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# Lecture Note Sketches Hermann Riecke - ESAM Home Page

Lecture Note Sketches Hermann Riecke - ESAM Home Page

## The second integral

The second integral converges since the lowest exponent appearing is −ā − 1 + N + 1 = N − ā > −1 � x t −ā−1 ∞� (−t) n dt = n! ∞� (−1) n xn−ā (n − ā)n! (20) 0 n=N+1 n=N+1 The first integral converges at the lower limit for the same reason. It converges at the upper limit because the high powers are given by those of the exponential function. Evaluate it by repeated integration by parts to eliminate the sum iteratively by taking derivatives � ∞ t 0 −ā−1 � e −t N� (−t) − n=0 n � dt = − n! 1 ā t−ā � e −t N� (−t) − n=0 n ��∞ ���� n! 0 � ∞ − − 1 ā t−ā � −e −t N−1 � (−t) − n � dt n! The boundary terms vanish: • at t → ∞ because ā > N • at t = 0 because all powers inside the curly braces have exponents N + 1 > ā and higher Thus after N + 1 integrations by parts � ∞ t 0 −ā−1 � e −t N� (−t) − n=0 n � dt = (−1) n! N+1 � ∞ 1 t ā (ā − 1)...(ā − N) 0 −ā+N e −t dt � �� � Notes: • The definition Γ(a) = � ∞ 0 t a−1 e −t dt is valid only for a > 0. • In rewriting the integral as Γ (N − ā + 1) we used that N + 1 > ā. Now use Thus Combining with (20) we get � ∞ x 0 n=0 Γ(N−ā+1) Γ (N − ā + 1) = (N − ā)(N − ā − 1) . . .(N − ā − N) Γ (−ā) = t −ā−1 e −t dt = Γ (a) − = (−1) N+1 (ā − N) (ā − N + 1)...(ā)Γ(−ā) � �� � Γ(a) � ∞ t −ā−1 � e −t − 0 n=0 N� (−t) n � dt = Γ (a) n! n=0 ∞� (−1) n xn−ā (n − ā) n! 46 = Γ (a) − ∞� (−1) n xn+a (n + a) n! n=0

as in the case a > 0. But: now the series contains also negative powers of x. iii) a < 0 integer using similar techniques one gets (cf. Bender&Orszag 6.2 example 4) Γ (−N, x) ∼ (−1)N+1 � N� � 1 γ − + ln x − N! n ∞� (−1) n xn−N n! (n − N) 2.2 Integration by Parts 2.2.1 x in the Limits n=1 n=0,n�=N Consider integration by parts of an integral of the form � x a tn f(t)dt. There are two ways to integrate by parts i) � x t n f(t)dt = 1 n + 1 tn+1 � � f(t) � � ii) � x a a x a � x 1 − n + 1 tn+1df dt dt t n f(t)dt = t n F(t)| x a − � x nt a n−1 F(t)dt with dF(t) dt In each step the power of the polynomial • increases in i): suggests using it for approximations for small x • decreases in ii): suggests using it for approximations for large x Example 1: Taylor series and remainder � x f(x) = f(0) + f 0 ′ (t)dt = f(0) + (t − x) f ′ (t)| t=x t=0 − � x (t − x) f 0 ′′ (t) dt = f(0) + xf ′ � x (0) + (x − t) f ′′ (t) dt 0 a = f(t) Note: in this case it is more useful to use t − x rather than t as the antiderivative of 1 Repeated integration by parts yields f(x) = N� n=0 x n n! f(n) (0) + 1 N! So far no approximation has been made: 47 � x 0 (x − t) N f (N+1) (t) dt

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