Asymptotic Evaluation Techniques in Integral Calculus

Learn about asymptotic evaluation of integrals through techniques like integration by parts and the stationary-phase method. Understand how to handle integrals involving real functions, and grasp the significance of concepts like the Riemann-Lebesgue lemma and small o notation. Delve into the physical interpretations and applications of these methods for solving complex integrals.

• Integral Calculus
• Asymptotic Evaluation
• Integration by Parts
• Stationary-Phase Method
• Riemann-Lebesgue Lemma

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1. ECE 6341 Spring 2016 Prof. David R. Jackson ECE Dept. Notes 28 1

2. Asymptotic Evaluation of Integral Goal: Evaluate the following integral as becomes large: b ( ) ( ) f x e ( ) = j g x I dx g(x) is a real function here. a ( ) ( ) g x 0, , x a b Integration by Parts Case 1: ( ) g x = 0, , x a b Stationary-Phase Method Case 2: 0 0 2

3. Integration by Parts b ( ) ( ) f x e ( ) = j g x I dx a ( ) g x 0 , x a b Note that d dx ( ) ( ) j g x e = j g x j g x ( ) ( ) e So that we may write 1 g x dx d b ( ) = ( ) f x j g x ( ) I e dx ( ) j a 3

4. Integration by Parts 1 g x dx d b ( ) = ( ) f x j g x ( ) I e dx ( ) j a v u d dx d dx b b ( ) u x ( ) v x dx ( ) ( ) u x v x ( ) v x ( ) u x dx b = a a a Hence ( ) f x g x ( ) = b j g x ( ) I e ( ) j a ( ) ( ) f x 1 j d dx g x b j g x ( ) e dx a 4

5. Integration by Parts (cont.) Riemann-LebesgueLemma: b aF x e ( ) ( ) g x j 0 dx as (The integrand oscillates faster, so the integral tends to zero) Hence ( ) f x g x ( ) b j g x = + ( ) 1/ I e o ( ) j a or Small o notation: It decays faster than this as gets large. ( ) f x g x b j g x ( ) ~ I e ( ) j a 5

6. Integration by Parts (cont.) Physical interpretation ( ) f x g x b ( ) ( ) f x e ( ) = j g x I dx b j g x ( ) ~ I e ( ) j a a ( ) ( ) f x = ( ) g x j Re f x e Endpoint contributions remain ( ) ( ) cos g x x Interior parts cancel 6

7. Stationary-Phase Method b ( ) f x e ( ) = ( ) g x ( ) g x j I dx a ( ) = 0 , x a b 0 0 ( ) ( ) x 0 f x Assume 0 0 g 0 1 2 ( ) ( ) g x ( )( 0 g x ) ( )( 0 x ) 2 + + g x x x g x x 0 0 0 ( ) ( ) f x = ( ) g x j Re f x e ( ) ( ) Stationary-phase point g x cos x 0x b a 7

8. Stationary-Phase Method (cont.) Conjecture: + x j g x 0 ( ) ~ ( ) f x e I dx x 0 If 0 ( ) slowenough To ensure that the stationary phase point region dominates over the outside regions, we define: 1 g x dx d + + x x ( ) = ( ) f x e ( ) f x j g x j g x = 0 0 ( ) ( ) I dx e dx ( ) SPP j x x 0 0 1 g x dx d x x ( ) = ( ) f x e ( ) f x j g x j g x = 0 0 ( ) ( ) I dx e dx ( ) left j a a 1 g x dx d b b ( ) = ( ) f x e ( ) f x j g x j g x = ( ) ( ) I dx e dx ( ) right j + + x x 0 0 8

9. Stationary-Phase Method (cont.) We have that A A I 1 I 2 ( ) ( ) g x g x left right + 0 0 For some constants A1 and A2. Big O notation: It behaves like this as gets large. O 1 = I We will show later that SPP 9

10. Stationary-Phase Method (cont.) Hence, we require that 1 1 I I left right Approximating the first derivative of g by a Taylor series, ( ) 0 g x ( ) ( 0 g x ) ( ) x + g 0 We require: 1 1 ( ) x g 0 1 Hence (This needs to be one of our assumptions.) 10

11. Stationary-Phase Method (cont.) Hence, we assume that: This ensures that 0 + x j g x 0 ( ) ~ ( ) f x e I dx x 0 x 0x b a 2 11

12. Stationary-Phase Method (cont.) 0 Since ( ) f x ( ) f x Assume 0 + x j g x 0 ( ) ~ ( ) f x I e dx Hence 0 x 0 12

13. Stationary-Phase Method (cont.) 1 2 ( ) ( ) g x ( )( 0 g x ) ( )( 0 x ) 2 + + g x x x g x x 0 0 0 1 2 ( )2 + x x x ( ) j g x ( ) ( ) f x e j g x 0 ( ) 0 0 I e dx 0 0 x 0 or 1 2 ( )2 + x x x ( ) j g x ( ) ( ) f x e 0 ( ) 0 0 j g x I e dx 0 0 x 0 where ( ) x ( ) x = g g 0 0 13

14. Stationary-Phase Method (cont.) Let ( ) 2 ( ) 2 g x g x ( ) 0 0 = = s x x ds dx 0 Then 2 + S ( ) ( ) f x e 2 j g x ( ) L js I e ds 0 ( ) x 0 g S L 0 where ( ) 2 g x 0 = S L , S As since L 14

15. Stationary-Phase Method (cont.) Therefore + + S 2 2 L js js I e ds e ds L S L Let + 2 2 + + = = js js I e ds e ds 1 L C C Ims 1 Deform the path: 45 Res C C C 1 15

16. Stationary-Phase Method (cont.) Let j = = 2 2 2 s t e jt 2 j = s te 4 j = ds dte 4 Hence ( ) + 2 j j jt = I e e dt 4 1 L + j 2 = t e e dt 4 j = e 4 16

17. Stationary-Phase Method (cont.) Similarly, + j 2 = js LI e ds e 4 2 Then 2 j ( ) j g x ( ) f x e I e 4 0 ( ) x 0 g 0 O 1 = Note: I 17

18. Stationary-Phase Method (cont.) Hence 2 j ( ) j g x ( ) I f x e e 4 0 ( ) x 0 g 0 ( ) ( ) x + , 0 g x 0 , 0 g 0 18

19. Example 1 ( ) = ( ) d cos sin J 0 0 1Re = d sin j e 0 ( ) ( ) ( ) = = 1 f sin g ( ) g = = cos 0 0 0 = +n 0 2 19

20. Example (cont.) / 2 / 2 ( ) ( ) ( ) = sin = = g ( ) g = = sin 1 0 cos g 0 0 sin g Hence 1 2 j ( ) ( ) 1 j ~ Re J e e 4 0 1 20

21. Example (cont.) 1 2 j ( ) ( ) 1 j ~ Re J e e 4 0 1 Hence 2 ( ) ~ cos J 0 4 as 21