Baixe Arfken cap10 e outras Manuais, Projetos, Pesquisas em PDF para Física, somente na Docsity!
Chapter 10
The Gamma Function
(Factorial Function)
The gamma function appears in physical problems of all kinds, such as the normalization of Coulomb wave functions and the computation of probabilities in statistical mechanics. Its importance stems from its usefulness in developing other functions that have direct physical application. The gamma function, therefore, is included here. A discussion of the numerical evaluation of the gamma function appears in Section 10.3. Closely related functions, such as the error integral, are presented in Section 10.4.
10.1 Definitions and Simple Properties
At least three different, convenient definitions of the gamma function are in common use. Our first task is to state these definitions, to develop some simple, direct consequences, and to show the equivalence of the three forms.
Infinite Limit (Euler)
The first definition, due to Euler, is
( z ) ≡ lim n →∞
1 · 2 · 3 · · · n z ( z + 1)( z + 2) · · · ( z + n )
nz , z = 0, −1, −2, −3,.... (10.1)
This definition of ( z ) is useful in developing the Weierstrass infinite-product form of ( z ) [Eq. (10.17)] and in obtaining the derivative of ln ( z ) (Sec- tion 10.2). Here and elsewhere in this chapter, z may be either real or complex.
523
524 Chapter 10 The Gamma Function (Factorial Function)
Replacing z with z + 1, we have
( z + 1) = lim n →∞
1 · 2 · 3 · · · n ( z + 1)( z + 2)( z + 3) · · · ( z + n + 1)
nz +^1
= lim n →∞
nz z + n + 1
1 · 2 · 3 · · · n z ( z + 1)( z + 2) · · · ( z + n )
nz
= z ( z ). (10.2)
This is the basic functional relation for the gamma function. It should be noted that it is a difference equation. The gamma function is one of a general class of functions that do not satisfy any differential equation with rational coef- ficients. Specifically, the gamma function is one of the very few functions of mathematical physics that does not satisfy any of the ordinary differential equations (ODEs) common to physics. In fact, it does not satisfy any useful or practical differential equation. Also, from the definition
(1) = (^) n lim→∞
1 · 2 · 3 · · · n 1 · 2 · 3 · · · n ( n + 1)
n = 1. (10.3)
Now, application of Eq. (10.2) gives
(2) = 1, (3) = 2 (2) = 2,... (10.4) ( n ) = 1 · 2 · 3 · · · ( n − 1) = ( n − 1)!
We see that the gamma function interpolates the factorials by a continuous function that returns the factorials at integer arguments.
Definite Integral (Euler)
A second definition, also frequently called the Euler integral, is
( z ) ≡
0
e − t^ t z −^1 dt , ( z ) > 0. (10.5)
The restriction on z is necessary to avoid divergence of the integral at t = 0. When the gamma function does appear in physical problems, it is often in this form or some variation, such as
( z ) = 2
0
e − t
2 t^2 z −^1 dt , ( z ) > 0 (10.6)
or
( z ) =
0
[
ln
t
)] z − 1 dt , ( z ) > 0.
526 Chapter 10 The Gamma Function (Factorial Function)
from the definition of the exponential
lim n →∞ F ( z , n ) = F ( z , ∞) =
0
e − t^ t z −^1 dt ≡ ( z ) (10.12)
by Eq. (10.5). Returning to F ( z , n ), we evaluate it in successive integrations by parts. For convenience let u = t / n. Then
F ( z , n ) = nz
0
(1 − u ) n^ u z −^1 du. (10.13)
Integrating by parts, we obtain for z > 0,
F ( z , n ) nz^
= (1 − u ) n^
uz z
1
0
n z
0
(1 − u ) n −^1 u z^ du. (10.14)
Repeating this, with the integrated part vanishing at both end points each time, we finally get
F ( z , n ) = nz^
n ( n − 1) · · · 1 z ( z + 1) · · · ( z + n − 1)
0
uz + n −^1 du
1 · 2 · 3 · · · n z ( z + 1)( z + 2) · · · ( z + n )
nz. (10.15)
This is identical to the expression on the right side of Eq. (10.1). Hence,
lim n →∞ F ( z , n ) = F ( z , ∞) ≡ ( z ) (10.16)
by Eq. (10.1), completing the proof. Using the functional equation (10.2) we can extend ( z ) from positive to negative arguments. For example, starting with Eq. (10.2) we define
π,
and Eq. (10.2) for z → 0 implies that ( z → 0) → ∞. Moreover, z ( z ) → 1 for z → 0 [Eq. (10.2)] shows that ( z ) has a simple pole at the origin. Similarly, we find simple poles of ( z ) at all negative integers.
