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Gaussian Quadrature for Numerically Computing Improper Integrals 📂Numerical Analysis

Gaussian Quadrature for Numerically Computing Improper Integrals

Definition 1

Gauss-Chebyshev Quadrature

1111x2f(x)dxi=1nwif(xi) \int_{-1}^{1} {{ 1 } \over { \sqrt{1 - x^2 } }} f(x) dx \approx \sum_{i=1}^{n} w_{i} f( x_{i} ) wi=πn w_{i} = {{ \pi } \over { n }} Here, xix_{i}s are the Chebyshev nodes that satisfy Tn(x)=0T_{n}(x) = 0.

Gauss-Laguerre Quadrature

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

Gauss-Hermite Quadrature

ex2f(x)dxi=1nwif(xi) \int_{-\infty}^{\infty} e^{-x^2} f(x) dx \approx \sum_{i=1}^{n} w_{i} f( x_{i} ) wi=2n1n!πn2[Hn1(xi)]2 w_{i} = {{ 2^{n-1} n! \sqrt{ \pi } } \over { n^2 \left[ H_{n-1} (x_{i} ) \right]^2 }} Here, xix_{i}s are the Hermite nodes that satisfy Hn(x)=0H_{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 ff does not contain 11x2\displaystyle {{ 1 } \over { \sqrt{1-x^2} }}, exe^{-x}, ex2e^{-x^2}, you can use the trick of making gg as follows. 0f(x)dx=0f(x)exexdx=0g(x)exdx \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. xi=cos(2i12nπ) 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. dx=πsinπtdt=π1x2dt \begin{align*} dx =& - \pi \sin \pi t dt \\ =& - \pi \sqrt{1 - x^2 } dt \end{align*} When performing such substitution integration, 1111x2f(x)dx=π1011x2f(x)1x2dt=π01f(cos(πt))dtπ1ni=1nf(cos(2i12nπ)) \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.


  1. Atkinson. (1989). An Introduction to Numerical Analysis(2nd Edition): p308. ↩︎