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By Reinhard Diestel

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This can now be compressed into one formula: f (t) = (1 − t2 )(1 − H(t + 2)) + (t + 2)(H(t + 2)−H(t − 1)) + (1 − t)H(t − 1) = (1 − t2 ) + (−1 + t2 + t + 2)H(t + 2) + (−t − 2 + 1 − t)H(t − 1) = 1 − t2 + (t2 + t + 1)H(t + 2) − (2t + 1)H(t − 1). Heaviside’s function is connected with the δ function via the formula t H(t) = δ(u) du. −∞ A very bold differentiation of this formula would give the result H (t) = δ(t). 5) 30 2. 5) is reasonable for t = 0. What is new is that the “derivative” of the jump discontinuity of H should be considered to be the “pulse” of δ.

0 It is easy to see that this integral converges for positive x. It is also easy to see that Γ(1) = 1. Integrating by parts we find ∞ Γ(x + 1) = 0 ux e−u du = −ux e−u ∞ 0 ∞ +x ux−1 e−u du = xΓ(x). 0 From this we deduce that Γ(2) = 1 · Γ(1) = 1, Γ(3) = 2, and, by induction, Γ(n + 1) = n! for integral n. Thus, this function can be viewed as an interpolation of the factorial. Now we let f (t) = ta , where a > −1. It is then clear that f has a Laplace transform, and we find, for s > 0, ∞ f (s) = ta e−st dt 0 = ∞ 1 sa+1 0 st = u dt = du/s ∞ = 0 u a −u du e s s Γ(a + 1) ua e−u du = .

A Here we used that the left-hand member is real and thus equal to its own real part. 20 2. 7 Compute the derivative of f (t) = eit by separating into real and imaginary parts. Compare the result with that obtained by using the chain rule, as if everything were real. 8 Show that the chain rule holds for the expression f (g(t)), where g is realvalued and f is complex-valued, and t is a real variable. 9 Compute the integral π eint dt, −π where n is an arbitrary integer (positive, negative, or zero).

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