chapter the gamma function (factorial...

CHAPTER 8

THE GAMMA FUNCTION

(FACTORIAL FUNCTION)

The gamma function appears occasionally in physical problems such as the normalizationof Coulomb wave functions and the computation of probabilities in statistical mechanics.In general, however, it has less direct physical application and interpretation than, say, theLegendre and Bessel functions of Chapters 11 and 12. Rather, its importance stems from itsusefulness in developing other functions that have direct physical application. The gammafunction, therefore, is included here.

8.1 DEFINITIONS, 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, andto show the equivalence of the three forms.

Infinite Limit (Euler)

The first definition, named after Euler, is

Ŵ(z)≡ limn→∞

1 · 2 · 3 · · ·nz(z+ 1)(z+ 2) · · · (z+ n)

nz, z = 0,−1,−2,−3, . . . . (8.1)

This definition ofŴ(z) is useful in developing the Weierstrass infinite-product form ofŴ(z), Eq. (8.16), and in obtaining the derivative of lnŴ(z) (Section 8.2). Here and else-

499

500 Chapter 8 Gamma–Factorial Function

where in this chapterz may be either real or complex. Replacingz with z+ 1, we have

Ŵ(z+ 1) = limn→∞

1 · 2 · 3 · · ·n(z+ 1)(z+ 2)(z+ 3) · · · (z+ n+ 1)

nz+1

= limn→∞

nz

z+ n+ 1· 1 · 2 · 3 · · ·nz(z+ 1)(z+ 2) · · · (z+ n)

nz

= zŴ(z). (8.2)

This is the basic functional relation for the gamma function. It should be noted that itis a difference equation. It has been shown that the gamma function is one of a generalclass of functions that do not satisfy any differential equation with rational coefficients.Specifically, the gamma function is one of the very few functions of mathematical physicsthat does not satisfy either the hypergeometric differential equation (Section 13.4) or theconfluent hypergeometric equation (Section 13.5).

Also, from the definition,

Ŵ(1)= limn→∞

1 · 2 · 3 · · ·n1 · 2 · 3 · · ·n(n+ 1)

n= 1. (8.3)

Now, application of Eq. (8.2) gives

Ŵ(2) = 1,

Ŵ(3) = 2Ŵ(2)= 2, . . . (8.4)

Ŵ(n) = 1 · 2 · 3 · · · (n− 1)= (n− 1)!.

Definite Integral (Euler)

A second definition, also frequently called the Euler integral, is

Ŵ(z)≡∫ ∞

0e−t tz−1dt, ℜ(z) > 0. (8.5)

The restriction onz is necessary to avoid divergence of the integral. When the gammafunction does appear in physical problems, it is often in this form or some variation, suchas

Ŵ(z) = 2∫ ∞

0e−t

2t2z−1dt, ℜ(z) > 0. (8.6)

Ŵ(z) =∫ 1

0

[ln

(1

t

)]z−1

dt, ℜ(z) > 0. (8.7)

Whenz= 12 , Eq. (8.6) is just the Gauss error integral, and we have the interesting result

Ŵ(1

2

)=√π. (8.8)

Generalizations of Eq. (8.6), the Gaussian integrals, are considered in Exercise 8.1.11. Thisdefinite integral form ofŴ(z), Eq. (8.5), leads to the beta function, Section 8.4.

8.1 Definitions, Simple Properties 501

To show the equivalence of these two definitions, Eqs. (8.1) and (8.5), consider thefunction of two variables

F(z,n)=∫ n

0

(1− t

n

)n

tz−1dt, ℜ(z) > 0, (8.9)

with n a positive integer.1 Since

limn→∞

(1− t

n

)n

≡ e−t , (8.10)

from the definition of the exponential

limn→∞

F(z,n)= F(z,∞)=∫ ∞

0e−t tz−1dt ≡ Ŵ(z) (8.11)

by Eq. (8.5).Returning toF(z,n), we evaluate it in successive integrations by parts. For convenience

let u= t/n. Then

F(z,n)= nz∫ 1

0(1− u)nuz−1du. (8.12)

Integrating by parts, we obtain

F(z,n)

nz= (1− u)n

uz

z

∣∣∣∣1

0+ n

z

∫ 1

0(1− u)n−1uz du. (8.13)

Repeating this with the integrated part vanishing at both endpoints each time, we finallyget

F(z,n) = nzn(n− 1) · · ·1

z(z+ 1) · · · (z+ n− 1)

∫ 1

0uz+n−1du

= 1 · 2 · 3 · · ·nz(z+ 1)(z+ 2) · · · (z+ n)

nz. (8.14)

This is identical with the expression on the right side of Eq. (8.1). Hence

limn→∞

F(z,n)= F(z,∞)≡ Ŵ(z), (8.15)

by Eq. (8.1), completing the proof.

Infinite Product (Weierstrass)

The third definition (Weierstrass’ form) is

1

Ŵ(z)≡ zeγ z

∞∏

n=1

(1+ z

n

)e−z/n, (8.16)

1The form ofF(z,n) is suggested by the beta function (compare Eq. (8.60)).


whereγ is the Euler–Mascheroni constant,

γ = 0.5772156619. . . . (8.17)

This infinite-product form may be used to develop the reflection identity, Eq. (8.23), andapplied in the exercises, such as Exercise 8.1.17. This form can be derived from the originaldefinition (Eq. (8.1)) by rewriting it as

Ŵ(z)= limn→∞

1 · 2 · 3 · · ·nz(z+ 1) · · · (z+ n)

nz = limn→∞

1

z

n∏

m=1

(1+ z

m

)−1

nz. (8.18)

Inverting Eq. (8.18) and using

n−z = e(− lnn)z, (8.19)

we obtain

1

Ŵ(z)= z lim

n→∞e(− lnn)z

n∏

m=1

(1+ z

m

). (8.20)

Multiplying and dividing by

exp

[(1+ 1

2+ 1

3+ · · · + 1

n

)z

]=

n∏

m=1

ez/m, (8.21)

we get

1

Ŵ(z)= z

{limn→∞

exp

[(1+ 1

2+ 1

3+ · · · + 1

n− lnn

)z

]}

×[

limn→∞

n∏

m=1

(1+ z

m

)e−z/m

]. (8.22)

As shown in Section 5.2, the parenthesis in the exponent approaches a limit, namelyγ , theEuler–Mascheroni constant. Hence Eq. (8.16) follows.

It was shown in Section 5.11 that the Weierstrass infinite-product definition ofŴ(z) leddirectly to an important identity,

Ŵ(z)Ŵ(1− z)= π

sinzπ. (8.23)

Alternatively, we can start from the product of Euler integrals,

Ŵ(z+ 1)Ŵ(1− z) =∫ ∞

0sze−sds

∫ ∞

0t−ze−t dt

=∫ ∞

0vz

dv

(v + 1)2

∫ ∞

0e−uudu= πz

sinπz,

transforming from the variabless, t to u= s + t, v = s/t , as suggested by combining theexponentials and the powers in the integrands. The Jacobian is

J =−∣∣∣∣1 11t− s

t2

∣∣∣∣=s + t

t2= (v+ 1)2

u,


where(v + 1)t = u. The integral∫∞

0 e−uudu = 1, while that overv may be derived bycontour integration, giving πz

sinπz .This identity may also be derived by contour integration (Example 7.1.6 and Exer-

cises 7.1.18 and 7.1.19) and the beta function, Section 8.4. Settingz = 12 in Eq. (8.23),

we obtain

Ŵ(1

2

)=√π (8.24a)

(taking the positive square root), in agreement with Eq. (8.8).Similarly one can establishLegendre’s duplication formula,

Ŵ(1+ z)Ŵ(z+ 1

2

)= 2−2z√πŴ(2z+ 1). (8.24b)

The Weierstrass definition shows immediately thatŴ(z) has simple poles atz =0,−1,−2,−3, . . . and that[Ŵ(z)]−1 has no poles in the finite complex plane, which meansthatŴ(z) has no zeros. This behavior may also be seen in Eq. (8.23), in which we note thatπ/(sinπz) is never equal to zero.

