HP 15C CE integration

[Albert Chan]

(09-21-2024 06:15 PM)carey Wrote:  Since \(\text{erf(65)} \approx 1 \) and \(\text{erf(3)} \approx 0.99997791 \), the integral becomes:

\[ \int_{3}^{65} e^{-t^2} \,dt = \frac{\sqrt\pi}{2} \times (\text{1 - 0.99997791}) \]

Perhaps it is better to skip subtraction, to avoid cancellation errors.
We may also use normal distribution functions.
Many scientific calculators had them, under statistics.

lua> x = 3
lua> sqrt(pi)/2 * erfc(x)
1.9577193236779756e-05
lua> sqrt(pi) * cdf(-sqrt(2)*x)
1.95771932367797e-05

65 treated as ∞, because ∫(e^(-t^2), t=65 .. ∞) ≈ e^(-65^2)/(2*65) ≈ 10^(-1837)
see A Continued Fraction for Error Function by Ramanujan, equation 3

Instead of treating upper limit as ∞, we may do opposite, make it as small as possible.

e^(-3^2)/(2*3) ≈ 2e-5
e^(-6^2)/(2*6) ≈ 2e-17

For 12 digits accuracy, integrate from 3 to 6 suffice.

>integ(3,6,1e-10,exp(-ix*ix))
1.95771932367E-5