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One can show by other

One can show by other methods that Note: I(x) ∼ ln x x x → ∞ • Integration by parts can only generate series in integer powers of x. Whenever the asymptotic series is not a power series integration by parts must fail. Therefore, no kind of integration by parts can work for the Stieltjes integral for large x. 2.3 Laplace Integrals Laplace integrals have the form � b I(x) = f(t)e xφ(t) dt a For large x the contributions to the integral for t near the maximal value(s) of φ(t) dominate exponentially. Therefore it is sufficient to restrict the integral to a vicinity of the maximum (maxima). If φ(t) has a maximum at c ∈ (a, b) consider the approximation � c+ǫ I(x; ǫ) = f(t)e xφ(t) dt If the maximum of φ(t) occurs at a consider the approximation and analogously if the maximum is at b. c−ǫ � a+ǫ I(x; ǫ) = f(t)e xφ(t) dt a To evaluate I(x; ǫ) it is then sufficient to use an approximation for f(t) that is valid in the vicinity of the maximum. For this to work one needs that • the final approximation does not depend on ǫ • the asymptotic expansions of I(x) and I(x; ǫ) are identical This is actually the case, because typically I(x) and I(x; ǫ) differ only by terms that are exponentially small in x. To show this estimate I(x) − I(x; ǫ) and I(x; ǫ). For simplicity assume the maximum occurs at c = 0, which could be inside the interval or at its boundary. 50

First consider maximum at the lower boundary, c = 0 = a, i.e. φ(t) is decreasing from a to b �� � b |I(x) − I(x; ǫ)| = � � f(t)e xφ(t) � �� � � � b � dt� � < exφ(ǫ) � � f(t)dt� � = exφ(c) e x(φ(ǫ)−φ(c)) �� � � b � � � f(t)dt� � Compare with ǫ � ǫ I(x; ǫ) = f(t)e xφ(t) dt 0 In the interval [0, ǫ] φ(t) can be approximated by φ(t) ≈ φ(c) − γtm , γ > 0, and f(t) by f(t) = f(c) + δtm′ . If φ does not have a local maximum at t = c one has m = 1 I(x; ǫ) ≈ e xφ(c) � ǫ 0 � f(c) + δt m′� e −xγt dt = e xφ(c) ǫ � f(c) � −γxǫ 1 − e γx � + δ 1 (−γ) m′ ǫ dm′ dxm′ � 1 � −γxǫ 1 − e γx ��� Independent of the value of m ′ , for large x the terms inside the curly braces amount to polynomial terms in x −1 with exponentially small corrections. Therefore one has I(x, ; ǫ) ≈ C x −ρ e xφ(c) for some ρ > 0 and some x-independent constant C. Thus |I(x) − I(x; ǫ)| |I(x; ǫ| ≈ � � � � b ǫ f(t)dt � � � C x ρ e x(φ(ǫ)−φ(c)) ≈ x → ∞ � � � � b ǫ f(t)dt � � � C x ρ e −ǫγx ���� → exponentially 0 (21) If the c ∈ (a, b) one has a local maximum and m ≥ 2. Then one has to estimate an integral of the form � ǫ � f(c) + δt m′� e −xγtm dt 0 With some more work one finds that they also decay with x less than exponentially and one obtains again an estimate like (21). Example 1: I(x) = � a 0 (1 − t)−1 e −xt dt Here φ(t) = −t. It is maximal at t = 0. Expand therefore f(t) around t = 0. � ǫ I(x; ǫ) ∼ 0 (1 + . . .) e −xt dt = 1 � −ǫx 1 − e x � The term containing ǫ is exponentially smaller than the other term (it is subdominant): Go to higher orders in the expansion � ǫ ∞� I(x; ǫ) ∼ 0 n=0 I(x) ∼ 1 x (−t) n e −xt dt = 51 x → ∞ ∞� � ǫ n=0 0 (−t) n e −xt dt

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