Actually the infinite-product definition ofŴ(z) may be derived from the Weierstrassfactorization theorem with the specification that[Ŵ(z)]−1 have simple zeros atz =0,−1,−2,−3, . . . . The Euler–Mascheroni constant is fixed by requiringŴ(1) = 1. Seealso the products expansions of entire functions in Section 7.1.

In probability theory the gamma distribution (probability density) is given by

f (x)=

1

βαŴ(α)xα−1e−x/β , x > 0

0, x ≤ 0.(8.24c)

The constant[βαŴ(α)]−1 is chosen so that the total (integrated) probability will be unity.Forx→E, kinetic energy,α→ 3

2 , andβ→ kT , Eq. (8.24c) yields the classical Maxwell–Boltzmann statistics.

Factorial Notation

So far this discussion has been presented in terms of the classical notation. As pointed outby Jeffreys and others, the−1 of thez− 1 exponent in our second definition (Eq. (8.5)) isa continual nuisance. Accordingly, Eq. (8.5) is sometimes rewritten as

∫ ∞

0e−t tz dt ≡ z!, ℜ(z) >−1, (8.25)

to define a factorial functionz!. Occasionally we may still encounter Gauss’ notation,∏(z), for the factorial function:

∏(z)= z! = Ŵ(z+ 1). (8.26)

TheŴ notation is due to Legendre. The factorial function of Eq. (8.25) is related to thegamma function by

Ŵ(z)= (z− 1)! or Ŵ(z+ 1)= z!. (8.27)


FIGURE 8.1 The factorialfunction — extension to negative

arguments.

If z= n, a positive integer (Eq. (8.4)) shows that

z! = n! = 1 · 2 · 3 · · ·n, (8.28)

the familiar factorial. However, it should be noted that sincez! is now defined by Eq. (8.25)(or equivalently by Eq. (8.27)) the factorial function is no longer limited to positive integralvalues of the argument (Fig. 8.1). The difference relation (Eq. (8.2)) becomes

(z− 1)! = z!z. (8.29)

This shows immediately that

0! = 1 (8.30)

and

n! = ±∞ for n, anegativeinteger. (8.31)

In terms of the factorial, Eq. (8.23) becomes

z!(−z)! = πz

sinπz. (8.32)

By restricting ourselves to the real values of the argument, we find thatŴ(x+1) definesthe curves shown in Figs. 8.1 and 8.2. The minimum of the curve is

Ŵ(x + 1)= x! = (0.46163. . .)! = 0.88560. . . . (8.33a)


FIGURE 8.2 The factorial function and the first two derivatives ofln(Ŵ(x + 1)).

Double Factorial Notation

In many problems of mathematical physics, particularly in connection with Legendre poly-nomials (Chapter 12), we encounter products of the odd positive integers and products ofthe even positive integers. For convenience these are given special labels as double facto-rials:

1 · 3 · 5 · · · (2n+ 1) = (2n+ 1)!!2 · 4 · 6 · · · (2n) = (2n)!!.

(8.33b)

Clearly, these are related to the regular factorial functions by

(2n)!! = 2nn! and (2n+ 1)!! = (2n+ 1)!2nn! . (8.33c)

We also define(−1)!! = 1, a special case that does not follow from Eq. (8.33c).

Integral Representation

An integral representation that is useful in developing asymptotic series for the Besselfunctions is ∫

C

e−zzν dz=(e2πiν − 1

)Ŵ(ν + 1), (8.34)

whereC is the contour shown in Fig. 8.3. This contour integral representation is onlyuseful whenν is not an integer,z= 0 then being abranch point. Equation (8.34) may be


FIGURE 8.3 Factorial function contour.

FIGURE 8.4 The contour of Fig. 8.3 deformed.

readily verified forν > −1 by deforming the contour as shown in Fig. 8.4. The integralfrom∞ into the origin yields−(ν!), placing the phase ofz at 0. The integral out to∞ (inthe fourth quadrant) then yieldse2πiνν!, the phase ofz having increased to 2π . Since thecircle around the origin contributes nothing whenν >−1, Eq. (8.34) follows.

It is often convenient to cast this result into a more symmetrical form:∫

C

e−z(−z)ν dz= 2iŴ(ν + 1)sin(νπ). (8.35)

This analysis establishes Eqs. (8.34) and (8.35) forν > −1. It is relatively simple toextend the range to include all nonintegralν. First, we note that the integral exists forν < −1 as long as we stay away from the origin. Second, integrating by parts we findthat Eq. (8.35) yields the familiar difference relation (Eq. (8.29)). If we take the differencerelation to define the factorial function ofν <−1, then Eqs. (8.34) and (8.35) are verifiedfor all ν (except negative integers).

Exercises

8.1.1 Derive the recurrence relations

Ŵ(z+ 1)= zŴ(z)

from the Euler integral (Eq. (8.5)),

Ŵ(z)=∫ ∞

0e−t tz−1dt.


8.1.2 In a power-series solution for the Legendre functions of the second kind we encounterthe expression

(n+ 1)(n+ 2)(n+ 3) · · · (n+ 2s − 1)(n+ 2s)

2 · 4 · 6 · 8 · · · (2s − 2)(2s) · (2n+ 3)(2n+ 5)(2n+ 7) · · · (2n+ 2s + 1),

in which s is a positive integer. Rewrite this expression in terms of factorials.

8.1.3 Show that, ass − n→ negative integer,

(s − n)!(2s − 2n)! →

(−1)n−s(2n− 2s)!(n− s)! .

Heres andn are integers withs < n. This result can be used to avoid negative facto-rials, such as in the series representations of the spherical Neumann functions and theLegendre functions of the second kind.

8.1.4 Show thatŴ(z) may be written

Ŵ(z) = 2∫ ∞

0e−t

2t2z−1dt, ℜ(z) > 0,

Ŵ(z) =∫ 1

0

[ln

(1

t

)]z−1

dt, ℜ(z) > 0.

8.1.5 In a Maxwellian distribution the fraction of particles with speed betweenv andv + dv

is

dN

N= 4π

(m

2πkT

)3/2

exp

(−mv2

2kT

)v2dv,

N being the total number of particles. The average or expectation value ofvn is definedas〈vn〉 =N−1

∫vn dN . Show that

⟨vn⟩=(

2kT

m

)n/2Ŵ(n+3

2

)

Ŵ(3/2).

8.1.6 By transforming the integral into a gamma function, show that

−∫ 1

0xk lnx dx = 1

(k + 1)2, k >−1.

8.1.7 Show that∫ ∞

0e−x

4dx = Ŵ

(5

4

).

8.1.8 Show that

limx→0

(ax − 1)!(x − 1)! =

1

a.

8.1.9 Locate the poles ofŴ(z). Show that they are simple poles and determine the residues.

8.1.10 Show that the equationx! = k, k = 0, has an infinite number of real roots.

8.1.11 Show that


(a)∫ ∞

0x2s+1 exp

(−ax2)dx = s!

2as+1.

(b)∫ ∞

0x2s exp

(−ax2)dx = (s − 1

2)!2as+1/2

= (2s − 1)!!2s+1as

√π

a.

These Gaussian integrals are of major importance in statistical mechanics.

8.1.12 (a) Develop recurrence relations for(2n)!! and for(2n+ 1)!!.(b) Use these recurrence relations to calculate (or to define) 0!! and(−1)!!.

ANS. 0!! = 1, (−1)!! = 1.

8.1.13 For s a nonnegative integer, show that

(−2s − 1)!! = (−1)s

(2s − 1)!! =(−1)s2ss!

(2s)! .

8.1.14 Express the coefficient of thenth term of the expansion of(1+ x)1/2

(a) in terms of factorials of integers,(b) in terms of the double factorial (!!) functions.

ANS. an = (−1)n+1 (2n− 3)!22n−2n!(n− 2)! = (−1)n+1 (2n− 3)!!

