Computation of $\int^{b}_{a} x^k e^{-x^2} \mathrm{d} x $ I am interested to know if there are methods to compute the definite integral $$\int^{b}_{a} x^k e^{-x^2} \mathrm{d} x$$ in closed form for some extended real numbers $a,b$ ($a<b$) and non-negative integer $k$.
It is also fine if the above integral can be expressed in terms of a function like $\mathrm{erf}$ whose values are widely known (or can be returned very fast by a computer), and hence is in closed form for 'practical' purposes.
I have answers for some very special cases:
(1) For $k = 0$ the integral, up to a scaling factor, is $\mathrm{erf}(b) - \mathrm{erf}(a)$.
(2) For $k = 1$ and $(a,b) \neq (-\infty,+\infty)$ the integral is $\frac{1}{2}(e^{-b^2} - e^{-a^2})$ since the derivative of $\frac{1}{2} e^{-x^2}$ w.r.t. $x$ is $x e^{-x^2}$.
(3) For the special case $a = -\infty, b = +\infty$, the above integral is the $k$th central moment of the standard normal distribution (up to a scaling factor) for which closed form formulas are known to exist.
 A: The derivative of $x^{k-1} e^{-x^2}$ is $((k-1)x^{k-2}-2x^k)e^{-x^2}$. That means that, given that we have primitives for $e^{-x^2}$ (the error function) and $xe^{-x^2}$, we can express the integral for a given $k$ by descending:
$$\int x^k e^{-x^2}dx = -\frac12x^{k-1} e^{-x^2} + C + \int\frac{k-1}{2}x^{k-2}e^{-x^2}dx$$
e.g. for $k=3$ this leads to
$$\int x^3 e^{-x^2}dx = -\frac12x^2 e^{-x^2} + C + \int xe^{-x^2}dx$$
$$=-\frac12x^2 e^{-x^2} + C + -\frac12e^{-x^2}$$
which is an elementary function (by induction, we only need elementary functions for all odd $k$), and for even $k$, the error function is needed in addition to elementary functions.
A: As said in comments, using $$x=\sqrt t\implies  dx=\frac{dt}{2 \sqrt{t}}$$ you have 
$$\int x^k e^{-x^2} \,dx=\frac{1}{2}\int  t^{\frac{k-1}{2}}e^{-t}\,dt=-\frac{1}{2} \Gamma \left(\frac{k+1}{2},t\right)$$ where appears the incomplete gamma function.
For the case of $$\int_{-\infty}^\infty x^k e^{-x^2} \,dx=\frac{1}{2} \left(1+(-1)^k\right)\, \Gamma \left(\frac{k+1}{2}\right)$$
A: If $k$ is odd, $k=2n+1$, then we can substitute $t=x^2$ to obtain
$$\int x^ke^{-x^2}\,dx=\int x^{2n+1}e^{-x^2}\,dx=\frac{1}{2}\int t^ne^{-t}\,dt,$$
which then can be integrated by parts $n$ times. It's not too difficult to figure out an inductive formula or maybe even an explicit formula for this case from here.
If $k$ is even, $k=2n$, we can let
$$I_{n}=\int x^{2n}e^{-x^2}\,dx,$$
and then apply integration by parts to derive inductive formulas for such integrals. We can set
$$u=x^{2k-2}e^{-x^2} \qquad \text{and} \qquad dv=x^2\,dx,$$
which will give us (skipping intermediate details) that
$$I_n=\frac{1}{3}x^{2n+1}e^{-x^2}-\frac{2n-2}{3}I_n+\frac{2}{3}I_{n+1},$$
or
$$I_{n+1}=-\frac{1}{2}x^{2n+1}e^{-x^2}+\frac{2n+1}{2}I_n.$$
This will lead us inductively down to the known error function integral $I_0=\int e^{-x^2}\,dx$.
