M next 1 m i1 i3 f d and 1 i2 m m f

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Unformatted text preview: besgue Lemma to prove the Fourier integral theorem: 1 f (x) = lim 2π N →∞ N f (λ)eixλdλ = F −1 [f ]. −N ∞ Note that sin θ dθ = π. −∞ θ Answer 1 2π 1 = π N ixλ f (λ)e −N 1 dλ = 2π ∞ N eiλ(x−ξ ) dλ f (ξ )dξ −∞ ∞ sin N (x − ξ ) 1 f (ξ ) dξ = x−ξ π −∞ 2 −N ∞ f (η + x) −∞ sin(N η ) dη η −M 1 = π ∞ M + + −∞ −M = I1 + I2 + I3 . M Next, ∞ 1 πM I1 , I3 ≤ |f (η )|dη −∞ and 1 I2 = π M −M f (η + x) − f (x) f (x) sin...
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