(2n)!! , n= 2,3, . . . .

8.1.15 Express the coefficient of thenth term of the expansion of(1+ x)−1/2

(a) in terms of the factorials of integers,(b) in terms of the double factorial (!!) functions.

ANS. an = (−1)n(2n)!

22n(n!)2 = (−1)n(2n− 1)!!(2n)!! , n= 1,2,3, . . . .

8.1.16 The Legendre polynomial may be written as

Pn(cosθ) = 2(2n− 1)!!(2n)!!

{cosnθ + 1

1· n

2n− 1cos(n− 2)θ

+ 1 · 31 · 2

n(n− 1)

(2n− 1)(2n− 3)cos(n− 4)θ

+ 1 · 3 · 51 · 2 · 3

n(n− 1)(n− 2)

(2n− 1)(2n− 3)(2n− 5)cos(n− 6)θ + · · ·

}.

Let n= 2s + 1. Then

Pn(cosθ)= P2s+1(cosθ)=s∑

m=0

am cos(2m+ 1)θ.

Findam in terms of factorials and double factorials.


8.1.17 (a) Show that

Ŵ(1

2 − n)Ŵ(1

2 + n)= (−1)nπ,

wheren is an integer.(b) ExpressŴ(1

2 + n) andŴ(12 − n) separately in terms ofπ1/2 and a!! function.

ANS.Ŵ(12 + n)= (2n− 1)!!

2nπ1/2.

8.1.18 From one of the definitions of the factorial or gamma function, show that∣∣(ix)!

∣∣2= πx

sinhπx.

8.1.19 Prove that

∣∣Ŵ(α + iβ)∣∣=

∣∣Ŵ(α)∣∣∞∏

n=0

[1+ β2

(α + n)2

]−1/2

.

This equation has been useful in calculations of beta decay theory.

8.1.20 Show that

∣∣(n+ ib)!∣∣=

(πb

sinhπb

)1/2 n∏

s=1

(s2+ b2)1/2

for n, a positive integer.

8.1.21 Show that

|x!| ≥∣∣(x + iy)!

∣∣

for all x. The variablesx andy are real.

8.1.22 Show that∣∣Ŵ(1

2 + iy)∣∣2= π

coshπy.

8.1.23 The probability density associated with the normal distribution of statistics is given by

f (x)= 1

σ(2π)1/2exp

[− (x −µ)2

2σ 2

],

with (−∞,∞) for the range ofx. Show that

(a) the mean value ofx, 〈x〉 is equal toµ,(b) the standard deviation(〈x2〉 − 〈x〉2)1/2 is given byσ .

8.1.24 From the gamma distribution

f (x)=

1

βαŴ(α)xα−1e−x/β , x > 0,

0, x ≤ 0,

show that

(a) 〈x〉 (mean)= αβ, (b) σ 2 (variance)≡ 〈x2〉 − 〈x〉2= αβ2.


8.1.25 The wave function of a particle scattered by a Coulomb potential isψ(r, θ). At theorigin the wave function becomes

ψ(0)= e−πγ/2Ŵ(1+ iγ ),

whereγ = Z1Z2e2/h̄v. Show that

∣∣ψ(0)∣∣2= 2πγ

e2πγ − 1.

8.1.26 Derive the contour integral representation of Eq. (8.34),

2iν!sinνπ =∫

C

e−z(−z)ν dz.

8.1.27 Write a function subprogramFACT(N) (fixed-point independent variable) that will cal-culateN !. Include provision for rejection and appropriate error message ifN is nega-tive.Note.For small integerN , direct multiplication is simplest. For largeN , Eq. (8.55),Stirling’s series would be appropriate.

8.1.28 (a) Write a function subprogram to calculate the double factorial ratio(2N − 1)!!/(2N)!!. Include provision forN = 0 and for rejection and an error message ifN isnegative. Calculate and tabulate this ratio forN = 1(1)100.

(b) Check your function subprogram calculation of 199!!/200!! against the value ob-tained from Stirling’s series (Section 8.3).

ANS.199!!200!! = 0.056348.

8.1.29 Using either the FORTRAN-supplied GAMMA or a library-supplied subroutine forx! or Ŵ(x), determine the value ofx for which Ŵ(x) is a minimum(1≤ x ≤ 2) andthis minimum value ofŴ(x). Notice that although the minimum value ofŴ(x) may beobtained to about six significant figures (single precision), the corresponding value ofx

is much less accurate. Why this relatively low accuracy?

8.1.30 The factorial function expressed in integral form can be evaluated by the Gauss–Laguerre quadrature. For a 10-point formula the resultantx! is theoretically exact forx an integer, 0 up through 19. What happens ifx is not an integer? Use the Gauss–Laguerre quadrature to evaluatex!, x = 0.0(0.1)2.0. Tabulate the absolute error as afunction ofx.

Check value.x!exact− x!quadrature= 0.00034 for x = 1.3.

8.2 DIGAMMA AND POLYGAMMA FUNCTIONS

Digamma Functions

As may be noted from the three definitions in Section 8.1, it is inconvenient to deal withthe derivatives of the gamma or factorial function directly. Instead, it is customary to take

8.2 Digamma and Polygamma Functions 511

the natural logarithm of the factorial function (Eq. (8.1)), convert the product to a sum, andthen differentiate; that is,

Ŵ(z+ 1)= zŴ(z)= limn→∞

n!(z+ 1)(z+ 2) · · · (z+ n)

nz (8.36)

and

lnŴ(z+ 1) = limn→∞

[ln(n!)+ z lnn− ln(z+ 1)

− ln(z+ 2)− · · · − ln(z+ n)], (8.37)

in which the logarithm of the limit is equal to the limit of the logarithm. Differentiatingwith respect toz, we obtain

d

dzlnŴ(z+ 1)≡ψ(z+ 1)= lim

n→∞

(lnn− 1

z+ 1− 1

z+ 2− · · · − 1

z+ n

), (8.38)

which definesψ(z + 1), the digamma function. From the definition of the Euler–Mascheroni constant,2 Eq. (8.38) may be rewritten as

ψ(z+ 1) = −γ −∞∑

n=1

(1

z+ n− 1

n

)

= −γ +∞∑

n=1

z

n(n+ z). (8.39)

One application of Eq. (8.39) is in the derivation of the series form of the Neumann function(Section 11.3). Clearly,

ψ(1)=−γ =−0.577 215 664 901. . . .3 (8.40)

Another, perhaps more useful, expression forψ(z) is derived in Section 8.3.

Polygamma Function

The digamma function may be differentiated repeatedly, giving rise to the polygammafunction:

ψ (m)(z+ 1) ≡ dm+1

dzm+1ln(z!)

= (−1)m+1m!∞∑

n=1

1

(z+ n)m+1, m= 1,2,3, . . . . (8.41)

2Compare Sections 5.2 and 5.9. We add and substract∑n

s=1 s−1.

3γ has been computed to 1271 places by D. E. Knuth,Math. Comput.16: 275 (1962), and to 3566 decimal places byD. W. Sweeney,ibid. 17: 170 (1963). It may be of interest that the fraction 228/395 givesγ accurate to six places.


A plot of ψ(x + 1) andψ ′(x + 1) is included in Fig. 8.2. Since the series in Eq. (8.41)defines the Riemann zeta function4 (with z= 0),

ζ(m)≡∞∑

n=1

1

nm, (8.42)

we have

ψ (m)(1)= (−1)m+1m!ζ(m+ 1), m= 1,2,3, . . . . (8.43)

The values of the polygamma functions of positive integral argument,ψ (m)(n+ 1), maybe calculated by using Exercise 8.2.6.

