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By I. D. Faux, Michael J. Pratt

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N)! 5 f¯(s) = e−as g¯(s) where g¯(s) = L{g(t)}. We know that f (t) = H(t − a)g(t − a) and for t ≥ a we can proceed to get g(t) as in previous examples and then replace t by t − a. 4) that if f¯(s) = e−as /(s2 + 1) we have f (t) = 0, t < a, (t − a)3 (t − a)5 = (t − a) − + − ··· , 3! 5! t ≥ a. 7) We conclude this section with an example which will be used in Chapter 9 to test the efficacy of various numerical methods of inversion. 6 Determine a series expansion for f (t) when f¯(s) = s1/2 1 . 1, 1 L−1 sν+1 = tν , Γ(ν + 1) ν > −1 we obtain f (t) = t−1/2 1 t1/6 t1/3 (−1)n tn/6 + ··· .

1 The Uniqueness Property We mentioned in the last Chapter that the Laplace transform is unique in the sense that if f¯(s) = g¯(s) and f (t) and g(t) are continuous functions then f (t) = g(t). This result was proved originally by Lerch [125] and the proof given here follows that in Carslaw and Jaeger [31]. 1 (Lerch’s theorem). 1). Proof. We require the following lemma. 1 Let ψ(x) be a continuous function in [0, 1] and let 1 xn−1 ψ(x)dx = 0, for n = 1, 2, · · · . 2) in 0 ≤ x ≤ 1. 3) 0 Then ψ(x) ≡ 0, Proof.

0 t−u I= Now let u = t + θ, F(θ) = e−γθ f (t + θ). The right hand side of the integral then becomes 1 ∞ sin T θ F(θ) dθ. 9) into two parts ∞ F(θ) 0 sin T θ dθ = F(0) θ X + δ δ 0 sin T θ dθ + θ sin T θ F(θ) dθ + θ δ 0 0 . −t We write F(θ) − F(0) sin T θdθ θ ∞ X and sin T θ F(θ) dθ. 2. THE BROMWICH INVERSION THEOREM 27 We can choose δ small and X large so that δ 0 F(θ) − F(0) sin T θdθ < , θ and ∞ F(θ) X for all T . Next consider sin T θ dθ < , θ X F(θ) δ sin T θ dθ. θ Integrating by parts X F(θ) δ sin T θ cos T θ dθ = − F(θ) θ Tθ X + δ 1 T X cos T θ δ d dθ F(θ) θ dθ, = O(1/T ), since each term involves 1/T and the integral is bounded.

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