Infinite Product (Weierstrass)
The third definition (Weierstrass’s form) is
1 ( z )
≡ ze γ^ z
∏^ ∞
n = 1
z n
e − z / n , (10.17)
where γ is the Euler–Mascheroni constant [Eq. (5.27)]
γ = lim n →∞
∑ n
m = 1
m
− ln n
10.1 Definitions and Simple Properties 527
This infinite product form may be used to develop the reflection identity, Eq. (10.24a), and applied in the exercises, such as Exercise 10.1.17. This form can be derived from the original definition [Eq. (10.1)] by rewriting it as
( z ) = lim n →∞
1 · 2 · 3 · · · n z ( z + 1) · · · ( z + n )
nz^ = lim n →∞
z
∏ n
m = 1
z m
nz. (10.19)
Inverting Eq. (10.19) and using
n − z^ = e − z^ ln^ n , (10.20)
we obtain
1 ( z )
= z lim n →∞ e (−^ ln^ n ) z
∏ n
m = 1
z m
Multiplying and dividing by
exp
[(
n
z
]
∏^ n
m = 1
e z / m , (10.22)
we get
1 ( z )
= z
lim n →∞ exp
[(
n
− ln n
z
]}
×
[
lim n →∞
∏^ n
m = 1
z m
e − z / m
]
As shown in Section 5.2, the infinite series in the exponent converges and defines γ , the Euler–Mascheroni constant. Hence, Eq. (10.17) follows. The Weierstrass infinite product definition of ( z ) leads directly to an important identity,
( z )(1 − z ) =
π sin z π
, (10.24a)
using the infinite product formulas for ( z ), (1 − z ) and sin z [Eq. (7.60)]. Alternatively, we can start from the product of Euler integrals
( z + 1)(1 − z ) =
0
s z^ e − s^ ds
0
t − z^ e − t^ dt
0
v z^
d v (v + 1)^2
0
e − u^ u du =
π z sin π z
transforming from the variables s , t to u = s + t , v = s / t , as suggested by com- bining the exponentials and the powers in the integrands. The Jacobian is
J = −
t
s t^2
s + t t^2
(v + 1)^2 u
where (v + 1) t = u. The integral
0 e
− u (^) u du = 1, whereas that over v may be
derived by contour integration (Exercise 7.2.18), giving (^) sinπ z π z. Similarly, one
10.1 Definitions and Simple Properties 529
to define a factorial function z !. Occasionally, we may even encounter Gauss’s notation,
( z ), for the factorial function ∏ ( z ) = z !. (10.26)
The notation is due to Legendre. The factorial function of Eq. (10.25a) is, of course, related to the gamma function by
( z ) = ( z − 1)!, or ( z + 1) = z !. (10.27)
If z = n , a positive integer [Eq. (10.4)] shows that
z! = n! = 1 · 2 · 3 · · · n , (10.28)
the familiar factorial. However, it should be noted that since z! is now defined by Eq. (10.25b) [or equivalently by Eq. (10.27)] the factorial function is no longer limited to positive integral values of the argument (Fig. 10.1). The difference relation [Eq. (10.2)] becomes
( z − 1)! =
z! z
This shows immediately that
0! = 1 (10.30)
x!
5 4 3 2 1
x –5 –3 –1 1 2 3 4 5
Figure 10.
The Factorial Function---Extension to Negative Arguments
530 Chapter 10 The Gamma Function (Factorial Function)
6 5 4 3 2 1 0
–1. –1.0 0 1.0 2.0 3.0 4.
x!
x!
d^2 dx^2 ln ( x !)
d^2 dx^2 ln ( x !)
d dx ln ( x !)
d dx ln ( x !)
x
Figure 10. The Factorial Function and the First Two Derivatives of ln( x !)
and
n! = ±∞ for n , a negative integer. (10.31)
In terms of the factorial, Eq. (10.24a) becomes
z !(− z )! =
π z sin π z
because
( z )(1 − z ) = ( z − 1)!(1 − z − 1)! = ( z − 1)!(− z )! =
z
z !(− z )!.