In terms of the perhaps more commonŴ notation,

dn+1

dzn+1lnŴ(z)= dn

dznψ(z)=ψ (n)(z). (8.44a)

Maclaurin Expansion, Computation

It is now possible to write a Maclaurin expansion for lnŴ(z+ 1):

lnŴ(z+ 1)=∞∑

n=1

zn

n!ψ(n−1)(1)=−γ z+

∞∑

n=2

(−1)nzn

nζ(n) (8.44b)

convergent for|z| < 1; for z = x, the range is−1< x ≤ 1. Alternate forms of this seriesappear in Exercise 5.9.14. Equation (8.44b) is a possible means of computingŴ(z+ 1) forreal or complexz, but Stirling’s series (Section 8.3) is usually better, and in addition, anexcellent table of values of the gamma function for complex arguments based on the useof Stirling’s series and the recurrence relation (Eq. (8.29)) is now available.5

Series Summation

The digamma and polygamma functions may also be used in summing series. If the generalterm of the series has the form of a rational fraction (with the highest power of the index inthe numerator at least two less than the highest power of the index in the denominator), itmay be transformed by the method of partial fractions (compare Section 15.8). The infiniteseries may then be expressed as a finite sum of digamma and polygamma functions. Theusefulness of this method depends on the availability of tables of digamma and polygammafunctions. Such tables and examples of series summation are given in AMS-55, Chapter 6(see Additional Readings for the reference).

4See Section 5.9. Forz = 0 this series may be used to define a generalized zeta function.5Table of the Gamma Function for Complex Arguments, Applied Mathematics Series No. 34. Washington, DC: National Bureauof Standards (1954).


Example 8.2.1 CATALAN’S CONSTANT

Catalan’s constant, Exercise 5.2.22, orβ(2) of Section 5.9 is given by

K = β(2)=∞∑

k=0

(−1)k

(2k + 1)2. (8.44c)

Grouping the positive and negative terms separately and starting with unit index (to matchthe form ofψ (1), Eq. (8.41)), we obtain

K = 1+∞∑

n=1

1

(4n+ 1)2− 1

9−

∞∑

n=1

1

(4n+ 3)2.

Now, quoting Eq. (8.41), we get

K = 89 + 1

16ψ(1)(1+ 1

4

)− 1

16ψ(1)(1+ 3

4

). (8.44d)

Using the values ofψ (1) from Table 6.1 of AMS-55 (see Additional Readings for thereference), we obtain

K = 0.91596559. . . .

Compare this calculation of Catalan’s constant with the calculations of Chapter 5, eitherdirect summation or a modification using Riemann zeta function values. �

Exercises

8.2.1 Verify that the following two forms of the digamma function,

ψ(x + 1)=x∑

r=1

1

r− γ

and

ψ(x + 1)=∞∑

r=1

x

r(r + x)− γ,

are equal to each other (forx a positive integer).

8.2.2 Show thatψ(z+ 1) has the series expansion

ψ(z+ 1)=−γ +∞∑

n=2

(−1)nζ(n)zn−1.

8.2.3 For a power-series expansion of ln(z!), AMS-55 (see Additional Readings for reference)lists

ln(z!)=− ln(1+ z)+ z(1− γ )+∞∑

n=2

(−1)n[ζ(n)− 1]zn

n.


(a) Show that this agrees with Eq. (8.44b) for|z|< 1.(b) What is the range of convergence of this new expression?

8.2.4 Show that

1

2ln

(πz

sinπz

)=

∞∑

n=1

ζ(2n)

2nz2n, |z|< 1.

Hint. Try Eq. (8.32).

8.2.5 Write out a Weierstrass infinite-product definition of ln(z!). Without differentiating,show that this leads directly to the Maclaurin expansion of ln(z!), Eq. (8.44b).

8.2.6 Derive the difference relation for the polygamma function

ψ (m)(z+ 2)=ψ (m)(z+ 1)+ (−1)mm!

(z+ 1)m+1, m= 0,1,2, . . . .

8.2.7 Show that if

Ŵ(x + iy)= u+ iv,

then

Ŵ(x − iy)= u− iv.

This is a special case of the Schwarz reflection principle, Section 6.5.

8.2.8 The Pochhammer symbol(a)n is defined as

(a)n = a(a + 1) · · · (a + n− 1), (a)0= 1

(for integraln).

(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

[ψ(a + n)−ψ(a)

].

(c) Show that

(a)n+k = (a + n)k · (a)n.

8.2.9 Verify the following special values of theψ form of the di- and polygamma functions:

ψ(1)=−γ, ψ (1)(1)= ζ(2), ψ (2)(1)=−2ζ(3).

8.2.10 Derive the polygamma function recurrence relation

ψ (m)(1+ z)=ψ (m)(z)+ (−1)mm!/zm+1, m= 0,1,2, . . . .

8.2.11 Verify

(a)∫ ∞

0e−r ln r dr =−γ .


(b)∫ ∞

0re−r ln r dr = 1− γ .

(c)∫ ∞

0rne−r ln r dr = (n− 1)! + n

∫ ∞

0rn−1e−r ln r dr, n= 1,2,3, . . . .

Hint. These may be verified by integration by parts, three parts, or differentiating theintegral form ofn! with respect ton.

8.2.12 Dirac relativistic wave functions for hydrogen involve factors such as[2(1−α2Z2)1/2]!where α, the fine structure constant, is1137 and Z is the atomic number. Expand[2(1− α2Z2)1/2]! in a series of powers ofα2Z2.

8.2.13 The quantum mechanical description of a particle in a Coulomb field requires a knowl-edge of the phase of the complex factorial function. Determine the phase of(1+ ib)!for smallb.

8.2.14 The total energy radiated by a blackbody is given by

u= 8πk4T 4

c3h3

∫ ∞

0

x3

ex − 1dx.

Show that the integral in this expression is equal to 3!ζ(4).[ζ(4)= π4/90= 1.0823. . .] The final result is the Stefan–Boltzmann law.

8.2.15 As a generalization of the result in Exercise 8.2.14, show that∫ ∞

0

xs dx

ex − 1= s!ζ(s + 1), ℜ(s) > 0.

8.2.16 The neutrino energy density (Fermi distribution) in the early history of the universe isgiven by

ρν =4π

h3

∫ ∞

0

x3

exp(x/kT )+ 1dx.

Show that

ρν =7π5

30h3(kT )4.

8.2.17 Prove that∫ ∞

0

xs dx

ex + 1= s!

(1− 2−s

)ζ(s + 1), ℜ(s) > 0.

Exercises 8.2.15 and 8.2.17 actually constitute Mellin integral transforms (compare Sec-tion 15.1).

8.2.18 Prove that

ψ (n)(z)= (−1)n+1∫ ∞

0

tne−zt

1− e−tdt, ℜ(z) > 0.


8.2.19 Using di- and polygamma functions, sum the series

(a)∞∑

n=1

1

n(n+ 1), (b)

∞∑

n=2

1

n2− 1.

Note.You can use Exercise 8.2.6 to calculate the needed digamma functions.

8.2.20 Show that

∞∑

n=1

1

(n+ a)(n+ b)= 1

(b− a)

{ψ(1+ b)−ψ(1+ a)

},

wherea = b and neithera nor b is a negative integer. It is of some interest to comparethis summation with the corresponding integral,

∫ ∞

1

dx

(x + a)(x + b)= 1

b− a

{ln(1+ b)− ln(1+ a)

}.

The relation betweenψ(x) and lnx is made explicit in Eq. (8.51) in the next section.

8.2.21 Verify the contour integral representation ofζ(s),

ζ(s)=− (−s)!2πi

∫

C

(−z)s−1

ez − 1dz.

The contourC is the same as that for Eq. (8.35). The pointsz=±2nπi, n= 1,2,3, . . . ,are all excluded.

8.2.22 Show thatζ(s) is analytic in the entire finite complex plane except ats = 1, where ithas a simple pole with a residue of+1.Hint. The contour integral representation will be useful.

8.2.23 Using the complex variable capability of FORTRAN calculateℜ(1+ ib)!, ℑ(1+ ib)!,|(1+ ib)!| and phase(1+ ib)! for b= 0.0(0.1)1.0. Plot the phase of(1+ ib)! versusb.Hint. Exercise 8.2.3 offers a convenient approach. You will need to calculateζ(n).

