Integral with conditions Compute
$$\displaystyle\min_{a,b,c} \displaystyle\int_{0}^{\infty} \left | x^3-a-bx-cx^2 \right |^2e^{-x}\, dx$$
Please, any suggestions are welcome.
Thanks you all.
 A: Hint:
with the scalar product (for instance, of the space of polynomial functions defined on $(0,\infty  )$)
$$
f,g \to \int_0^\infty f(x)g(x) e^{-x}dx
$$
$x\to a+bx+cx^2$ is the orthogonal projection of $x\to x^3$ on $\text{span }\{
x\to 1, x\to x, x\to x^2
\}$

to solve this, see that if $\pi f$ is the orthogonal projection of $f$,
\begin{align}
\langle \pi f - f, x\to x^2\rangle &=0\\
\langle \pi f - f, x\to x\rangle &=0\\
\langle \pi f - f, x\to 1\rangle &=0\\
\end{align}
which are three equations for 3 unknowns:
\begin{align}
\int_0^\infty e^{-x}[x^3 - a - bx - cx^2] x^2dx &=0\\
\int_0^\infty e^{-x}[x^3 - a - bx - cx^2] x dx &=0\\
\int_0^\infty e^{-x}[x^3 - a - bx - cx^2] dx &=0\\
\end{align}
which you can simplify using $$
\int_0^\infty x^n e^{-x}dx = n!
$$to give
\begin{align}
120 - 2a - 6b- 24c &=0\\
24 - a - 2b- 6c &=0\\
3 -  a - b- 2c &=0\\
\implies a=-3, b=-9, c&= 7.5
\end{align}
A: Expand the square (the absolute value disappears), and you will obtain a polynomial times a negative exponential. Use the result on the Gamma function for integer arguments.
You'll get
$$6!-2c5!+c^24!-2b4!+2bc3!-2a3!+b^22!+2ac2!+2ab1!+a^20!.$$
Cancel the gradient and solve the $3\times 3$ linear system
$$2a+2b+4c-12=0$$
$$2a+4b+12c-48=0$$
$$4a+12b+48c-240=0$$
