How to compute $\min_{\{a, b, c\}}\int^{\infty}_0|x^3-a-bx-cx^2|^2e^{-x}\,dx$ I have no idea with $\min_{\{a, b, c\}}\int^{\infty}_0|x^3-a-bx-cx^2|^2e^{-x}\,dx$, especially the $e^{-x}$ part.

$$F(a,b,c) = \int_0^{\infty} \left(x^3 - a -bx - cx^2\right)^2 \exp(-x) dx$$
$$\dfrac{\partial F}{\partial a} = \int_0^{\infty} -2\left(x^3 - a -bx - cx^2\right) \exp(-x) dx = 0$$
$$\dfrac{\partial F}{\partial b} = \int_0^{\infty} -2x\left(x^3 - a -bx - cx^2\right) \exp(-x) dx = 0$$
$$\dfrac{\partial F}{\partial c} = \int_0^{\infty} -2x^2\left(x^3 - a -bx - cx^2\right) \exp(-x) dx = 0$$
This gives us
$$\Gamma(4) - a \Gamma(1) - b \Gamma(2) - c \Gamma(3) = 0$$
$$\Gamma(5) - a \Gamma(2) - b \Gamma(3) - c \Gamma(4) = 0$$
$$\Gamma(6) - a \Gamma(3) - b \Gamma(4) - c \Gamma(5) = 0$$
where $\Gamma(n) = (n-1)!$. Now solve the linear system to get your $a,b$ and $c$.
