How do we calculate this exponential integral if we change limit from $\infty$ to $x_1$

Question

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}

Answer 1

Let

$$I(a) = \int_0^{x_1} dx \, e^{-a x^2} = \frac12 \sqrt{\frac{\pi}{a}} \text{erf}(\sqrt{a} x_1)$$

Then the integral in question is $-\partial I/\partial a$, which is

$$\frac14 \sqrt{\frac{\pi}{a^3}} \text{erf}(\sqrt{a} x_1) - e^{-a x_1^2} \frac{x_1}{2 a}$$

This may be used for even $n$, using successively higher derivatives with respect to $a$. For odd $n$, you may rather use derivatives with respect to $a$ of the integral

$$\int_0^{x_1} dx \, x \, e^{-a x^2} = \frac12 \frac{1-e^{-a x_1^2}}{a}$$

Answer 2

$\int_0^{x_1}x^2e^{-ax^2}~dx$

$=\int_0^{x_1}x^2\sum\limits_{n=0}^\infty\dfrac{(-1)^na^nx^{2n}}{n!}dx$

$=\int_0^{x_1}\sum\limits_{n=0}^\infty\dfrac{(-1)^na^nx^{2n+2}}{n!}dx$

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

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^na^n{x_1}^{2n+3}}{n!(2n+3)}$