8.3 STIRLING’S SERIES

For computation of ln(z!) for very largez (statistical mechanics) and for numerical com-putations at nonintegral values ofz, a series expansion of ln(z!) in negative powers ofz isdesirable. Perhaps the most elegant way of deriving such an expansion is by the method ofsteepest descents (Section 7.3). The following method, starting with a numerical integra-tion formula, does not require knowledge of contour integration and is particularly direct.

8.3 Stirling’s Series 517

Derivation from Euler–Maclaurin Integration Formula

The Euler–Maclaurin formula for evaluating a definite integral6 is∫ n

0f (x)dx = 1

2f (0)+ f (1)+ f (2)+ · · · + 12f (n)

− b2[f ′(n)− f ′(0)

]− b4

[f ′′′(n)− f ′′′(0)

]− · · · , (8.45)

in which theb2n are related to the Bernoulli numbersB2n (compare Section 5.9) by

(2n)!b2n = B2n, (8.46)

B0= 1, B6 = 142,

B2= 16, B8 =− 1

30,

B4=− 130, B10= 5

66, and so on.

(8.47)

By applying Eq. (8.45) to the definite integral∫ ∞

0

dx

(z+ x)2= 1

z(8.48)

(for z not on the negative real axis), we obtain

1

z= 1

2z2+ψ (1)(z+ 1)− 2!b2

z3− 4!b4

z5− · · · . (8.49)

This is the reason for using Eq. (8.48). The Euler–Maclaurin evaluation yieldsψ (1)(z+1),which isd2 lnŴ(z+ 1)/dz2.

Using Eq. (8.46) and solving forψ (1)(z+ 1), we have

ψ (1)(z+ 1)= d

dzψ(z+ 1) = 1

z− 1

2z2+ B2

z3+ B4

z5+ · · ·

= 1

z− 1

2z2+

∞∑

n=1

B2n

z2n+1. (8.50)

Since the Bernoulli numbers diverge strongly, this series does not converge. It is a semi-convergent, or asymptotic, series, useful if one retains a small enough number of terms(compare Section 5.10).

Integrating once, we get the digamma function

ψ(z+ 1) = C1+ ln z+ 1

2z− B2

2z2− B4

4z4− · · ·

= C1+ ln z+ 1

2z−

∞∑

n=1

B2n

2nz2n. (8.51)

Integrating Eq. (8.51) with respect toz from z−1 toz and then lettingz approach infinity,C1, the constant of integration, may be shown to vanish. This gives us a second expressionfor the digamma function, often more useful than Eq. (8.38) or (8.44b).

6This is obtained by repeated integration by parts, Section 5.9.


Stirling’s Series

The indefinite integral of the digamma function (Eq. (8.51)) is

lnŴ(z+ 1)= C2+(z+ 1

2

)ln z− z+ B2

2z+ · · · + B2n

2n(2n− 1)z2n−1+ · · · , (8.52)

in which C2 is another constant of integration. To fixC2 we find it convenient to use thedoubling, or Legendre duplication, formula derived in Section 8.4,

Ŵ(z+ 1)Ŵ(z+ 1

2

)= 2−2zπ1/2Ŵ(2z+ 1). (8.53)

This may be proved directly whenz is a positive integer by writingŴ(2z+ 1) as a productof even terms times a product of odd terms and extracting a factor of 2 from each term(Exercise 8.3.5). Substituting Eq. (8.52) into the logarithm of the doubling formula, wefind thatC2 is

C2= 12 ln 2π, (8.54)

giving

lnŴ(z+ 1)= 1

2ln 2π +

(z+ 1

2

)ln z− z+ 1

12z− 1

360z3+ 1

1260z5− · · · . (8.55)

This is Stirling’s series, an asymptotic expansion. The absolute value of the error is lessthan the absolute value of the first term omitted.

The constants of integrationC1 andC2 may also be evaluated by comparison with thefirst term of the series expansion obtained by the method of “steepest descent.” This iscarried out in Section 7.3.

To help convey a feeling of the remarkable precision of Stirling’s series forŴ(s + 1),the ratio of the first term of Stirling’s approximation toŴ(s + 1) is plotted in Fig. 8.5.A tabulation gives the ratio of the first term in the expansion toŴ(s + 1) and the ratio ofthe first two terms in the expansion toŴ(s + 1) (Table 8.1). The derivation of these formsis Exercise 8.3.1.

Exercises

8.3.1 Rewrite Stirling’s series to giveŴ(z+ 1) instead of lnŴ(z+ 1).

ANS.Ŵ(z+ 1)=√

2πzz+1/2e−z(

1+ 1

12z+ 1

288z2− 139

51,840z3+ · · ·

).

8.3.2 Use Stirling’s formula to estimate 52!, the number of possible rearrangements of cardsin a standard deck of playing cards.

8.3.3 By integrating Eq. (8.51) fromz− 1 to z and then lettingz→∞, evaluate the constantC1 in the asymptotic series for the digamma functionψ(z).

8.3.4 Show that the constantC2 in Stirling’s formula equals12 ln2π by using the logarithm ofthe doubling formula.

8.3 Stirling’s Series 519

FIGURE 8.5 Accuracy of Stirling’s formula.

Table 8.1

s1

Ŵ(s + 1)

√2πss+1/2e−s

1

Ŵ(s + 1)

√2πss+1/2e−s

(1+ 1

12s

)

1 0.92213 0.998982 0.95950 0.999493 0.97270 0.999724 0.97942 0.999835 0.98349 0.999886 0.98621 0.999927 0.98817 0.999948 0.98964 0.999959 0.99078 0.99996

10 0.99170 0.99998

8.3.5 By direct expansion, verify the doubling formula forz= n+ 12 ; n is an integer.

8.3.6 Without using Stirling’s series show that

(a) ln(n!) <∫ n+1

1lnx dx, (b) ln(n!) >

∫ n

1lnx dx; n is an integer≥ 2.

Notice that the arithmetic mean of these two integrals gives a good approximation forStirling’s series.

8.3.7 Test for convergence

∞∑

p=0

[(p− 1

2)!p!

]2

× 2p+ 1

2p+ 2= π

∞∑

p=0

(2p− 1)!!(2p+ 1)!!(2p)!!(2p+ 2)!! .


This series arises in an attempt to describe the magnetic field created by and enclosedby a current loop.

8.3.8 Show that

limx→∞

xb−a(x + a)!(x + b)! = 1.

8.3.9 Show that

limn→∞

(2n− 1)!!(2n)!! n1/2= π−1/2.

8.3.10 Calculate the binomial coefficient(2nn

)to six significant figures forn= 10, 20, and 30.

Check your values by

(a) a Stirling series approximation through terms inn−1,(b) a double precision calculation.

ANS.(20

10

)= 1.84756× 105,

(4020

)= 1.37846× 1011,

(6030

)= 1.18264× 1017.

8.3.11 Write a program (or subprogram) that will calculate log10(x!) directly from Stirling’sseries. Assume thatx ≥ 10. (Smaller values could be calculated via the factorial re-currence relation.) Tabulate log10(x!) versusx for x = 10(10)300. Check your resultsagainst AMS-55 (see Additional Readings for this reference) or by direct multiplication(for n= 10, 20, and 30).

Check value. log10(100!)= 157.97.

8.3.12 Using the complex arithmetic capability of FORTRAN, write a subroutine that will cal-culate ln(z!) for complexz based on Stirling’s series. Include a test and an appropriateerror message ifz is too close to a negative real integer. Check your subroutine againstalternate calculations forz real,z pure imaginary, andz= 1+ ib (Exercise 8.2.23).

Check values. |(i0.5)!| = 0.82618phase(i0.5)! = −0.24406.

8.4 THE BETA FUNCTION

Using the integral definition (Eq. (8.25)), we write the product of two factorials as theproduct of two integrals. To facilitate a change in variables, we take the integrals over afinite range:

m!n! = lima2→∞

∫ a2

0e−uum du

∫ a2

0e−vvn dv,

ℜ(m) >−1,ℜ(n) >−1.

