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2.4 Generalized Fourier

2.4 Generalized Fourier Integral In general the Laplace integral could involve a complex φ(t). Consider here the case of purely imaginar φ � b φ(t) = iψ(t) I(x) = f(t)e ixψ(t) dt Note: • I(x) is a generalized Fourier integral Often one can use integration by parts, which results in I(x) = f(t) �b � � � ixψ ′ (t) eixψ(t) a a − 1 � b � d f(t) ix a dt ψ ′ � e (t) ixψ(t) dt (28) For ψ(t) = t the integral term can be neglected under quite general conditions: Riemann-Lebesgue Lemma: Notes: � b f(t)e ixt � b dt → 0 for x → ∞ if |f(t)|dt exists a • in the context of Fourier transformations the Riemann-Lebesgue lemma states that under very general conditions the amplitudes of the highest Fourier modes of a function go to 0. • the integral vanishes since the high-frequency oscillations lead to a cancellation of the smooth parts of f(t) Example: I(x) = � 1 0 e ixt 1+t dt Using integration by parts we get I(x) = 1 1 1 + t ix eixt �1 � � � 0 � �� � i i − x 2xeix − 1 ix � 1 −1 0 (1 + t) 2eixtdt � �� � Riemann-Lebesgue →0 To confirm the Riemann-Lebesgue result that the integral term vanishes more rapidly than the boundary term (the integral itself goes to 0 in addition to the prefactor x−1 ) perform an additional integration by parts � 1 −1 0 (1 + t) 2eixt −1 dt = ix (1 + t) 2eixt �1 � � � − 0 � �� � 1 � 1 2 ix 0 (1 + t) 3eixtdt →0 x→∞ 60 a

Bound the integral term ���� � 1 1 ix 0 2 (1 + t) 3eixt � � dt� 1 2 � ≤ 1 → 0 x → ∞ x 1 To get higher-order approximations one could repeat the integration by parts. Example: I(x) = � 1 √ ixt te 0 Again integration by parts √ t I(x) = ix eixt �1 � � � − 0 � �� � 1 � 1 1 ix 0 2 √ t eixtdt − i x eix By Riemann-Lebesgue the integral remainder term can be neglected. If we wanted to check this explicitly we could not use an additional integration by parts: the boundary terms 1 (ix) 2 1 2 √ t eixt �1 � � � does not exist, nor does the integral because of the divergence at t = 0. Rescale using s = xt � 1 0 � 1 1 2 √ t eixtdt = 1 � x √ x 0 0 1 e ixt dt t 3 2 0 1 2 √ s eisds ∼ 1 � ∞ √ x 0 Deform the integration contour and separate it into two contours: C1, 0 ≤ s ≡ iu < i∞, and CR, s = Reiθ , 0 ≤ θ ≤ π, with R → ∞, 2 � 1 � 1 e 2 −u i 1 1 π i du = √ e 4 Γ( x x 1 ) = 2 C1 2 � iu x The integral over CR vanishes by Jordan’s Lemma, � �� � � 1 iR(cos θ+i � � e sinθ)R � CR R x x ieiθ � � � dθ� � � ≤ � � π R 2 e x 0 −R sinθ dθ � � δ R ≤ e x 0 −Rθ � R dθ + x � R 1 � −Rδ ≤ 1 − e x R � + Thus Note: I(x) = −i x eix � + O 61 x −3 2 1 2 √ s eis ds � π π ei 4 (29) x � π 2 e δ −R sinδ dθ � R sinδ π e−R x 2 � → 0 for R → ∞

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