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• A series � n fn (x

• A series � n fn (x − x0) αn is called asymptotic to f(x), if for any fixed N � � � �f(x) − � N� n f(x) ∼ � n fn (x − x0) αn fn (x − x0) αn , � � � αN � ≪ |x − x0| � for x → x0 • The series need not be an expansion in integer powers of x − x0. • If a series is asymptotic to f(x) it does not have to converge to f(x) for fixed x when N → ∞. In fact, the asymptotic series typically approaches f(x) as N is increased up to some intermediate value of N and then it diverges when N is increased further. Example 1 Consider small-x behavior of I(x) = � 1 0 The series � 1 converges uniformly to 1 sin xt t an+1 an for all x ∈ [0, 1]. = 0 1 sin xt dt t � 1 1 1 sin xt dt = t 0 t 1 (2n − 1)(2n − 2) x2 < Therefore we integrate term by term to get � 1 0 � xt − 1 3! (xt)3 � + . . . dt x − 1 3! x3 t 2 + 1 5! x5 t 4 + . . . 1 (2n − 1) (2n − 2) 1 1 1 sin xt dt ∼ x − t 3! 3 x3 + 1 1 5! 5 x5 . . . Example 2 I(x) = � ∞ x ta−1 e −t dt Determine the behavior of the incomplete Γ-function for small x � ∞ Γ(a, x) = x t a−1 e −t dt → 0 for n → ∞ We would like to expand the exponential and then integrate term by term. This would lead to terms � ∞ a−1 1 t n! (−t)n x 44

which diverge at the upper limit for large enough n. To treat the upper limit we need to keep the exponential. Depending on a we need to do this in different ways: i) a > 0 For a > 0 the lower limit x → 0 poses no difficulty. Use it to rewrite � ∞ x t a−1 e −t dt = � ∞ 0 = Γ(a) − t a−1 e −t dt − n=0 � x t a−1 0 ∞� (−t) n dt (19) n! n=0 ∞� (−1) n xn+a (n + a) n! since the second term can be integrated term by term ii) a ≡ −ā < 0 but not an integer Now we also have to worry about the divergence at x = 0 in the first term in (19). Divide the integral into two parts: • one in which the upper limit t → ∞ does not require to keep the exponential function include in this integral as many terms of the expansion of the exponential as possible need t −ā+N → 0 for t → ∞ i.e. N < ā • one in which the exponential has to be kept Choose N as the largest integer smaller than ā: Then � ∞ x t a−1 e −t dt = � ∞ t −ā−1 x N < ā < N + 1 N� (−t) n � ∞ dt + t n! −ā−1 n=0 x ∞� n=N+1 By construction the first integral converges at the upper limit t → ∞. � ∞ t −ā−1 x N� (−t) n dt = − n! n=0 N� (−1) n xn−ā (n − ā) n! n=0 (−t) n n! dt The second integral cannot be done term by term because of the upper limit. But it poses no problem at the lower limit t = x → 0: the term in the integrand with the lowest power in t is t −ā−1+N+1 with N − ā > −1 Exploit the integral over (0, ∞) and retain again the exponential function strategically � ∞ t −ā−1 ∞� (−t) n � ∞ dt = t n! −ā−1 � e −t N� (−t) − n � � x dt − t n! −ā−1 ∞� (−t) n n! dt. x n=N+1 0 45 n=0 0 n=N+1

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