Adaptive Simpson and Romberg integration methods

[Albert Chan]

(03-25-2021 11:57 PM)Valentin Albillo Wrote:  Now, as announced above, a few examples from my own suite, I1 to I7, for you to try ...

All integrals is hitting the assumption that end-points does not matter.
After u-transformation, integrand end-points mean (t1) is still not zero.
First 5 can be fixed with *another* round of u-transfomation.

∫(f(x), x=0..1) = ∫(6*u*(1-u)*f(u*u*(3-2*u)), u=0..1)

> 10 DEF FNU(U) @ N=N+1 @ FNU=6*U*(1-U)*FNF(U*U*(3-2*U)) @ ENDDEF
> 20 N=0 @ DISP INTEGRAL(0,1,P,FNU(IVAR)),N

> 1 DEF FNF(X)=COS(X)*LN(X) @ REM I1 = -Si(1)
> RADIANS @ P=1E-8 @ RUN
-.946083070248       511

> 1 DEF FNF(X)=1/SQRT(X) @ REM I2 = 1/0.5 = 2
> P=1E-10 @ RUN
2                             127

> 1 DEF FNF(X)=LN(GAMMA(X)) @ REM I3 = ln(2*pi)/2
>P=1E-9 @ RUN
.918938533197       1023

>1 DEF FNF(X)=1/SQRT(-LN(X)) @ REM I4 = sqrt(pi)
>P=1E-6 @ RUN
1.77245405564       31

>1 DEF FNF(X)=LN(SIN(X)) @ REM I5 = re(1j/2*Li2(exp(2j)))-ln(2)
>P=1E-9 @ RUN
-1.05672020598      1023

Last 2 still failed with extra round of u-transformation ...

>1 DEF FNF(X)=X^(-0.8) @ REM I6 = 1/0.2 = 5
>P=1E-8 @ RUN
4.99943927847      65535

>1 DEF FNF(X)=X^(-0.99) @ REM I7 = 1/0.01 = 100
>P=1E-7 @ RUN
35.9675524685      65535