(8.56a)

Replacingu with x2 andv with y2, we obtain

m!n! = lima→∞

4∫ a

0e−x

2x2m+1dx

∫ a

0e−y

2y2n+1dy. (8.56b)

8.4 The Beta Function 521

FIGURE 8.6 Transformation fromCartesian to polar coordinates.

Transforming to polar coordinates gives us

m!n! = lima→∞

4∫ a

0e−r

2r2m+2n+3dr

∫ π/2

0cos2m+1 θ sin2n+1 θ dθ

= (m+ n+ 1)!2∫ π/2

0cos2m+1 θ sin2n+1 θ dθ. (8.57)

Here the Cartesian area elementdx dy has been replaced byr dr dθ (Fig. 8.6). The lastequality in Eq. (8.57) follows from Exercise 8.1.11.

The definite integral, together with the factor 2, has been named the beta function:

B(m+ 1, n+ 1) ≡ 2∫ π/2

0cos2m+1 θ sin2n+1 θ dθ

= m!n!(m+ n+ 1)! . (8.58a)

Equivalently, in terms of the gamma function and noting its symmetry,

B(p,q)= Ŵ(p)Ŵ(q)

Ŵ(p+ q), B(q,p)= B(p,q). (8.58b)

The only reason for choosingm+ 1 andn+ 1, rather thanm andn, as the arguments ofBis to be in agreement with the conventional, historical beta function.

Definite Integrals, Alternate Forms

The beta function is useful in the evaluation of a wide variety of definite integrals. Thesubstitutiont = cos2 θ converts Eq. (8.58a) to7

B(m+ 1, n+ 1)= m!n!(m+ n+ 1)! =

∫ 1

0tm(1− t)n dt. (8.59a)

7The Laplace transform convolution theorem provides an alternate derivation of Eq. (8.58a), compare Exercise 15.11.2.


Replacingt by x2, we obtain

m!n!2(m+ n+ 1)! =

∫ 1

0x2m+1(1− x2)n dx. (8.59b)

The substitutiont = u/(1+ u) in Eq. (8.59a) yields still another useful form,

m!n!(m+ n+ 1)! =

∫ ∞

0

um

(1+ u)m+n+2du. (8.60)

The beta function as a definite integral is useful in establishing integral representations ofthe Bessel function (Exercise 11.1.18) and the hypergeometric function (Exercise 13.4.10).

Verification of πα/sinπα Relation

If we takem= a, n=−a,−1< a < 1, then∫ ∞

0

ua

(1+ u)2du= a!(−a)!. (8.61)

By contour integration this integral may be shown to be equal toπa/sinπa (Exer-cise 7.1.18), thus providing another method of obtaining Eq. (8.32).

Derivation of Legendre Duplication Formula

The form of Eq. (8.58a) suggests that the beta function may be useful in deriving thedoubling formula used in the preceding section. From Eq. (8.59a) withm = n = z andℜ(z) >−1,

z!z!(2z+ 1)! =

∫ 1

0tz(1− t)z dt. (8.62)

By substitutingt = (1+ s)/2, we have

z!z!(2z+ 1)! = 2−2z−1

∫ 1

−1

(1− s2)z ds = 2−2z

∫ 1

0

(1− s2)z ds. (8.63)

The last equality holds because the integrand is even. Evaluating this integral as a betafunction (Eq. (8.59b)), we obtain

z!z!(2z+ 1)! = 2−2z−1 z!(−1

2)!(z+ 1

2)!. (8.64)

Rearranging terms and recalling that(−12)! = π1/2, we reduce this equation to one form

of the Legendre duplication formula,

z!(z+ 1

2

)! = 2−2z−1π1/2(2z+ 1)!. (8.65a)

Dividing by (z+ 12), we obtain an alternate form of the duplication formula:

z!(z− 1

2

)! = 2−2zπ1/2(2z)!. (8.65b)


Although the integrals used in this derivation are defined only forℜ(z) >−1, the results(Eqs. (8.65a) and (8.65b) hold for all regular pointsz by analytic continuation.8

Using the double factorial notation (Section 8.1), we may rewrite Eq. (8.65a) (withz=n, an integer) as

(n+ 1

2

)! = π1/2(2n+ 1)!!/2n+1. (8.65c)

This is often convenient for eliminating factorials of fractions.

Incomplete Beta Function

Just as there is an incomplete gamma function (Section 8.5), there is also an incompletebeta function,

Bx(p, q)=∫ x

0tp−1(1− t)q−1dt, 0≤ x ≤ 1, p > 0, q > 0 (if x = 1). (8.66)

Clearly,Bx=1(p, q) becomes the regular (complete) beta function, Eq. (8.59a). A power-series expansion ofBx(p, q) is the subject of Exercises 5.2.18 and 5.7.8. The relation tohypergeometric functions appears in Section 13.4.

The incomplete beta function makes an appearance in probability theory in calculatingthe probability of at mostk successes inn independent trials.9

Exercises

8.4.1 Derive the doubling formula for the factorial function by integrating(sin 2θ)2n+1 =(2 sinθ cosθ)2n+1 (and using the beta function).

8.4.2 Verify the following beta function identities:

(a) B(a, b)= B(a + 1, b)+B(a, b+ 1),

(b) B(a, b)= a + b

bB(a, b+ 1),

(c) B(a, b)= b− 1

aB(a + 1, b− 1),

(d) B(a, b)B(a + b, c)= B(b, c)B(a, b+ c).

8.4.3 (a) Show that

∫ 1

−1

(1− x2)1/2

x2n dx =

π/2, n= 0

π(2n− 1)!!(2n+ 2)!! , n= 1,2,3, . . . .

8If 2z is a negative integer, we get the valid but unilluminating result∞=∞.9W. Feller,An Introduction to Probability Theory and Its Applications, 3rd ed. New York: Wiley (1968), Section VI.10.


(b) Show that

∫ 1

−1

(1− x2)−1/2

x2n dx =

π, n= 0

π(2n− 1)!!(2n)!! , n= 1,2,3, . . . .

8.4.4 Show that

∫ 1

−1

(1− x2)n dx =

22n+1 n!n!(2n+ 1)! , n >−1

2(2n)!!

(2n+ 1)!! , n= 0,1,2, . . . .

8.4.5 Evaluate∫ 1−1(1+ x)a(1− x)b dx in terms of the beta function.

ANS. 2a+b+1B(a + 1, b+ 1).

8.4.6 Show, by means of the beta function, that∫ z

t

dx

(z− x)1−α(x − t)α= π

sinπα, 0< α < 1.

8.4.7 Show that the Dirichlet integral∫∫

xpyq dx dy = p!q!(p+ q + 2)! =

B(p+ 1, q + 1)

p+ q + 2,

where the range of integration is the triangle bounded by the positivex- andy-axes andthe linex + y = 1.

8.4.8 Show that ∫ ∞

0

∫ ∞

0e−(x

2+y2+2xy cosθ) dx dy = θ

2 sinθ.

What are the limits onθ?Hint. Consider obliquexy-coordinates.

ANS.−π < θ < π .

8.4.9 Evaluate (using the beta function)

(a)∫ π/2

0cos1/2 θ dθ = (2π)3/2

16[(14)!]2

,

(b)∫ π/2

0cosn θ dθ =

∫ π/2

0sinn θ dθ =

√π[(n− 1)/2]!

2(n/2)!

=

(n− 1)!!n!! for n odd,

π

2· (n− 1)!!

n!! for n even.


8.4.10 Evaluate∫ 1

0 (1− x4)−1/2dx as a beta function.

ANS.[(1

4)!]2 · 4(2π)1/2

= 1.311028777.

8.4.11 Given

Jν(z)=2

π1/2(ν − 12)!

(z

2

)ν ∫ π/2

0sin2ν θ cos(zcosθ) dθ, ℜ(ν) >−1

2,

show, with the aid of beta functions, that this reduces to the Bessel series

Jν(z)=∞∑

s=0

(−1)s1

s!(s + ν)!

