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## • Although the series

• Although the series coming from the binomial theorem only converges for | 1t| < 1, 2 it can be used in the integral over [0, ∞) because all dominant contributions to the integral come from t ∈ [0, ǫ]. Use Watson’s lemma with α = −1 , β = 1, and 2 to get K0(x) = e −x = e −x an = (−1)n 2n+1/2 Γ � 1 2 ∞� n=0 ∞� n=0 2.3.2 General Laplace Integrals (−1) n 2n+1/2 Γ � 1 (−1) n 2 n+1/2 + n� n!Γ � � 1 2 2 + n� Γ � −1 + n + 1� 2 n!Γ � � 1 − x 2 1 2 +n+1 � � 1 Γ 2 + n�� 2 n!Γ � 1 � 1 2 xn+1 x → ∞ 2 Obtain the leading-order behavior of the general Laplace integral � b I(x) = f(t)e xφ(t) dt a As indicated before one has to distinguish the cases when φ(t) has local maxima and when the maxima are attained at the boundaries of the interval. In each case φ(t) and f(t) are expanded around the maximum t = c. i) Maximum at endpoint, c = a Expand Thus Note: � a+ǫ I(x; ǫ) ≈ a = f(a)e xφ(a) � a+ǫ φ(t) = φ(a) + (t − a)φ ′ (a) with φ ′ (a) < 0 [f(a) + f ′ (a) (t − a)]e xφ(a)+xφ′ (a)(t−a) dt xφ(a) −1 = f(a)e xφ ′ (a) = −f(a) exφ(a) a e xφ′ (a)(t−a) dt + f ′ (a)e xφ(a) xφ ′ (a) + f ′ (a)e xφ(a) � a+ǫ � 1 − e xφ′ � (a)ǫ + f ′ xφ(a) 1 (a)e 1 I(x) ∼ −f(a) exφ(a) xφ ′ (a) a (t − a) e xφ′ (a)(t−a) dt φ ′ d (a) dx � −1 xφ ′ � 1 − e (a) xφ′ � (a)ǫ � x2φ ′ + exponentially small terms (a) 2 54 x → ∞ (23)

• since we kept only the first non-trivial term in the expansion of φ(t) only the leadingorder term in the result can be trusted. For higher-order terms see Sec.2.3.3. ii) Maximum inside the interval Now the maximum is a local maximum Using we get φ(t) = φ(c) + 1 p! φ(p) (c) (t − c) p I(x; ǫ) = � c+ǫ c−ǫ p even φ (p) (c) < 0 (24) [f(c) + . . .]e xφ(c) 1 x e p! φ(p) (c)(t−c) p dt ∼ e xφ(c) � (p) −xφ (c) f(c) p! � ∞ −∞ Γ I(x) ∼ 2 e −sp Γ ds = 2 � � 1 p p �− 1 p � ∞ � � 1 p p � − p! xφ (p) � 1 p −∞ f(c) e xφ(c) e −sp ds In the generic case of a quadratic maximum, p = 2, this reduces (using Γ( 1 2 ) = √ π) to I(x) ∼ � 2π −xφ ′′ (c) f(c)exφ(c) Example I(x) = � π/2 0 e −x sin2 t dt Here φ(t) has its maximum at t = 0. Expect that sin 2 t can be replaced by t 2 since only small t contribute. If that is the case one gets � ǫ I(x : ǫ) ∼ Can again extend the integral to +∞ � ∞ I(x; ǫ) ∼ 0 0 e −xt2 dt e −xt2 dt = 1 � π 2 x Is the replacement sin 2 t → t 2 justified? We need to show for small ǫ that � ǫ 0 e −xsin2 t dt ∼ � ǫ 0 e −xt2 dt Clearly, ex sin2 t xt → e 2 for sufficiently small t at fixed x. But with increasing x the values of t need to go to 0. How small do they need to be? 55 (25) (26)

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