📂Numerical Analysis

Gaussian Quadrature for Numerically Computing Improper Integrals

Definition ¹

Gauss-Chebyshev Quadrature

$$ \int_{-1}^{1} {{ 1 } \over { \sqrt{1 - x^2 } }} f(x) dx \approx \sum_{i=1}^{n} w_{i} f( x_{i} ) $$ $$ w_{i} = {{ \pi } \over { n }} $$ Here, $x_{i}$s are the Chebyshev nodes that satisfy $T_{n}(x) = 0$.

Gauss-Laguerre Quadrature

$$ \int_{0}^{\infty} e^{-x} f(x) dx \approx \sum_{i=1}^{n} w_{i} f( x_{i} ) $$ $$ w_{i} = {{ x_{i} } \over { (n+1)^2 \left[ L_{n+1} (x_{i} ) \right]^2 }} $$ Here, $x_{i}$s are the Laguerre nodes that satisfy $L_{n}(x) = 0$.

Gauss-Hermite Quadrature

$$ \int_{-\infty}^{\infty} e^{-x^2} f(x) dx \approx \sum_{i=1}^{n} w_{i} f( x_{i} ) $$ $$ w_{i} = {{ 2^{n-1} n! \sqrt{ \pi } } \over { n^2 \left[ H_{n-1} (x_{i} ) \right]^2 }} $$ Here, $x_{i}$s are the Hermite nodes that satisfy $H_{n} (x) = 0$.

Explanation

The Gauss quadrature is excellent in itself, and it can perform calculations even when the integration range is infinitely given, by choosing nodes wisely.

For improper integrals, if $f$ does not contain $\displaystyle {{ 1 } \over { \sqrt{1-x^2} }}$, $e^{-x}$, $e^{-x^2}$, you can use the trick of making $g$ as follows. $$ \begin{align*} \int_{0}^{\infty} f(x) dx =& \int_{0}^{\infty} f(x) e^{x} e^{-x} dx \\ =& \int_{0}^{\infty} g(x) e^{-x} dx \end{align*} $$ Let’s look at the motive of the Gauss-Chebyshev quadrature. Chebyshev nodes are given as follows. $$ x_{i} = \cos \left( {{ 2i-1 } \over { 2n }} \pi \right) $$ Of course, these nodes are discrete, but if viewed continuously, it can be accepted as variable substitution as follows. $$ \begin{align*} dx =& - \pi \sin \pi t dt \\ =& - \pi \sqrt{1 - x^2 } dt \end{align*} $$ When performing such substitution integration, $$ \begin{align*} \int_{-1}^{1} {{ 1 } \over { \sqrt{1 - x^2 } }} f(x) dx =& - \pi \int_{1}^{0} {{ 1 } \over { \sqrt{1 - x^2 } }} f(x) \sqrt{1 - x^2 } dt \\ =& \pi \int_{0}^{1} f \left( \cos ( \pi t ) \right) dt \\ \approx & \pi {{ 1 } \over { n }} \sum_{i=1}^{n} f \left( \cos \left( {{ 2 i - 1 } \over { 2n }} \pi \right) \right) \end{align*} $$ Of course, this is not an accurate derivation, and it is not necessarily related to the Gauss quadrature either. The purpose of the above formula development is to verify that the idea of choosing special nodes can be similar to substitution.