(z

2

)2s+ν,

identifying the initialJν as an integral representation of the Bessel function,Jν (Sec-tion 11.1).

8.4.12 Given the associated Legendre function

Pmm (x)= (2m− 1)!!

(1− x2)m/2

,

Section 12.5, show that

(a)∫ 1

−1

[Pmm (x)

]2dx = 2

2m+ 1(2m)!, m= 0,1,2, . . . ,

(b)∫ 1

−1

[Pmm (x)

]2 dx

1− x2= 2 · (2m− 1)!, m= 1,2,3, . . . .

8.4.13 Show that

(a)∫ 1

0x2s+1(1− x2)−1/2

dx = (2s)!!(2s + 1)!! ,

(b)∫ 1

0x2p(1− x2)q dx = 1

2

(p− 12)!q!

(p+ q + 12)!

.

8.4.14 A particle of massm moving in a symmetric potential that is well described byV (x)=A|x|n has a total energy12m(dx/dt)2+ V (x)= E. Solving fordx/dt and integratingwe find that the period of motion is

τ = 2√

2m∫ xmax

0

dx

(E −Axn)1/2,

wherexmax is a classical turning point given byAxnmax=E. Show that

τ = 2

n

√2πm

E

(E

A

)1/nŴ(1/n)

Ŵ(1/n+ 12)

.

8.4.15 Referring to Exercise 8.4.14,


(a) Determine the limit asn→∞ of

2

n

√2πm

E

(E

A

)1/nŴ(1/n)

Ŵ(1/n+ 12)

.

(b) Find limn→∞

τ from the behavior of the integrand(E −Axn)−1/2.

(c) Investigate the behavior of the physical system (potential well) asn→∞. Obtainthe period from inspection of this limiting physical system.

8.4.16 Show that∫ ∞

0

sinhα x

coshβ xdx = 1

2B

(α+ 1

2,β − α

2

), −1< α < β.

Hint. Let sinh2x = u.

8.4.17 The beta distribution of probability theory has a probability density

f (x)= Ŵ(α + β)

Ŵ(α)Ŵ(β)xα−1(1− x)β−1,

with x restricted to the interval (0, 1). Show that

(a) 〈x〉(mean)= α

α + β.

(b) σ 2(variance)≡ 〈x2〉 − 〈x〉2= αβ

(α + β)2(α + β + 1).

8.4.18 From

limn→∞

∫ π/20 sin2n θ dθ

∫ π/20 sin2n+1 θ dθ

= 1

derive the Wallis formula forπ :

π

2= 2 · 2

1 · 3 ·4 · 43 · 5 ·

6 · 65 · 7 · · · .

8.4.19 Tabulate the beta functionB(p,q) for p andq = 1.0(0.1)2.0 independently.

Check value.B(1.3,1.7)= 0.40774.

8.4.20 (a) Write a subroutine that will calculate the incomplete beta functionBx(p, q). For0.5< x ≤ 1 you will find it convenient to use the relation

Bx(p, q)= B(p,q)−B1−x(q,p).

(b) TabulateBx(32,

32). Spot check your results by using the Gauss–Legendre quadra-

ture.

8.5 Incomplete Gamma Function 527

8.5 THE INCOMPLETE GAMMA FUNCTIONS AND RELATED

FUNCTIONS

Generalizing the Euler definition of the gamma function (Eq. (8.5)), we define the incom-plete gamma functions by the variable limit integrals

γ (a, x)=∫ x

0e−t ta−1dt, ℜ(a) > 0

and

Ŵ(a, x)=∫ ∞

x

e−t ta−1dt. (8.67)

Clearly, the two functions are related, for

γ (a, x)+ Ŵ(a, x)= Ŵ(a). (8.68)

The choice of employingγ (a, x) or Ŵ(a, x) is purely a matter of convenience. If the para-metera is a positive integer, Eq. (8.67) may be integrated completely to yield

γ (n, x) = (n− 1)!(

1− e−xn−1∑

s=0

xs

s!

)

Ŵ(n,x) = (n− 1)!e−xn−1∑

s=0

xs

s! , n= 1,2, . . . .

(8.69)

For nonintegrala, a power-series expansion ofγ (a, x) for smallx and an asymptotic ex-pansion ofŴ(a, x) (denoted asI (x,p)) are developed in Exercise 5.7.7 and Section 5.10:

γ (a, x) = xa∞∑

n=0

(−1)nxn

n!(a + n), |x| ∼ 1 (smallx),

Ŵ(a, x) = xa−1e−x∞∑

n=0

(a − 1)!(a − 1− n)! ·

1

xn(8.70)

= xa−1e−x∞∑

n=0

(−1)n(n− a)!(−a)! ·

1

xn, x≫ 1 (largex).

These incomplete gamma functions may also be expressed quite elegantly in terms of con-fluent hypergeometric functions (compare Section 13.5).

Exponential Integral

Although the incomplete gamma functionŴ(a, x) in its general form (Eq. (8.67)) is onlyinfrequently encountered in physical problems, a special case is quite common and very


FIGURE 8.7 The exponential integral,E1(x)=−Ei(−x).

useful. We define the exponential integral by10

−Ei(−x)≡∫ ∞

x

e−t

tdt =E1(x). (8.71)

(See Fig. 8.7.) Caution is needed here, for the integral in Eq. (8.71) diverges logarithmicallyasx→ 0. To obtain a series expansion for smallx, we start from

E1(x)= Ŵ(0, x)= lima→0

[Ŵ(a)− γ (a, x)

]. (8.72)

We may split the divergent term in the series expansion forγ (a, x),

E1(x)= lima→0

[aŴ(a)− xa

a

]−

∞∑

n=1

(−1)nxn

n · n! . (8.73)

Using l’Hôpital’s rule (Exercise 5.6.8) and

d

da

{aŴ(a)

}= d

daa! = d

daeln(a!) = a!ψ(a + 1), (8.74)

and then Eq. (8.40),11 we obtain the rapidly converging series

E1(x)=−γ − lnx −∞∑

n=1

(−1)nxn

n · n! . (8.75)

An asymptotic expansionE1(x) ≈ e−x[ 1x− 1!

x2 + · · · ] for x →∞ is developed in Sec-tion 5.10.

10The appearance of the two minus signs in−Ei(−x) is a historical monstrosity. AMS-55, Chapter 5, denotes this integral asE1(x). See Additional Readings for the reference.11dxa/da = xa lnx.


FIGURE 8.8 Sine and cosine integrals.

Further special forms related to the exponential integral are the sine integral, cosineintegral (Fig. 8.8), and logarithmic integral, defined by12

si(x) = −∫ ∞

x

sint

tdt

Ci(x) = −∫ ∞

x

cost

tdt (8.76)

li(x) =∫ x

0

du

lnu= Ei(lnx)

for their principal branch, with the branch cut conventionally chosen to be along the nega-tive real axis from the branch point at zero. By transforming from real to imaginary argu-ment, we can show that

si(x)= 1

2i

[Ei(ix)−Ei(−ix)

]= 1

2i

[E1(ix)−E1(−ix)

], (8.77)

whereas

Ci(x)= 1

2

[Ei(ix)+Ei(−ix)

]=−1

2

[E1(ix)+E1(−ix)

], |argx|< π

2. (8.78)

Adding these two relations, we obtain

Ei(ix)=Ci(x)+ isi(x), (8.79)

to show that the relation among these integrals is exactly analogous to that amongeix ,cosx, and sinx. Reference to Eqs. (8.71) and (8.78) shows that Ci(x) agrees with thedefinitions of AMS-55 (see Additional Readings for the reference). In terms ofE1,

E1(ix)=−Ci(x)+ isi(x).

Asymptotic expansions of Ci(x) and si(x) are developed in Section 5.10. Power-seriesexpansions about the origin for Ci(x), si(x), and li(x) may be obtained from those for

12Another sine integral is given by Si(x)= si(x)+ π/2.


