# Gaussian quadrature, with no exact result over polynomial, but on inverse functions

Generally, a Gaussian quadrature of degree $$n$$ over an interval $$I$$ is defined so that it integrates exactly polynomials up to degree $$2n - 1$$. The main tool are the orthogonal polynomials.

When $$I$$ is unbounded, the approximation of a continuous function $$f$$ over $$I$$ by polynomials could be bad. It seems that the vector space of polynomials where the integration over $$I$$ is exact could be changed to another vector space, so that the approximation over $$I$$ of $$f$$ by another family of functions is better than this with polynomials.

For example, the space $$\mathbb{R} \left [\dfrac{1}{X} \right ]$$ could be useful to develop quadratures over $$[1 ; + \infty[$$, i.e. a quadrature rule $$\int _1 ^{+ \infty} \omega(x) f(x) \ dx \approx \sum _{i = 1} ^n \omega_i f(x_i)$$ such that $$\int _1 ^{+ \infty} \dfrac{\omega(x)}{x^k} \ dx = \sum _{i = 1} ^n \dfrac{\omega_i}{x_i^k}$$ for all $$k \in \{0 ; \cdots ; 2n - 1\}$$, where $$\omega : I \longrightarrow \mathbb{R}^+$$ is a nonnegative continuous function, integrable over $$[1 ; +\infty[$$.

Another example could be to use trigonometric polynomials instead of polynomials.

Does someone know some references using such a change of point of view?

• If I am not missing something we can rewrite the integral with the substitution $y=1/x$ to get a standard Gauss quadrature problem with weight factor $y^2$. – user35593 Jan 19 at 19:23
• @user35593, you're right, we can perform a substitution to come back to a Gauss-Legendre standart rule. But this does not define a new quadrature rule. – MathTolliob Jan 19 at 23:13
• @user35593, in addition to my previous comment, the change of variable you suggest could creates some sigularities... The problem is that Gauss quadratures are known not to be the best quadrature scheme to take account of singularities... – MathTolliob Jan 26 at 12:39
• You can get a "new" quadrature rule by transforming back the nodes and using the same weights. I agree that there might be problem with a singularity. – user35593 Jan 26 at 12:57

## 1 Answer

The book

Stroud A. H., Secrest D., "Gaussian Quadrature Formulas". Prentice-Hall, Englewood Cliffs, N.J., 1966

gives some answer to my question.

See Section 3.2.2 "Finite to semi-infinite segment", where two quadratures over $$[0 ; + \infty[$$ which are exact for functions defined respectively by $$t \longmapsto \sum _{k = 1} ^M \dfrac{a_k}{(1 + t)^k} \ .$$ $$t \longmapsto \sum _{k = 1} ^M a_k e^{- k t} \ .$$

In section 3.2.3 "Semi-infinite to Finite segment", a quadrature over $$]0, 1]$$ is construted to be exact for functions $$t \longrightarrow \displaystyle { \sum _{k = 1} ^M a_k (\ln x)^k } \ .$$