By restricting ourselves to the real values of the argument, we find that x! defines the curve shown in Fig. 10.2. The minimum of the curve in Fig. 10.1 is
x! = (0. 46163 · · ·)! = 0. 88560 · · ·. (10.33a)
Double Factorial Notation
In many problems of mathematical physics, particularly in connection with Legendre polynomials (Chapter 11), we encounter products of the odd positive integers and products of the even positive integers. For convenience, these are given special labels as double factorials:
1 · 3 · 5 · · · (2 n + 1) = (2 n + 1)!! 2 · 4 · 6 · · · (2 n ) = (2 n )!!.
(10.33b)
Clearly, these are related to the regular factorial functions by
(2 n )!! = 2 n^ n! and (2 n + 1)!! =
(2 n + 1)! 2 n^ n!
. (10.33c)
532 Chapter 10 The Gamma Function (Factorial Function)
function of ν < −1, then Eqs. (10.34) and (10.35) are verified for all ν (except negative integers).
EXERCISES
10.1.1 Derive the recurrence relations
( z + 1) = z ( z )
from the Euler integral [Eq. (10.5)]
( z ) =
0
e − t^ t z −^1 dt.
10.1.2 In a power series solution for the Legendre functions of the second kind, we encounter the expression
( n + 1)( n + 2)( n + 3) · · · ( n + 2 s − 1)( n + 2 s ) 2 · 4 · 6 · 8 · · · (2 s − 2)(2 s ) · (2 n + 3)(2 n + 5)(2 n + 7) · · · (2 n + 2 s + 1)
in which s is a positive integer. Rewrite this expression in terms of factorials. 10.1.3 Show that ( s − n )! (2 s − 2 n )!
(−1) n − s (2 n − 2 s )! ( n − s )!
where s and n are integers with s < n. This result can be used to avoid negative factorials such as in the series representations of the spher- ical Neumann functions and the Legendre functions of the second kind. 10.1.4 Show that ( z ) may be written
( z ) = 2
0
e − t
2 t^2 z −^1 dz , ( z ) > 0,
( z ) =
0
[
ln
t
)] z − 1 dt , ( z ) > 0.
10.1.5 In a Maxwellian distribution the fraction of particles with speed between v and v + d v is
dN N
= 4 π
m 2 π kT
exp
m v^2 2 kT
v^2 d v,
where N is the total number of particles. The average or expectation value of v n^ is defined as 〈v n 〉 = N −^1
v n^ dN. Show that
〈v n 〉 =
2 kT m
) n / 2 ( n + 1 2
10.1 Definitions and Simple Properties 533
10.1.6 By transforming the integral into a gamma function, show that
0
x k^ ln x dx =
( k + 1)^2
, k > − 1.
10.1.7 Show that ∫ (^) ∞
0
e − x 4 dx =
10.1.8 Show that
lim x → 0
( ax − 1)! ( x − 1)!
a
10.1.9 Locate the poles of ( z ). Show that they are simple poles and deter- mine the residues.
10.1.10 Show that the equation x! = k , k = 0, has an infinite number of real roots.
10.1.11 Show that
(a)
0
x^2 s +^1 exp(− ax^2 ) dx =
s! 2 as +^1
(b)
0
x^2 s^ exp(− ax^2 ) dx =
( s − 12 )! 2 as +^1 /^2
(2 s − 1)!! 2 s +^1 as
π a
These Gaussian integrals are of major importance in statistical mechanics.
10.1.12 (a) Develop recurrence relations for (2 n )!! and for (2 n + 1)!!. (b) Use these recurrence relations to calculate (or define) 0!! and (−1)!!.
ANS. 0!! = 1, (−1)!! = 1.
10.1.13 For s a nonnegative integer, show that
(− 2 s − 1)!! =
(−1) s (2 s − 1)!!
(−1) s 2 s^ s! (2 s )!
