Introduction to Gamma Function and Equivalent Integral Forms
The Gamma function is a versatile mathematical function that generalizes the factorial function to non-integer and complex values. It has various integral definitions such as the Euler-integral form. The proof of the factorial property of the Gamma function is demonstrated through analytical continuation. Equivalence between different integral forms is explored, showcasing the interconnectedness of these representations.
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ECE 6382 ( ) z y x z = 0 Fall 2023 David R. Jackson Notes 14 The Gamma Function Notes are from D. R. Wilton, Dept. of ECE 1
The Gamma Function The Gamma function appears in many expressions, including Bessel functions, etc. It generalizes the factorial function n! to non-integer values and even complex values. It appears in the method of steepest descent (a method for obtaining the asymptotic expansion of a class of integrals). 2
Definition 1 Definition # 1 1 2 3 1)( z n z 0, 1, 2, ( ) z lim n , n z + + + ( 2) ( ) z z z n This definition gives the Gamma function a nice property for z = n (a positive integer), as proven on the next slide: ( ) = ( ) n 1 ! n (factorial property) 3
Definition 1 (cont.) Proof of factorial property: 1 2 3 1)( z n z 0, 1, 2, ( ) z lim n , n z + + + ( 2) 1 2 3 z + ( ) z z z n n z nz n + 1 z + = = ( 1) lim n ( ) lim z z n + + + = + + ( 1)( 2) ( + 1) ( 1) z n z n ( 1) ( ) z z z 1 2 3 (4) n n = = = = Note that , and (1) lim n 1 (2) 1 (1) 1, n + 1 2 3 ( (3) 1) = n = 2 1, = = = 4 3 2 1, = (3) 2 (2) 3 3 2 1, (5) 4 (4) . etc Hence ( ) + = ( 1) ! = n n ( ) n 1 ! n or n = n = 0,1,2, 1,2,3 4
Definition 2 Definition # 2 1 t z ( ) z , Re 0 e t dt z 0 This is the Euler-integral form of the definition. Note: Leonard Euler ( ) ( ) iy 1 1 1 1 z x x x = = = ln ln iy t iy t t t t t e t e 1 1 z x = 0 t t x for the integral to converge at = 0 t Note: Definition 1 is the analytic continuation of definition 2 from the right-half plane into the entire complex plane (except at zero and the negative integers). 5
Equivalent Integral Forms The following three integral definitions are all equivalent: 1 t z ( ) z , Re 0 e t dt z 0 2 2 1 2 s z = = (let ) ( ) z 2 , Re 0 e s ds z t s 0 1 z 1 1 s ( ) = = (let ) ( ) z ln , Re 0 ln 1/ ds z t s 0 6
Equivalence of Definitions 1 and 2 Equivalence of definitions #1 and #2 n t n t Use , lim 1 n e n n t n ( ) z 1 1 z t z = Define ( , ) F z n 1 ; ( , ) F z n t dt e t dt 2 0 0 n t n = Letting and integrating by parts t imes , w n Factor appearing in Definition #1 ( ) + 1 1 1 2 3 1)( + 1 z n n n ( ) z n + 1 1 z z z = = ( , ) F z n 1 n w w dw n w dw + ( 2) ( 1) z z z n 0 0 1 + = Hence lim ( , n ) ( ) z F z n z n 1 (Please see next slide.) 1 2 3 1)( z n z 0, 1, 2, ( ) z lim n , n z 1 + + + ( 2) ( ) z z z n 7
Equivalence of Definitions 1 and 2 (cont.) Integration by parts development: 1 n ( ) 1 z 1 I w w dw 0 dv dw u Integrate by parts once: 1 1 z z w w 1 n n ( ) ( ) = 1 1 I w n w dw z z 0 0 1 z w 1 n ( ) = + 0 1 n w dw z 0 1 n z 1 n ( ) z = 1 w w dw 0 8
