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Jordan Lemma Proof H. Vic Dannon with ρ →∞. On the semi-circle, e iz ∫ dz → 0 , as ρ →∞. 2 z + 1 Therefore, by the Residue Theorem, x =∞ ∫ x =−∞ ∩ ix iz e ⎧ e ⎫ dx = 2πiRes ⎪ ⎨ ⎪ ⎬ 2 x + 1 ⎪⎩ ( z + i )( z −i ) ⎪⎭ ⎧ iz e ⎫ = 2πi lim⎪ ⎨ ⎪ ⎬ z→i ⎪⎩z + i ⎪⎭ = πe − It follows that this is the value of the desired integral. The vanishing of the integral on the infinite semi-circle, holds under the conditions of a general result attributed by Whittaker and Watson, to Jordan. 1 . z= i z= i If f () z is an analytic function in the upper-half complex plane, so that f () z → 0 , uniformly on any upper semi-circle C with ρ radius ρ →∞, centered at the origin, 2
Jordan Lemma Proof H. Vic Dannon ∫ imz Then, for m > 0 , e f() z dz → 0, as ρ →∞. C ρ We show here that the proof of the Lemma by Jordan, and the proof by Whittaker and Watson are incomplete. Whittaker and Watson proof permeates many if not all textbooks on complex variables, and has to be corrected. 1. Jordan’s Proof In Jordan’s proof, m = 1. On the semi-circle, C ρ We have z i e θ = ρ , 0 ≤ θ ≤ π. iz iz ∫ e f() z dz ≤ ∫ e f() z dz . C ρ C ρ Substituting iz iρcos θ−ρsin θ −ρsin θ e = e = e , dz i θ = ρe dθ, dz = ρθ d , θ= π iz ∫ e f() z dz ≤ max f() z ρ∫ −ρsin θ e dθ. C C θ= 0 ρ ρ Since 3
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Jordan Lemma Proof - Gauge-institute.org

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