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Jordan Lemma Proof H. Vic Dannon 3. θ= ∫ π 2 θ= 0 e −mρcos θ d θ, and θ= ∫ π 2 θ= 0 e −mρsin θ d θ by MAPLE Using in MAPLE, we obtain Using in MAPLE, we obtain Thus, an equality of areas under the graphs may be due to symmetry of the graphs. But we cannot confirm it, because do not know how to evaluate the integrals. 6
Jordan Lemma Proof H. Vic Dannon In MAPLE, we confirm that and θ= ∫ π 2 θ= 0 θ= ∫ π 2 θ= 0 e e −2cosθ −2cosθ dθ dθ = = (1/2)*Pi*BesselI(0, 2)-(1/2)*Pi*StruveL(0, 2) (1/2)*Pi*BesselI(0, 2)-(1/2)*Pi*StruveL(0, 2) In the notations of Abramowitz, and Stegun BesselI(0, 2)=Modified Bessel Function I 0(2) StruveL(0, 2)=Struve Function L (2) . That is, both integrals equal π ( 0(2) 0(2) ) 2 I − L . This confirms Whittaker’s guess of equality. Nevertheless, it seems easier to complete the Jordan Lemma Proof by bounding θ= ∫ π 2 θ= 0 e −mρcosθ 0 d θ, then by trying to prove the equality. 4. Completed Proof To complete the Jordan Lemma Proof, we need to bound ρ θ= ∫ π 2 θ= 0 e −ρcos θ dθ, similarly to the bounding of ρ π 2 θ= −ρsin θ ∫ e dθ. θ= 0 7
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