Adaptive Simpson and Romberg integration methods

[Albert Chan]

(03-24-2021 08:54 PM)robve Wrote:  Improper integrals require integration over an open interval to avoid evaluating the endpoints, for example using Romberg with midpoint quadratures, see e.g. Numerical Recipes 2nd Ed. Ch.4.4 qromo and as implemented in the HP-71B Math Pac.

No. closed-intervals is fine, if we can assume end-points really is zero.
Normally, u-transformation will achieve it, but not always ...

A counter-example, from one of my old thread. Expected: \(\displaystyle \int _{-1}^1 {dx \over \sqrt{1-x^2}} = asin(x)\bigg| _{-1}^1 = \pi\)

(02-06-2020 11:16 PM)Albert Chan Wrote:  A simple test showed that HP71B INTEGRAL do extrapolations from trapezoids, not mid-points rule.

>10 DISP INTEGRAL(-1, 1, 1E-5, 1/SQRT(1-IVAR^2)), IBOUND
>RUN
3.14156045534              -3.1415397954E-5

this failure to converge (65535 sample points !) is due to missing end points evaluation.

\(\large \int _{-1} ^1 f(x) dx
= \int _{-1} ^ 1 {3 \over 2}(1-u^2) f \left({ u (3-u^2) \over 2} \right) du
= \int _{-1} ^ 1 g(u) du \)

\(f(x) = {1\over\sqrt{1-x^2}}\quad → \quad
g(u) = {3 (1-u^2) \over \sqrt{(4-u^2)(1-u^2)^2}}\)

\(\displaystyle{\lim_{u^2 \to 1^-} g(u)} = \displaystyle{\lim_{u^2 \to 1^-}{3\over\sqrt{4-u^2}}} = \sqrt3 ≠ 0\)