Equivalence of Definitions 1 and 2 (cont.) 1 n ( ) 1 z Integrate by parts twice: 1 I w w dw 0 dv dw 1 u n z 1 n ( ) z = 1 I w w dw 0 1 1 + + 1 1 z z n z w z n z w z 1 2 n n ( ) ( )( ) ( ) = 1 1 1 1 w n w dw + + 1 1 0 0 ( ( ) ) 1 + 1 1 n n z z 2 n ( ) + 1 z = + 0 1 w w dw 0 ( ( ) ) 1 + 1 1 1 n n z z 2 n ( ) + z n + 1 z 1 = 1 w w dw w dw 0 0 After n times: ( )( z ) + 1 1 2 3 2 1 n n n z z n ( ) z n + n n 1 = 1 I w w dw + + ( 1)( 2) ( 1) z 0 9
Definition 3 Definition # 3 The Weierstrass product form can be shown to be equivalent to definitions #1 and #2. z n 1 ( ) z z n z = + 1 ze e = 1 n = where is the Euler -Mascheroni constant. 0.5772156619 10
Euler Reflection Formula Euler Reflection Formula ( ) (1 z = ) z (Proof omitted.) y sin z x = 1/ 2 z Geometric interpretation of reflection formula: In the horizontal direction, the two points are reflections about the x = 1/2 line. ( ) 1 1/ 2 1/ 2 x x x 1 1 z z Note: We can use this formula along with definition #2 to find (z) for Re(z)< 0. 1 = 0, 1, 2, ( ) z , Re 0, z z sin (1 ) z z 11
Euler Reflection Formula (cont.) A special result that occurs frequently is (1/2). To calculate this, use the reflection formula: ( ) (1 z = ) z sin z Set z = 1/2: = (1/2) 12
Summary of Factorial Properties Summary of Factorial Generalization ( )( ) ( )( )( ) 3 2 1 = ! 1 2 n n n n Integers 1,2,3 n = ( ) = Real numbers x t x = + ! 1 x x e t dt 1 0 ( ) Complex numbers ( ) Re z = t z = + ! 1 z z e t dt 1 0 13
Summary of Factorial Properties (cont.) Summary of Factorial Generalization (cont.) ( ) ( ) t z = + = ! 1 Re 1 z z e t dt z 0 Complex numbers z + 1, 2 1 = ( ) z sin (1 ) z z 14
Pole Behavior Simple poles of (z) are at n = 0, -1, -2, -3, + = = ( 1) ( ) z & (1) 1 z z Recall: (z) has simple pole at z = 0 Residue = 1 Use + ( 1) z + = = ( 1) ( ) z ( ) z z z z + + ( 2) 1 z ( ) + = + + + = ( 2) 1 ( 1) ( 1) z z z z z + + + + ( 2) 1 2) 1 z = ( ) z z (z) has simple pole at z = -1 Residue = -1 z ( z = ( ) z ( ) z z 15
Pole Behavior (cont.) + ( z 3) 2 ( z z z ( ) + = + + + = ( 3) 2 ( 2) ( 2) z z z z + + 3) 2 (z) has simple pole at z = -2 Residue = +1/2 ( )( ) 1 + = ( ) z z z + + + ( 3) z z = ( ) z ( )( ) + 1 2 z z + ( z 4) 3 z ( ) + = + + + = ( 4) 3 ( 3) ( 3) z z z z + + ( z 4) 3 z ( )( )( ) 1 (z) has simple pole at z = -3 Residue = -1/6 + + = 2 ( ) z z z z + + + ( 4) 2 z z = ( ) z ( )( )( ) + + 1 3 z z z 16
Pole Behavior (cont.) Residues at Poles In general (after n+1 steps), we will have: (z) has simple pole at z = -n + + ( 1) 3 z n z = ( ) z ( )( )( ) ( ) + + + + 1 2 z z z z n ( ) + + + ( z 1) 3 n n ( ) z ( ) = + Res Lim z z n ( )( )( ) ( ) + + + = z n 1 2 z z z z n n 1 z = ( )( )( 1 ) ( ) + + + + 1 2 3 1 z z z z n Hence = z n = ( ) 1 ! n n ( )( ) ) ( )( )( ) + 1 3 2 1 n n ( ) z = Res ( = z n n 1 = ( )( n ) ( )( )( ) 3 2 1 1 n 17
Plot of Gamma Function ( ) z y x z = 0 ( ) 1 ! n n ( ) z = Res Note: There are simple poles at z = 0, -1, -2, = z n 18
Plot of Gamma Function (cont.) (x) and 1 / (x) Note: (x) never goes to zero. In fact, 1 / (z) is analytic everywhere. 19
Asymptotic Form of Gamma Function Sterling s formula (asymptotic series for large argument): + + z as 2 1 1 139 571 ( ) z + z z 1 z e 2 3 4 12 288 51840 2488320 z z z z z w Taking the ln of both sides, we also have 1 2 1 1 1 z ( ) z + + + ln ln ln z z z 3 5 2 12 360 1260 z z z 2 3 w w ( ) + = + Note: ln 1 w w 2 3 20