HP 15C CE integration

[carey]

(09-21-2024 05:42 PM)Thomas Klemm Wrote:  

(09-21-2024 04:45 PM)AnnoyedOne Wrote:  FYI in high school I tried to solve the integral analytically. I never succeeded and to my knowledge no-one else ever has either.

That's no surprise due to Liouville's theorem:

Quote:The antiderivatives of certain elementary functions cannot themselves be expressed as elementary functions. These are called nonelementary antiderivatives. A standard example of such a function is \(e^{-x^{2}}\) whose antiderivative is (with a multiplier of a constant) the error function, familiar from statistics.

Yes, while there is no closed-form solution to the integral of \( e^{-x^2} \) in terms of the elementary functions that appear on standard calculator keys, a closed-form solution exists in terms of special functions (e.g., the error function in this case).

Step 1: Start with the definition of the error function: \[\text{erf(x)} = \frac{2}{\sqrt\pi} \int_{0}^{x} e^{-t^2} \,dt\]

Step 2: Multiply both sides by \(\frac{\sqrt\pi}{2}\) and switch sides: \[ \int_{0}^{x} e^{-t^2} \,dt = \frac{\sqrt\pi}{2}\text{erf(x)} \]

Step 3: Convert the definite integral to an indefinite integral by the addition of a constant C. \[ \int {e^{-t^2}} \,dt \ = \frac{\sqrt\pi}{2}\text{erf(x)} + C \]

This is the closed-form solution of the integral in terms of the error function.

To check, we can evaluate this closed-form solution between limits of 3 and 65 as in the original question:

Since \(\text{erf(65)} \approx 1 \) and \(\text{erf(3)} \approx 0.99997791 \), the integral becomes:

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

\[ = 1.958 \times 10^{-5} \]

as expected.