Gamma Function

29 Pages·2012·0.41 MB·English

Chapter 10
The Gamma Function
(Factorial Function)
The gamma function appears in physical problems of all kinds, such as the
normalizationofCoulombwavefunctionsandthecomputationofprobabilities
instatisticalmechanics.Itsimportancestemsfromitsusefulnessindeveloping
other functions that have direct physical application. The gamma function,
therefore, is included here. A discussion of the numerical evaluation of the
gammafunctionappearsinSection10.3.Closelyrelatedfunctions,suchasthe
errorintegral,arepresentedinSection10.4.
10.1 DefinitionsandSimpleProperties
At least three different, convenient definitions of the gamma function are in
commonuse.Ourfirsttaskistostatethesedefinitions,todevelopsomesimple,
directconsequences,andtoshowtheequivalenceofthethreeforms.
InfiniteLimit(Euler)
Thefirstdefinition,duetoEuler,is
1·2·3···n
(cid:2)(z)≡ lim nz, z=(cid:7) 0,−1,−2,−3,.... (10.1)
n→∞ z(z+1)(z+2)···(z+n)
Thisdefinitionof(cid:2)(z)isusefulindevelopingtheWeierstrassinfinite-product
form of (cid:2)(z) [Eq. (10.17)] and in obtaining the derivative of ln(cid:2)(z) (Sec-
tion10.2).Hereandelsewhereinthischapter,zmaybeeitherrealorcomplex.
523
524 Chapter10 TheGammaFunction(FactorialFunction)
Replacingzwithz+1,wehave
1·2·3···n
(cid:2)(z+1)= lim nz+1
n→∞(z+1)(z+2)(z+3)···(z+n+1)
nz 1·2·3···n
= lim · nz
n→∞ z+n+1 z(z+1)(z+2)···(z+n)
= z(cid:2)(z). (10.2)
Thisisthebasicfunctionalrelationforthegammafunction.Itshouldbenoted
thatitisadifferenceequation.Thegammafunctionisoneofageneralclass
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)commontophysics.Infact,itdoesnotsatisfyanyusefulor
practicaldifferentialequation.
Also,fromthedefinition
1·2·3···n
(cid:2)(1)= lim n=1. (10.3)
n→∞1·2·3···n(n+1)
Now,applicationofEq.(10.2)gives
(cid:2)(2)=1,
(cid:2)(3)=2(cid:2)(2)=2,... (10.4)
(cid:2)(n)=1·2·3···(n−1)=(n−1)!
We see that the gamma function interpolates the factorials by a continuous
functionthatreturnsthefactorialsatintegerarguments.
DefiniteIntegral(Euler)
Aseconddefinition,alsofrequentlycalledtheEulerintegral,is
(cid:2)
∞
(cid:2)(z)≡ e−ttz−1dt, (cid:10)(z)>0. (10.5)
0
Therestrictiononzisnecessarytoavoiddivergenceoftheintegralatt = 0.
Whenthegammafunctiondoesappearinphysicalproblems,itisofteninthis
formorsomevariation,suchas
(cid:2)
∞
(cid:2)(z)=2 e−t2t2z−1dt, (cid:10)(z)>0 (10.6)
0
or
(cid:2) (cid:3) (cid:4) (cid:5)(cid:6)
1 1 z−1
(cid:2)(z)= ln dt, (cid:10)(z)>0.
t
0
10.1 DefinitionsandSimpleProperties 525
EXAMPLE10.1.1 The Euler Integral Interpolates the Factorials The Euler integral for
positive integer argument, z = n+1 with n> 0, yields the factorials. Using
integrationbypartsrepeatedlywefind
(cid:2) (cid:7) (cid:2) (cid:2)
∞ (cid:7)∞ ∞ ∞
e−ttndt =−tne−t(cid:7)(cid:7) +n e−ttn−1dt =n e−ttn−1dt
0 (cid:3) 0 (cid:7) 0 (cid:2) 0 (cid:6)
(cid:7)∞ ∞
=n −tn−1e−t(cid:7)(cid:7) +(n−1) e−ttn−2dt
(cid:2) 0 0
∞
=n(n−1) e−ttn−2dt =···=n!
0
(cid:8)
because ∞e−tdt =1.Thus,theEulerintegralalsointerpolatesthefactorials.
