∫Calc Practice

Laplace transforms from the definition

Problem 4.1210 · medium

Use the definition \( \displaystyle F(s) = \int_0^\infty e^{-st} f(t)\,dt \) to find the Laplace transform of \( \displaystyle f(t) = \sin{\left(t \right)} \), for \( \displaystyle s > 0 \).
  1. \[ \frac{\partial}{\partial t} \frac{\left(- s \sin{\left(t \right)} - \cos{\left(t \right)}\right) e^{- s t}}{s^{2} + 1} = e^{- s t} \sin{\left(t \right)} \]
    An antiderivative of e^(−st) f(t), by differentiating it.✓ Proved
  2. For s > 0 the antiderivative tends to 0 as t → ∞ (the exponential decay wins).
    Reviewed
  3. \[ - \left. \frac{\left(- s \sin{\left(t \right)} - \cos{\left(t \right)}\right) e^{- s t}}{s^{2} + 1} \right|_{\substack{ t=0 }} = \frac{1}{s^{2} + 1} \]
    So the integral is 0 minus its value at t = 0.✓ Proved
Answer \( F(s) = \frac{1}{s^{2} + 1} \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify

The full receipt
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2Reviewedreviewers: gpt-oss:20b, qwen3.6:27b-mlxa sentence; read, not computed
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0the defining integral computed by mpmath at two values of s

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass — The solution correctly identifies the antiderivative, justifies the boundary condition at infinity for s > 0, and correctly evaluates the definite integral using the Fundamental Theorem of Calculus.
Every verdict on record (6)
  • qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly identifies the antiderivative, justifies the boundary condition at infinity for s > 0, and correctly evaluates the definite integral using the Fundamental Theorem of Calculus.
  • gpt-oss:20b: pass 2026-10-18
  • qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly applies the Fundamental Theorem of Calculus using the verified antiderivative. The limit justification for s > 0 is correct, and the evaluation at the lower bound is accurate.
  • gpt-oss:20b: pass 2026-10-16
  • qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly applies the Fundamental Theorem of Calculus using the verified antiderivative. The limit argument for s > 0 is valid, and the evaluation at the lower bound is correct.
  • 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.