Modified the Gauss–Laguerre quadrature in the case of $\int\limits_{x=0}^{+\infty}{{e^{-ax}}f(x)}dx$

laguerre-polynomialsnumerical methodsroots

1/ How to modified the integral Gauss–Laguerre quadrature rule so that we could approximate the following integral:

$I = \int\limits_{x = 0}^{ + \infty } {{e^{ – ax}}f\left( x \right)} dx$

The things that make me feel uneasy is that although the exponential term is scaled by a positive constant $a$.

In that case what should we do to modified the original Gauss–Laguerre quadrature that is

$I = \int\limits_{x = 0}^{ + \infty } {{e^{ – x}}f\left( x \right)} dx \approx \sum\limits_{i = 1}^{i = n} {{w_i}f\left( {{x_i}} \right)}$

Where ${{x_i}}$ is the i-th root of the Laguerre polynomial ${L_n}\left( x \right)$ and the weight is calculated by

${w_i} = \frac{{{x_i}}}{{{{\left( {n + 1} \right)}^2} \times {{\left[ {{L_{n + 1}}\left( {{x_i}} \right)} \right]}^2}}}$

2/ What does it really mean by saying the "the i-th root of the Laguerre polynomial". Suppose I was looking for the root of
${L_3}\left( x \right) = \frac{1}{6}\left( { – {x^3} + 9{x^2} – 18x + 6} \right)$ then between 3 of these root

$\begin{gathered}
x \approx 0.416 \hfill \\
x \approx 2.294 \hfill \\
x \approx 6.290 \hfill \\
\end{gathered} $

Which root would be consider as the ${1^{st}},{2^{nd}},{3^{rd}}$. In my opinion, the order seem to be quite irrelevant in this case

Please help me with this, thank you very much !

Best Answer

  1. The Gauss-Laguerre formula is for integrals of the form $\int_0^{+\infty}e^{-x} f(x) dx$. In your case, you just need to perform a change of variable $y = ax$ $$ \int_0^{+\infty} e^{-ax} f(x) dx = \frac 1a \int_0^{+\infty} e^{-y} \underbrace{f(y/a)}_{= \tilde f(y)} \,dy, $$ and apply the usual rule for $\tilde f$, i.e. $$ \int_0^{+\infty}e^{-ax} f(x) dx \approx \frac 1a \sum_{i=1}^n w_i f\left(\frac{x_i}{a}\right). $$

  2. The order of the points is irrelevant... as long as you assign the correct weights.

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