Laplace transforms from the definition
Problem 4.1118 · medium
Use the definition \( \displaystyle F(s) = \int_0^\infty e^{-st} f(t)\,dt \) to find the Laplace transform of \( \displaystyle f(t) = 1 \), for \( \displaystyle s > 0 \).
- \[ \frac{\partial}{\partial t} \left(- \frac{e^{- s t}}{s}\right) = e^{- s t} \]An antiderivative of e^(−st) f(t), by differentiating it.✓ Proved
- For s > 0 the antiderivative tends to 0 as t → ∞ (the exponential decay wins).Reviewed
- \[ - \left. \frac{\left(-1\right) e^{- s t}}{s} \right|_{\substack{ t=0 }} = \frac{1}{s} \]So the integral is 0 minus its value at t = 0.✓ Proved
Answer \( F(s) = \frac{1}{s} \)
Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the defining integral computed by mpmath at two values of s |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly applies the definition of the Laplace transform, identifies the antiderivative, and correctly evaluates the limits at infinity and zero for s > 0.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly applies the definition of the Laplace transform, identifies the antiderivative, and correctly evaluates the limits at infinity and zero for s > 0.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-14gpt-oss:20b: pass 2026-10-14qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution fails to explicitly state the definition of the Laplace transform as an integral, which is the core requirement of the problem ('Use the definition...'). It jumps straight to an antiderivative check without setting up the definite integral \int_0^\infty e^{-st} \cdot 1 \, dt, making the logical connection between the problem statement and the calculation opaque.gpt-oss:20b: pass 2026-10-14
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/laplace_transform, checked 2026-10-17 with SymPy 1.14.0.