The $n$-th Gauss-Laguerre quadrature scheme aims to approximate integral of exponentially decreassing function over $[0 ; \infty[$ by a finite sum, according to: $$ \int _0 ^{+ \infty} e^{-x} f(x) \ dx \approx \sum _{i = 1} ^n \omega_i f(x_i) \ , $$ where $x_1$, $\cdots$, $x_n$ are the roots of the $n$-th Laguerre polynomial $L_n$ and the weights $\omega_1$, $\cdots$, $\omega_n$ are chosen according to $\omega_i = \dfrac{1}{x_i \big ( L_n'(x_i) \big )^2}$, $1 \leq i \leq n$.
$\bullet$ Let us denotes by $E_n(f)$ the error of the quadrature: $E_n(f) = \displaystyle { \int _0 ^{+ \infty} e^{-x} f(x) \ dx } - \displaystyle { \sum _{i = 1} ^n \omega_i f(x_i) } $. The general estimation of the error of gaussian quadratures, specialized in Gauss-Laguerre scheme, is the following:
For all $n \in \mathbb{N}$, there exists $\xi \in ]0 ; \infty[$ such that $E_n(f) = \dfrac{n!^2}{(2n)!} f^{(2n)}(\xi)$.
Unfortunately, this is unusable in many case, since we know nothing on this $\xi$. Therefore, we shall consider functions $f$ with derivatives satisfyng $||f^{2n}||_{\infty, \mathbb{R}^+} = \mathcal{o}\left ( \dfrac{1}{n!^2} \right )$... This is a bit restrictive...
$\bullet$ We also know the Uspensky theorem:
$E_n(f) \underset{n \longrightarrow + \infty}{\longrightarrow} 0$ for functions $f$ satisfying $|f(x)| \leq c \dfrac{e^x}{x^{1 + \rho}}$ for large $x >> 1$, where some $\rho > 0$.
(See J. V. Uspenksy, On the convergence of quadrature formulas related to an infinite interval, Trans. Amer. Math. Soc. 30 (1928), 542-559)
But, this do not gives explicit result on the convergence speed to $0$, nor a upper bound of $E_n(f)$ going to $0$.
$\bullet$ So, what is the most precise result about the error term? In particular, how can we know which degree $n$ should we use to find a numerical approximation of $\displaystyle { \int _0 ^{+ \infty} e^{-x + \sqrt{x}} \ dx}$.