New Trapedoizal Tail integration method

[Albert Chan]

(05-13-2021 12:12 PM)Maximilian Hohmann Wrote:  But what if the runtime errors occur at those 1/3 points?
I guess that the probability for a runtime error will be the same for any point.

If singularity occurs inside integration limits, we *have* to break it up. (runtime errors or not)

This is an integral made famous by Richard Feynman.
(I saw this from Nahin's book "Inside Interesting Integrals", page 18)

\(\displaystyle \int _0 ^1 {dx \over [a\,x + b\,(1-x)]^2} = {1 \over a\,b}\)

If no singularity within the limits, integral is correct.
But, what if denominator goes zero ?

a*x + b*(1-x) = b + (a-b)*x = 0
x = b / (b-a)       // assumed a ≠ b, ok since a=b made integrand constant: 1/b^2 = 1/(ab)

If singularity occured inside integral limits:

0 < x < 1
0 < b < b-a

a < 0 and b > 0

Without break up integral to pieces, we have LHS > 0, RHS < 0, which make no sense.