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**Sample text**

If ∞ an t n f (t) n 0 converges for t ≥ 0, with Kαn , n! for all n sufﬁciently large and α > 0, K > 0, then |an | ≤ L f (t) ∞ an L(t n ) n 0 ∞ n 0 an n! s n +1 Re(s) > α . Proof. Since f (t) is represented by a convergent power series, it is continuous on [0, ∞). 6. Basic Properties of the Laplace Transform 19 N ≤ Lx f (t) − an t n n 0 ∞ −xt e h(t) dt, converges to zero as N → ∞, where Lx h(t) Re(s). To this end, x ∞ N f (t) − 0 an t n an t n n N +1 n 0 ∞ ≤K n (αt)n n! N +1 N (αt)n n! K eαt − n 0 ∞ n since ex n 0 x /n!.

This latter is considered in an entirely new but rigorous fashion from the standpoint of the Riemann–Stieltjes integral. 9) that L(t n ) n! s n +1 , n 1, 2, 3, . . 41 42 2. Applications and Properties In order to extend this result for non-integer values of n, consider ∞ L(t ν ) e−st t ν dt (ν > −1). 0 t ν is not piecewise Actually, for −1 < ν < 0, the function f (t) continuous on [0, ∞) since it becomes inﬁnite as t → 0+ . However, τ as the (improper) integral 0 t ν dt exists for ν > −1, and f (t) t ν is bounded for all large values of t, the Laplace transform, L(t ν ), exists.

As it turns out, F(s) → 0 as Re(s) → ∞ whenever the Laplace transform exists, that is, for all f ∈ L (cf. 2). As a consequence, any function F(s) without this behavior, say (s − 1)/(s + 1), es /s, or s2 , cannot be the Laplace transform of any function f . 6 1. Find L(2t + 3e2t + 4 sin 3t). ω . 2. Show that L(sinh ωt) 2 s − ω2 3. Compute (a) L(cosh2 ωt) (b) L(sinh2 ωt). 4. Find L(3 cosh 2t − 2 sinh 2t). 5. Compute L(cos ωt) and L(sin ωt) from the Taylor series representations ∞ cos ωt n 0 (−1)n (ωt)2n , (2n)!