0
(cid:2)
When z = 1, Eq. (10.6) contains the Gauss error integral, and we have the
2
interestingresult
(cid:9) (cid:10) √
(cid:2) 1 = π. (10.7)
2
Thevalue(cid:2)(1)canbederiveddirectlyfromthesquareofEq.(10.6)forz= 1
2 2
byintroducingplanepolarcoordinates(x2+y2 =ρ2,dxdy=ρdρdϕ)inthe
productofintegrals
(cid:4) (cid:5) (cid:2) (cid:2)
1 2 ∞ ∞
(cid:2) =4 e−x2−y2dxdy
2
0 0
(cid:2) (cid:2) (cid:7)
π/2 ∞ (cid:7)∞
=4 e−ρ2ρdρdϕ =−πe−ρ2(cid:7)(cid:7) =π.
ϕ=0 ρ=0 0
GeneralizationsofEq.(10.7),theGaussianintegrals,are
(cid:2)
∞ s!
x2s+1exp(−ax2)dx= , (10.8)
2as+1
0
(cid:2) (cid:11)
∞ (s− 1)! (2s−1)!! π
x2sexp(−ax2)dx= 2 = , (10.9)
2as+1/2 2s+1as a
0
whichareofmajorimportanceinstatisticalmechanics.Aproofislefttothe
readerinExercise10.1.11.ThedoublefactorialnotationisexplainedinExer-
cise5.2.12andEq.(10.33b).
To show the equivalence of the two definitions, Eqs. (10.1) and (10.5),
considerthefunctionoftwovariables
(cid:2) (cid:4) (cid:5)
n t n
F(z,n)= 1− tz−1dt, (cid:10)(z)>0, (10.10)
n
0
withnapositiveinteger.Since
(cid:4) (cid:5)
t n
lim 1− ≡e−t, (10.11)
n→∞ n
526 Chapter10 TheGammaFunction(FactorialFunction)
fromthedefinitionoftheexponential
(cid:2)
∞
lim F(z,n)= F(z,∞)= e−ttz−1dt ≡(cid:2)(z) (10.12)
n→∞ 0
byEq.(10.5).
ReturningtoF(z,n),weevaluateitinsuccessiveintegrationsbyparts.For
convenienceletu=t/n.Then
(cid:2)
1
F(z,n)=nz (1−u)nuz−1du. (10.13)
0
Integratingbyparts,weobtainfor(cid:10)(z)>0,
(cid:7) (cid:2)
F(z,n) =(1−u)nuz(cid:7)(cid:7)(cid:7)1+ n 1(1−u)n−1uzdu. (10.14)
nz z z
0 0
Repeating this, with the integrated part vanishing at both end points each
time,wefinallyget
(cid:2)
n(n−1)···1 1
F(z,n)=nz uz+n−1du
z(z+1)···(z+n−1)
0
1·2·3···n
= nz. (10.15)
z(z+1)(z+2)···(z+n)
ThisisidenticaltotheexpressionontherightsideofEq.(10.1).Hence,
lim F(z,n)= F(z,∞)≡(cid:2)(z) (10.16)
n→∞
byEq.(10.1),completingtheproof.
Usingthefunctionalequation(10.2)wecanextend(cid:2)(z)frompositiveto
negativearguments.Forexample,startingwithEq.(10.2)wedefine
(cid:4) (cid:5) (cid:4) (cid:5)
1 1 1 √
(cid:2) − =− (cid:2) =−2 π,
2 0.5 2
andEq.(10.2)forz→0impliesthat(cid:2)(z→0)→∞.Moreover,z(cid:2)(z)→1for
z → 0 [Eq. (10.2)] shows that (cid:2)(z)has a simple pole at the origin. Similarly,
wefindsimplepolesof(cid:2)(z)atallnegativeintegers.
InfiniteProduct(Weierstrass)
Thethirddefinition(Weierstrass’sform)is
(cid:4) (cid:5)
(cid:12)∞
1 z
≡zeγz 1+ e−z/n, (10.17)
(cid:2)(z) n
n=1
whereγ istheEuler–Mascheroniconstant[Eq.(5.27)]
(cid:13) (cid:15)
(cid:14)n 1
γ = lim −lnn =0.5772156···. (10.18)
n→∞ m
m=1
10.1 DefinitionsandSimpleProperties 527
This infinite product form may be used to develop the reflection identity,
Eq.(10.24a),andappliedintheexercises,suchasExercise10.1.17.Thisform
canbederivedfromtheoriginaldefinition[Eq.(10.1)]byrewritingitas
(cid:4) (cid:5)
1·2·3···n 1 (cid:12)n z −1
(cid:2)(z)= lim nz= lim 1+ nz. (10.19)
n→∞ z(z+1)···(z+n) n→∞ z m
m=1
InvertingEq.(10.19)andusing
n−z=e−zlnn, (10.20)
weobtain
1 (cid:12)n (cid:16) z(cid:17)
=z lim e(−lnn)z 1+ . (10.21)
(cid:2)(z) n→∞ m
m=1
Multiplyinganddividingby
(cid:3)(cid:4) (cid:5) (cid:6)
1 1 1 (cid:12)n
exp 1+ + +···+ z = ez/m, (10.22)
2 3 n
m=1
weget
(cid:18) (cid:3)(cid:4) (cid:5) (cid:6)(cid:19)
1 1 1 1
= z lim exp 1+ + +···+ −lnn z
(cid:2)(z) n→∞ 2 3 n
(cid:20) (cid:21)
(cid:12)n (cid:16) z(cid:17)
× lim 1+ e−z/m . (10.23)
n→∞ m
m=1
As shown in Section 5.2, the infinite series in the exponent converges and
definesγ,theEuler–Mascheroniconstant.Hence,Eq.(10.17)follows.
