+3)}\right]_0^{x_1}$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^na^n{x_1}^{2n+3}}{n!(2n+3)}$
