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Lecture Note Sketches Hermann Riecke - ESAM Home Page

Lecture Note Sketches Hermann Riecke - ESAM Home Page

Although all these

Although all these integrals can be done easily directly, it is yet easier to replace ǫ by ∞ I(x) ∼ ∞� � ∞ n=0 0 (−t) n e −xt dt = ∞� n=0 dn dxn � ∞ e 0 −xt dt = ∞� n=0 dn dxn 1 x x → ∞ Why can we replace ǫ by ∞? The approximation for (1 + t) −1 is only valid for small t, 0 ≤ t < 1. The difference is subdominant relative to I(x) (remember that x → ∞ for ǫ fixed), Note: � ∞ ǫ ∞� n=0 (−t) n e −xt dt = ∞� n=0 dn dxn � 1 x e−xǫ � • The contributions to I(x) all come from inside the interval [c − ǫ, c + ǫ]. There we need to approximate f(t) systematically to make the integrals doable. • The contributions from outside of [c − ǫ, c + ǫ] are only subdominant with respect to I(x); therefore it does not matter how we approximate f(t) outside that interval. In particular, we can replace ǫ by ∞, even if the approximation of f(t) is not valid there. 2.3.1 Watson’s Lemma For the less general integral � b I(x) = f(t)e −xt dt one can give a general expression if f(t) is given by an asymptotic series f(t) ∼ t α 0 ∞� ant βn n=0 For the integral to converge we need α > −1 and β > 0. Consider Inserting the asymptotic series one gets I(x; ǫ) ∼ t → 0 � ǫ I(x; ǫ) = f(t)e −xt dt ∼ = 0 � ǫ t α 0 � ∞ t α 0 ∞� n=0 an 52 ∞� n=0 ∞� n=0 ant βn e −xt dt ant βn e −xt dt Γ (α + βn + 1) x α+βn+1

Is this series for I(x; ǫ) asymptotic? The asymptotic series for f(t) satisfies � � � �f(t) − tα � Therefore � � � ǫ � �I(x; ǫ) − t � α 0 N� n=0 N� ant βn � � � � � n=0 ant βn e −xt dt � � � � � ≪ Ktα+βN � ǫ ≪ K 0 � ∞ < K = K 0 t α+βN e −xt dt t α+βN e −xt dt Γ (α + βN + 1) x α+βN+1 Thus for all N the error is small compared to the last term in the series and we have established the asymptotic series Note: I(x) ∼ ∞� n=0 an Γ (α + βn + 1) x α+βn+1 x → ∞ • Watson’s lemma will always generate a power series in an algebraic power of x because the maximum of φ(t) ≡ −xt is at t = 0. Example 2: Bessel function K0(x) = � ∞ 1 (s2 − 1) −1/2 e −xs ds To bring K0(x) in the correct form shift the limit t = s − 1 K0(x) = e −x � ∞ To expand around t = 0 rewrite 0 � t 2 + 2t � − 1 2 e −xt dt � � 1 2 −2 − t + 2t = (2t) 1 � 2 1 + 1 2 t � 1 −2 For small u Taylor series leads to the binomial theorem in the form 1 (1 − u) s s = 1 + (1 − u) s+1 � � � s (s + 1) � u + u=0 (1 − u) s+2 � � � 1 � 2! u=0 u2 ∞� 1 (s + n) + . . . = unΓ n! Γ (s) Insert to get Note: K0(x) = e −x ∞� � ∞ n=0 0 (2t) −1 2 53 � − 1 2 t � � n 1 Γ 2 n=0 + n� n!Γ � � e 1 2 −xt dt (22)

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