10.1.14 Express the coefficient of the n th term of the expansion of (1 + x )^1 /^2 (a) in terms of factorials of integers; and (b) in terms of the double factorial (!!) functions.
ANS. a (^) n = (−1) n +^1
(2 n − 3)! 22 n −^2 n !( n − 2)!
= (−1) n +^1
(2 n − 3)!! (2 n )!!
, n = 2, 3, · · ·.
10.1.15 Express the coefficient of the n th term of the expansion of (1 + x )^1 /^2 (a) in terms of the factorials of integers; and (b) in terms of the double factorial (!!) functions.
ANS. a (^) n = (−1) n^
(2 n )! 22 n ( n !)^2
= (−1) n^
(2 n − 1)!! (2 n )!!
, n = 1, 2, 3 · · ·.
10.2 Digamma and Polygamma Functions 535
10.1.23 The probability density associated with the normal distribution of statistics is given by
f ( x ) =
σ (2π)^1 /^2
exp
[
( x − μ)^2 2 σ 2
]
with (−∞, ∞) for the range of x. Show that (a) the mean value of x , 〈 x 〉 is equal to μ; and (b) the standard deviation (〈 x^2 〉 − 〈 x 〉^2 )^1 /^2 is given by σ. 10.1.24 From the gamma distribution
f ( x ) =
β α^ (α)
x α−^1 e − x /β^ , x > 0 0, x ≤ 0,
show that (a) 〈 x 〉 (mean) = αβ, (b) σ 2 (variance) ≡ 〈 x^2 〉 − 〈 x 〉^2 = αβ^2. 10.1.25 The wave function of a particle scattered by a Coulomb potential is ψ( r , θ). At the origin the wave function becomes
ψ(0) = e −πγ /^2 (1 + i γ ),
where γ = Z 1 Z 2 e^2 / h ¯v. Show that
|ψ(0)|^2 =
2 πγ e^2 πγ^ − 1
10.1.26 Derive the contour integral representation of Eq. (10.34)
(2 i )ν! sin νπ =
C
e − z (− z )ν^ dz.
10.2 Digamma and Polygamma Functions
Digamma Function
As may be noted from the definitions in Section 10.1, it is inconvenient to deal with the derivatives of the gamma or factorial function directly. Instead, it is customary to take the natural logarithm of the factorial function [Eq. (10.1)], convert the product to a sum, and then differentiate; that is, we start from
z! = z ( z ) = lim n →∞
n! ( z + 1)( z + 2) · · · ( z + n )
nz. (10.36)
Then, because the logarithm of the limit is equal to the limit of the logarithm, we have
ln( z !) = lim n →∞ [ln( n !) + z ln n − ln( z + 1)
− ln( z + 2) − · · · − ln( z + n )]. (10.37)
536 Chapter 10 The Gamma Function (Factorial Function)
Differentiating with respect to z , we obtain d dz
ln( z !) ≡ ψ( z + 1) = lim n →∞
ln n −
z + 1
z + 2
z + n
which defines ψ( z + 1), the digamma function. From the definition of the Euler–Mascheroni constant, 1 Eq. (10.38) may be rewritten as
ψ( z + 1) = −γ −
n = 1
z + n
n
= −γ +
n = 1
z n ( n + z )
Clearly,
ψ(1) = −γ = − 0 .577 215 664 901 · · ·.^2 (10.40)
Another, even more useful, expression for ψ( z ) is derived in Section 10.3.
Polygamma Function
The digamma function may be differentiated repeatedly, giving rise to the polygamma function:
ψ( m )^ ( z + 1) ≡
d m +^1 dz m +^1
ln( z !)
= (−1) m +^1 m!
