Calculator inputs
Example data table
| f(t) | Amplitude | s | a | b | Scenario | Expected estimate |
|---|---|---|---|---|---|---|
| A | 1 | 2.0 | 0 | ∞ | Constant | ≈ 0.5000 |
| sin(3t) | 1 | 1.0 | 0 | ∞ | Oscillatory with damping | ≈ 0.3000 |
| e^(0.4t) | 1 | 1.2 | 0 | ∞ | Growth slower than damping | ≈ 1.2500 |
| 3 + 2t | — | 1.5 | 0 | 6 | Finite interval polynomial | Numeric estimate |
These examples help you compare damping strength, interval choice, and numeric behavior before running your own case.
Formula used
The calculator evaluates a Laplace-type integral numerically. It supports finite intervals and an infinite upper limit approximated with a cutoff.
I(s) = ∫[a to b] e^(-st) · f(t) dtFor Simpson integration, the step size is:
h = (b - a) / nThe Simpson estimate is:
I ≈ (h / 3) · [y₀ + yₙ + 4(y₁ + y₃ + ...) + 2(y₂ + y₄ + ...)]The trapezoid estimate is:
I ≈ h · [0.5y₀ + y₁ + y₂ + ... + yₙ₋₁ + 0.5yₙ]Here, yᵢ = e-s tᵢf(tᵢ). A larger positive s usually increases damping and improves convergence on long intervals.
How to use this calculator
- Select the function form that matches your integrand source term.
- Enter the damping parameter s and the integration bounds.
- Check the infinite upper option when your interval extends to infinity.
- Set a cutoff value to approximate infinity in a controlled way.
- Choose Simpson for higher smooth-curve accuracy, or trapezoid for a simpler estimate.
- Fill amplitude, frequency, growth, or polynomial coefficients as needed.
- Press submit to show the result above the form, along with diagnostics and a graph.
- Use the CSV or PDF buttons to save results for reports or review.
FAQs
1) What does this calculator compute?
It estimates a Laplace-type integral of the form ∫e-stf(t)dt over finite limits or a cutoff approximation to infinity.
2) When should I use the infinite upper option?
Use it when the mathematical model extends to infinity. The tool replaces infinity with a practical cutoff and warns when convergence may be weak.
3) Why is positive damping important?
Positive damping usually suppresses tail values as t grows. That makes many infinite-interval integrals stable and easier to estimate accurately.
4) Which method is better, Simpson or trapezoid?
Simpson is usually more accurate for smooth curves. Trapezoid is simpler and still useful, especially when you want a quick estimate or a comparison.
5) What does the tail heuristic mean?
It is a rough indicator of the remaining contribution beyond the chosen cutoff. Smaller values suggest the missed tail area is less important.
6) Can I model oscillating functions?
Yes. Sine and cosine options are included. Damping helps control oscillatory tails and often improves the stability of the numeric estimate.
7) What if my function grows exponentially?
Then the damping parameter must dominate the growth rate for an infinite-interval case. Otherwise the integral may diverge or converge too slowly.
8) Why do more intervals matter?
More intervals create a finer mesh, which often improves precision. Extremely large counts may increase runtime without much added benefit.