The Weierstrass infinite product definition of (cid:2)(z) leads directly to an
importantidentity,
π
(cid:2)(z)(cid:2)(1−z)= , (10.24a)
sinzπ
usingtheinfiniteproductformulasfor(cid:2)(z),(cid:2)(1−z),andsinz[Eq.(7.60)].
Alternatively,wecanstartfromtheproductofEulerintegrals
(cid:2) (cid:2)
∞ ∞
(cid:2)(z+1)(cid:2)(1−z)= sze−sds t−ze−tdt
(cid:2)0 0 (cid:2)
∞ dv ∞ πz
= vz e−uudu= ,
(v+1)2 sinπz
0 0
transforming from the variables s,t to u = s+t,v = s/t, as suggested by
combiningtheexponentialsandthepowersintheintegrands.TheJacobianis
(cid:7) (cid:7)
J =−(cid:7)(cid:7)(cid:7)11 −1 s (cid:7)(cid:7)(cid:7)= st+2 t = (v+u1)2,
t t2
(cid:8)
where (v +1)t = u. The integral ∞e−uudu = 1, whereas that over v may
0
be derived by contour integration, giving πz . Similarly, one can establish
sinπz
528 Chapter10 TheGammaFunction(FactorialFunction)
Legendre’sduplicationformula
(cid:9) (cid:10) √
(cid:2)(1+z)(cid:2) z+ 1 =2−2z π(cid:2)(2z+1).
2
Settingz= 1 inEq.(10.24a),weobtain
2
(cid:9) (cid:10) √
(cid:2) 1 = π, (10.24b)
2
(takingthepositivesquareroot)inagreementwithEqs.(10.7)and(10.9).
The Weierstrass definition shows immediately that (cid:2)(z)has simple poles
atz= 0,−1,−2,−3,...,andthat[(cid:2)(z)]−1 hasnopolesinthefinitecomplex
plane,whichmeansthat(cid:2)(z)hasnozeros.Thisbehaviormayalsobeseenin
Eq.(10.24a),inwhichwenotethatπ/(sinπz)isneverequaltozero.
Actually, the infinite product definition of (cid:2)(z) may be derived from the
Weierstrass factorization theorem with the specification that [(cid:2)(z)]−1 have
simplezerosatz=0,−1,−2,−3,....TheEuler–Mascheroniconstantisfixed
byrequiring(cid:2)(1)= 1.Seealsotheproductexpansionsofentirefunctionsin
Chapter7.
In mathematical probability theory the gamma distribution (probability
density)isgivenby

 1 xα−1e−x/β, x>0
f(x)= βα(cid:2)(α) (10.25a)

0, x≤0.
Theconstant[βα(cid:2)(α)]−1 isincludedsothatthetotal(integrated)probability
will be unity. For x → E, kinetic energy, α → 3 and β → kT, Eq. (10.25a)
2
yieldstheclassicalMaxwell–Boltzmannstatistics.
BiographicalData
Weierstrass,KarlTheodorWilhelm. Weierstrass,aGermanmathemati-
cian,wasbornin1815inOstenfelde,Germany,anddiedin1897inBerlin.
Heobtainedadegreeinmathematicsin1841andstudiedAbel’sandJacobi’s
workonellipticalfunctionsandextendedtheirworkonanalyticfunctions
whilelivingasaschoolteacher.Afterbeingrecognizedasthefatherofmod-
ernanalysis,hebecameaprofessorattheUniversityofBerlinandamember
oftheAcademyofSciencesin1856.
FactorialNotation
So far, this discussion has been presented in terms of the classical notation.