∑^ ∞
n = 1
( z + n ) m +^1
, m = 1, 2, 3,.... (10.41)
A plot of ψ( x + 1) and ψ′( x + 1) is included in Fig. 10.2. Since the series in Eq. (10.41) defines the Riemann zeta function, 3 when z is set to zero,
ζ ( m ) ≡
n = 1
n m^
we have
ψ( m )^ (1) = (−1) m +^1 m !ζ ( m + 1), m = 1, 2, 3,.... (10.43)
The values of the polygamma functions of positive integral argument, ψ( m )^ ( n ), may be calculated using Exercise 10.2.6. In terms of the more common notation, d n +^1 dz n +^1
ln ( z ) =
d n dz n^
ψ( z ) = ψ( n )^ ( z ). (10.44a)
(^1) Compare Section 5.2, Eq. (5.27). We add and substract ∑ ns = 1 s − (^1). (^2) γ has been computed to 1271 places by D. E. Knuth, Math. Comput. 16 , 275 (1962) and to 3566 decimal places by D. W. Sweeney, Math. Comput. 17 , 170 (1963). It may be of interest that the fraction 228/395 gives γ accurate to six places. (^3) See Chapter 5. For z = 0 this series may be used to define a generalized zeta function.
538 Chapter 10 The Gamma Function (Factorial Function)
EXERCISES
10.2.1 Verify that the following two forms of the digamma function,
ψ( x + 1) =
∑^ x
r = 1
r
− γ
and
ψ( x + 1) =
r = 1
x r ( r + x )
− γ ,
are equal to each other (for x a positive integer). 10.2.2 Show that ψ( z + 1) has the series expansion
ψ( z + 1) = −γ +
n = 2
(−1) n ζ( n ) z n −^1.
10.2.3 For a power series expansion of ln( z !), AMS-55 lists
ln( z !) = − ln(1 + z ) + z (1 − γ ) +
n = 2
(−1) n [ζ ( n ) − 1] z n / n.
(a) Show that this agrees with Eq. (10.44c) for | z | < 1. (b) What is the range of convergence of this new expression? 10.2.4 Show that 1 2
ln
π z sin π z
n = 1
ζ (2 n ) 2 n
z^2 n , | z | < 1.
Hint. Try Eq. (10.32). 10.2.5 Write out a Weierstrass infinite product definition of z !. Without dif- ferentiating, show that this leads directly to the Maclaurin expansion of ln( z !) [Eq. (10.44b)]. 10.2.6 Derive the difference relation for the polygamma function
ψ( m )^ ( z + 2) = ψ( m )^ ( z + 1) + (−1) m^
m! ( z + 1) m +^1
, m = 0, 1, 2,....
10.2.7 Show that if ( x + iy ) = u + i v
then ( x − iy ) = u − i v.
This is a special case of the Schwarz reflection principle (Section 6.5). 10.2.8 The Pochhammer symbol ( a ) n is defined as ( a ) n = a ( a + 1) · · · ( a + n − 1), ( a ) 0 = 1
(for integral n ).
10.2 Digamma and Polygamma Functions 539
(a) Express ( a ) n in terms of factorials. (b) Find ( d / da )( a ) n in terms of ( a ) n and digamma functions.
ANS.
d da
( a ) n = ( a ) n [F( a + n − 1) − F( a − 1)]. (c) Show that ( a ) n + k = ( a + n ) k · ( a ) n.
10.2.9 Verify the following special values of the ψ form of the di- and poly- gamma functions: ψ(1) = −γ , ψ(1)^ (1) = ζ(2), ψ(2)^ (1) = − 2 ζ (3).
10.2.10 Derive the polygamma function recurrence relation
ψ( m )^ (1 + z ) = ψ( m )^ ( z ) + (−1) mm !/ z m +^1 , m = 0, 1, 2,....
10.2.11 Verify
(a)
0
e − r^ ln r dr = −γ.
(b)
0
re − r^ ln r dr = 1 − γ.
(c)
0
r n^ e − r^ ln r dr = ( n −1)! + n
0
r n −^1 e − r^ ln r dr , n = 1, 2, 3,....
Hint. These may be verified by integration by parts, three parts, or differentiating the integral form of n! with respect to n.
10.2.12 Dirac relativistic wave functions for hydrogen involve factors such as [2(1 − α^2 Z^2 )^1 /^2 ]!, where α, the fine structure constant, is 1371 , and Z is the atomic number. Expand [2(1 − α^2 Z^2 )^1 /^2 ]! in a series of powers of α^2 Z^2.
10.2.13 The quantum mechanical description of a particle in a Coulomb field requires a knowledge of the phase of the complex factorial function. Determine the phase of (1 + ib )! for small b.
10.2.14 The total energy radiated by a black body is given by
u =
8 π k^4 T^4 c^3 h^3
0
x^3 e x^ − 1
dx.
