Laplace transforms from the definition
Problem 4.1209 · medium
Use the definition \( \displaystyle F(s) = \int_0^\infty e^{-st} f(t)\,dt \) to find the Laplace transform of \( \displaystyle f(t) = t e^{- 2 t} \), for \( \displaystyle s > -2 \).
- \[ \frac{\partial}{\partial t} \frac{\left(t \left(- s - 2\right) - 1\right) e^{t \left(- s - 2\right)}}{\left(- s - 2\right)^{2}} = t e^{- 2 t} e^{- s t} \]An antiderivative of e^(−st) f(t), by differentiating it.✓ Proved
- For s > -2 the antiderivative tends to 0 as t → ∞ (the exponential decay wins).
- \[ - \left. \frac{\left(t \left(- s - 2\right) - 1\right) e^{t \left(- s - 2\right)}}{\left(- s - 2\right)^{2}} \right|_{\substack{ t=0 }} = \frac{1}{\left(s + 2\right)^{2}} \]So the integral is 0 minus its value at t = 0.✓ Proved
Answer \( F(s) = \frac{1}{\left(s + 2\right)^{2}} \)
Lines: 2 proved, 1 not checked. 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 | Not checked | — | 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: fail (misleading) — The evaluation at t=0 gives –1/(s+2)^2, not +1/(s+2)^2, so the conclusion that the integral equals (s+2)^(–2) is incorrect unless the sign error is corrected.qwen3.6:27b-mlx: pass — The solution correctly applies the Fundamental Theorem of Calculus using the provided antiderivative. The limit at infinity is justified by the condition s > -2, and the evaluation at t=0 is algebraically correct.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly applies the Fundamental Theorem of Calculus using the provided antiderivative. The limit at infinity is justified by the condition s > -2, and the evaluation at t=0 is algebraically correct.gpt-oss:20b: fail (misleading) 2026-10-18 — The evaluation at t=0 gives –1/(s+2)^2, not +1/(s+2)^2, so the conclusion that the integral equals (s+2)^(–2) is incorrect unless the sign error is corrected.qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly applies the Fundamental Theorem of Calculus using the provided antiderivative. The limit at infinity is correctly identified as 0 for s > -2, and the evaluation at t=0 yields the correct result.gpt-oss:20b: fail (misleading) 2026-10-16 — The sentence claims the antiderivative at t=0 equals (s+2)^{-2}, but it is actually -(s+2)^{-2}. This sign error would mislead a student about the evaluation of the integral.qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the antiderivative, justifies the boundary condition at infinity for the given domain, and correctly evaluates the definite integral.gpt-oss:20b: pass 2026-10-15
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 2026-10-18 with SymPy 1.14.0.