By Reinhard Diestel
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Extra info for A Fourier Analysis And Its Applications
Suppose that f ∈ Ek0 and f ∈ Ek1 , and take s to be larger than both k0 and k1 . Let T be a positive number. Integration by parts gives T 0 f (t) e−st dt = f (T ) e−sT − f (0+) e0 + s T f (t) e−st dt. 0 When T → ∞, the ﬁrst term in the right-hand member tends to zero, and the result is the desired formula. 3 will be used for solving diﬀerential equations. The following theorem states a few additional properties of the Laplace transform. 4 (a) If f ∈ E, then lim f (s) = 0. 3) (b) The initial value rule: If f (0+) exists, then lim sf (s) = f (0+).
A The proofs of these rules are easy. As an example we give the computations that yield rules 3 and 4: u= t−a ∞ du = dt f (t − a) e−st dt L[f (t − a)](s) = 0 t = 0 ⇔ u = −a 4. L[f (at)](s) = ∞ = f (u) e−s(u+a) du = e−as −a ∞ f (u) e−su du −a ∞ = e−as f (u) e−su du = e−as f (s); 0 L[f (at)](s) = = ∞ f (at) e−st dt 0 1 a ∞ 0 u = at du = a dt ∞ = f (u) e−s·u/a 0 1 s s . 4. 3 in the preceding section, we can ﬁnd the Laplace transforms of cos and sin: 1 s 1 + = 2 , s − ib s + ib s + b2 1 b 1 1 1 = 2 L[sin bt](s) = L[eibt − e−ibt ](s) = .
For example: if we state that lim f (t) = A, t→∞ 18 2. Preparations then we are asserting the following: for every positive number ε, there exists a number R such that as soon as t > R we are assured that |f (t) − A| < ε. If we split f (t) into real and imaginary parts, f (t) = u(t) + iv(t), u(t) and v(t) real, the following inequalities hold: |u(t)| ≤ |f (t)|, |v(t)| ≤ |f (t)|; |f (t)| ≤ |u(t)| + |v(t)|. 2) This should make it rather clear that convergence in a complex-valued setting is equivalent to the simultaneous convergence of real and imaginary parts.
A Fourier Analysis And Its Applications by Reinhard Diestel