Show that the integral in this expression is equal to 3!ζ (4) [ζ (4) = π^4 / 90 = 1. 0823.. .]. The final result is the Stefan–Boltzmann law.
10.2.15 As a generalization of the result in Exercise 10.2.14, show that ∫ (^) ∞
0
x s^ dx e x^ − 1
= s !ζ( s + 1), ( s ) > 0.
10.2.16 The neutrino energy density (Fermi distribution) in the early history of the universe is given by
ρν =
4 π h^3
0
x^3 exp( x / kT ) + 1
dx.
10.3 Stirling’s Series 541
in which the b 2 n are related to the Bernoulli numbers B 2 n by
(2 n )! b 2 n = B 2 n , (10.46)
B 0 = 1, B 6 =
B 2 =
, B 8 = −
B 4 = −
, B 10 =
, and so on.
By applying Eq. (10.45) to the elementary definite integral ∫ (^) ∞
0
dx ( z + x )^2
z
, f ( x ) =
( z + x )^2
(for z not on the negative real axis), we obtain for n → ∞, 1 z
2 z^2
2! b 2 z^3
4! b 4 z^5
This is the reason for using Eq. (10.48). The Euler–Maclaurin evaluation yields ψ′( z + 1), which is d^2 ln( z !)/ dz^2 =
n = 1
1 ( z + n )^2 from Eq. (10.41). Using Eq. (10.46) and solving for ψ(1)^ ( z + 1), we have
ψ′( z + 1) =
d dz
ψ( z + 1) =
z
2 z^2
B 2
z^3
B 4
z^5
z
2 z^2
n = 1
B 2 n z^2 n +^1
Since the Bernoulli numbers diverge strongly, this series does not converge. It is a semiconvergent or asymptotic series (Section 5.10), in which the sum has always a finite number of terms (compare Section 5.10). Integrating once, we get the digamma function
ψ( z + 1) = C 1 + ln z +
2 z
B 2
2 z^2
B 4
4 z^4
= C 1 + ln z +
2 z
n = 1
B 2 n 2 nz^2 n^
Integrating Eq. (10.51) with respect to z from z − 1 to z and then letting z approach infinity, C 1 , the constant of integration may be shown to vanish. This gives us a second expression for the digamma function, often more useful than Eq. (10.38).
Stirling’s Series
The indefinite integral of the digamma function [Eq. (10.51)] is
ln( z !) = C 2 +
z +
ln z − z +
B 2
2 z
B 2 n 2 n (2 n − 1) z^2 n −^1
542 Chapter 10 The Gamma Function (Factorial Function)
9
S 1 2 3 4 5 6 7 8
0.83% low
2 p s s +1/2^ e – s s!
s!
2 p s s +1/2^ e – s^ 1 +^121 s
Figure 10. Accuracy of Stirling’s Formula
in which C 2 is another constant of integration. To fix C 2 , we start from the asymptotic formula [Eq. (7.89) from Example 7.3.2] ( z )! ∼
2 π z z +^1 /^2 e − z. This gives for large enough | z |
ln z! ∼
ln 2π + ( z + 1 /2) ln z − z , (10.53)
and comparing with Eq. (10.52), we find that C 2 is
C 2 =
ln 2π, (10.54)
giving for | z | → ∞
ln( z !) =
ln 2π +
z +
ln z − z +
12 z
360 z^3
1260 z^5
This is Stirling’s series, an asymptotic expansion. The absolute value of the error is less than the absolute value of the first term neglected. For large enough | z |, the simplest approximation ln z! ∼ z ln z − z may be sufficient. To help convey a sense of the remarkable precision of Stirling’s series for s !, the ratio of the first term of Stirling’s approximation to s! is plotted in Fig. 10.5. A tabulation gives the ratio of the first term in the expansion to s! and the ratio of the first two terms in the expansion to s! (Table 10.1). The derivation of these forms is Exercise 10.3.1.
Numerical Computation
The possibility of using the Maclaurin expansion [Eq. (10.44b)] for the numer- ical evaluation of the factorial function is mentioned in Section 10.2. How- ever, for large x , Stirling’s series [Eq. (10.55)] gives much more accuracy. The