FIGURE 8.9 Error function, erfx.

the exponential integral,E1(x), or by direct integration, Exercise 8.5.10. The exponential,sine, and cosine integrals are tabulated in AMS-55, Chapter 5, (see Additional Readingsfor the reference) and can also be accessed by symbolic software such as Mathematica,Maple, Mathcad, and Reduce.

Error Integrals

The error integrals

erfz= 2√π

∫ z

0e−t

2dt, erfcz= 1− erfz= 2√

π

∫ ∞

z

e−t2dt (8.80a)

(normalized so that erf∞= 1) are introduced in Exercise 5.10.4 (Fig. 8.9). Asymptoticforms are developed there. From the general form of the integrands and Eq. (8.6) we ex-pect that erfz and erfcz may be written as incomplete gamma functions witha = 1

2 . Therelations are

erfz= π−1/2γ(1

2, z2), erfcz= π−1/2Ŵ

(12, z

2). (8.80b)

The power-series expansion of erfz follows directly from Eq. (8.70).

Exercises

8.5.1 Show that

γ (a, x)= e−x∞∑

n=0

(a − 1)!(a + n)!x

a+n

(a) by repeatedly integrating by parts.(b) Demonstrate this relation by transforming it into Eq. (8.70).

8.5.2 Show that

(a)dm

dxm

[x−aγ (a, x)

]= (−1)mx−a−mγ (a +m,x),


(b)dm

dxm

[exγ (a, x)

]= ex

Ŵ(a)

Ŵ(a −m)γ (a −m,x).

8.5.3 Show thatγ (a, x) andŴ(a, x) satisfy the recurrence relations

(a) γ (a + 1, x)= aγ (a, x)− xae−x ,(b) Ŵ(a + 1, x)= aŴ(a, x)+ xae−x .

8.5.4 The potential produced by a 1S hydrogen electron (Exercise 12.8.6) is given by

V (r)= q

4πε0a0

{1

2rγ (3,2r)+ Ŵ(2,2r)

}.

(a) Forr≪ 1, show that

V (r)= q

4πε0a0

{1− 2

3r2+ · · ·

}.

(b) Forr≫ 1, show that

V (r)= q

4πε0a0· 1

r.

Herer is expressed in units ofa0, the Bohr radius.Note.For computation at intermediate values ofr , Eqs. (8.69) are convenient.

8.5.5 The potential of a 2P hydrogen electron is found to be (Exercise 12.8.7)

V (r) = 1

4πε0· q

24a0

{1

rγ (5, r)+ Ŵ(4, r)

}

− 1

4πε0· q

120a0

{1

r3γ (7, r)+ r2Ŵ(2, r)

}P2(cosθ).

Herer is expressed in units ofa0, the Bohr radius.P2(cosθ) is a Legendre polynomial(Section 12.1).

(a) Forr≪ 1, show that

V (r)= 1

4πε0· qa0

{1

4− 1

120r2P2(cosθ)+ · · ·

}.

(b) Forr≫ 1, show that

V (r)= 1

4πε0· q

a0r

{1− 6

r2P2(cosθ)+ · · ·

}.

8.5.6 Prove that the exponential integral has the expansion∫ ∞

x

e−t

tdt =−γ − lnx −

∞∑

n=1

(−1)nxn

n · n! ,

whereγ is the Euler–Mascheroni constant.


8.5.7 Show thatE1(z) may be written as

E1(z)= e−z∫ ∞

0

e−zt

1+ tdt.

Show also that we must impose the condition|argz| ≤ π/2.

8.5.8 Related to the exponential integral (Eq. (8.71)) by a simple change of variable is thefunction

En(x)=∫ ∞

1

e−xt

tndt.

Show thatEn(x) satisfies the recurrence relation

En+1(x)=1

ne−x − x

nEn(x), n= 1,2,3, . . . .

8.5.9 With En(x) as defined in Exercise 8.5.8, show thatEn(0)= 1/(n− 1), n > 1.

8.5.10 Develop the following power-series expansions:

(a) si(x)=−π

2+

∞∑

n=0

(−1)nx2n+1

(2n+ 1)(2n+ 1)! ,

(b) Ci(x)= γ + lnx +∞∑

n=1

(−1)nx2n

2n(2n)! .

8.5.11 An analysis of a center-fed linear antenna leads to the expression∫ x

0

1− cost

tdt.

Show that this is equal toγ + lnx −Ci(x).

8.5.12 Using the relation

Ŵ(a)= γ (a, x)+ Ŵ(a, x),

show that ifγ (a, x) satisfies the relations of Exercise 8.5.2, thenŴ(a, x) must satisfythe same relations.

8.5.13 (a) Write a subroutine that will calculate the incomplete gamma functionsγ (n, x) andŴ(n,x) for n a positive integer. Spot checkŴ(n,x) by Gauss–Laguerre quadratures.

(b) Tabulateγ (n, x) andŴ(n,x) for x = 0.0(0.1)1.0 andn= 1, 2, 3.

8.5.14 Calculate the potential produced by a 1S hydrogen electron (Exercise 8.5.4) (Fig. 8.10).TabulateV (r)/(q/4πε0a0) for x = 0.0(0.1)4.0. Check your calculations forr≪ 1 andfor r≫ 1 by calculating the limiting forms given in Exercise 8.5.4.

8.5.15 Using Eqs. (5.182) and (8.75), calculate the exponential integralE1(x) for

(a)x = 0.2(0.2)1.0, (b)x = 6.0(2.0)10.0.

Program your own calculation but check each value, using a library subroutine if avail-able. Also check your calculations at each point by a Gauss–Laguerre quadrature.

8.5 Additional Readings 533

FIGURE 8.10 Distributed charge potential producedby a 1S hydrogen electron, Exercise 8.5.14.

You’ll find that the power-series converges rapidly and yields high precision for smallx. The asymptotic series, even forx = 10, yields relatively poor accuracy.

Check values. E1(1.0) = 0.219384E1(10.0) = 4.15697× 10−6.

8.5.16 The two expressions forE1(x), (1) Eq. (5.182), an asymptotic series and (2) Eq. (8.75),a convergent power series, provide a means of calculating the Euler–Mascheroni con-stantγ to high accuracy. Using double precision, calculateγ from Eq. (8.75), withE1(x) evaluated by Eq. (5.182).Hint. As a convenient choice takex in the range 10 to 20. (Your choice ofx will seta limit on the accuracy of your result.) To minimize errors in the alternating series ofEq. (8.75), accumulate the positive and negative terms separately.

ANS. Forx = 10 and “double precision,”γ = 0.57721566.

Additional Readings

Abramowitz, M., and I. A. Stegun, eds.,Handbook of Mathematical Functions with Formulas, Graphs, andMathematical Tables(AMS-55). Washington, DC: National Bureau of Standards (1972), reprinted, Dover(1974). Contains a wealth of information about gamma functions, incomplete gamma functions, exponentialintegrals, error functions, and related functions — Chapters 4 to 6.

Artin, E.,The Gamma Function(translated by M. Butler). New York: Holt, Rinehart and Winston (1964). Demon-strates that if a functionf (x) is smooth (log convex) and equal to(n − 1)! whenx = n = integer, it is thegamma function.

Davis, H. T.,Tables of the Higher Mathematical Functions. Bloomington, IN: Principia Press (1933). Volume Icontains extensive information on the gamma function and the polygamma functions.

Gradshteyn, I. S., and I. M. Ryzhik,Table of Integrals, Series, and Products. New York: Academic Press (1980).

Luke, Y. L.,The Special Functions and Their Approximations, Vol. 1. New York: Academic Press (1969).

Luke, Y. L., Mathematical Functions and Their Approximations. New York: Academic Press (1975). This isan updated supplement toHandbook of Mathematical Functions with Formulas, Graphs, and MathematicalTables(AMS-55). Chapter 1 deals with the gamma function. Chapter 4 treats the incomplete gamma functionand a host of related functions.

chapter the gamma function (factorial...

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