As pointed out by Jeffreys and others, the −1 of the z−1 exponent in our
seconddefinition[Eq.(10.5)]isacontinualnuisance.Accordingly,Eq.(10.5)
isrewrittenas
(cid:2)
∞
e−ttzdt ≡z!, (cid:10)(z)>1, (10.25b)
0
10.1 DefinitionsandSimpleProperties 529
todefine(cid:26)afactorialfunctionz!.Occasionally,wemayevenencounterGauss’s
notation, (z),forthefactorialfunction
(cid:12)
(z)=z!. (10.26)
The(cid:2)notationisduetoLegendre.ThefactorialfunctionofEq.(10.25a)is,
ofcourse,relatedtothegammafunctionby
(cid:2)(z)=(z−1)!, or (cid:2)(z+1)=z!. (10.27)
Ifz=n,apositiveinteger[Eq.(10.4)]showsthat
z!=n!=1·2·3···n, (10.28)
thefamiliarfactorial.However,itshouldbenotedthatsincez!isnowdefined
byEq.(10.25b)[orequivalentlybyEq.(10.27)]the factorial function is no
longer limited to positive integral values of the argument (Fig. 10.1).
Thedifferencerelation[Eq.(10.2)]becomes
z!
(z−1)!= . (10.29)
z
Thisshowsimmediatelythat
0!=1 (10.30)
Figure10.1
TheFactorial
x!
Function---Extension
toNegative
Arguments 5
4
3
2
1
x
–5 –3 –1 1 2 3 4 5
–1
–2
–3
–4
–5
530 Chapter10 TheGammaFunction(FactorialFunction)
Figure10.2
TheFactorial 6
Functionandthe
5
FirstTwo
Derivativesofln(x!) 4 x!
3
2 d2 ln (x!) ddxln (x!)
dx2
1 d2 ln (x!)
x! dx2
0
d
ln (x!)
–1.0 dx
–1.0 0 1.0 2.0 3.0 4.0
x
and
n!=±∞ forn,anegativeinteger. (10.31)
Intermsofthefactorial,Eq.(10.24a)becomes
πz
z!(−z)!= (10.32)
sinπz
because
1
(cid:2)(z)(cid:2)(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!
definesthecurveshowninFig.10.2.TheminimumofthecurveinFig.10.1is
x!=(0.46163···)!=0.88560···. (10.33a)
DoubleFactorialNotation
In many problems of mathematical physics, particularly in connection with
Legendrepolynomials(Chapter11),weencounterproductsoftheoddpositive
integersandproductsoftheevenpositiveintegers.Forconvenience,theseare
givenspeciallabelsasdoublefactorials:
1·3·5···(2n+1)=(2n+1)!!
(10.33b)
2·4·6···(2n)=(2n)!!.
Clearly,thesearerelatedtotheregularfactorialfunctionsby
(2n+1)!
(2n)!!=2nn! and (2n+1)!!= . (10.33c)
2nn!
10.1 DefinitionsandSimpleProperties 531
IntegralRepresentation
An integral representation that is useful in developing asymptotic series for
theBesselfunctionsis
(cid:2)
e−zzνdz=(e2πiν −1)ν!, (10.34)
C
whereCisthecontourshowninFig.10.3.Thiscontourintegralrepresentation
is only useful when ν is not an integer, z = 0 then being a branch point.
Equation (10.34) may be verified for ν > −1 by deforming the contour as
shown in Fig. 10.4. The integral from ∞ to (cid:14) > 0 in Fig. 10.4 yields −(ν!),
placingthephaseofzat0.Theintegralfrom(cid:14) to∞(inthefourthquadrant)
then yields e2πiνν!, the phase of z having increased to 2π. Since the circle
of radius (cid:14) around the origin contributes nothing as (cid:14) → 0, when ν > −1,
Eq.(10.34)follows.
Itisoftenconvenienttoputthisresultintoamoresymmetricalform
(cid:2)
e−z(−z)νdz=2isin(νπ)ν!, (10.35)
C
multiplyingbothsidesofEq.(10.34)by(−1)ν =e−πiν.
ThisanalysisestablishesEqs.(10.34)and(10.35)forν >−1.Itisrelatively
simpletoextendtherangetoincludeallnonintegralν.First,wenotethatthe
integralexistsforν < −1, aslongaswestayawayfromtheorigin.Second,
integratingbypartswefindthatEq.(10.35)yieldsthefamiliardifferencere-
lation [Eq. (10.29)]. If we take the difference relation to define the factorial
Figure10.3
y
FactorialFunction
Contour
C
x +∞
Figure10.4
y
TheContourof
Fig.10.3Deformed Cut line
e
x
+∞
0
532 Chapter10 TheGammaFunction(FactorialFunction)
functionofν <−1,thenEqs.(10.34)and(10.35)areverifiedforallν (except
negativeintegers).
EXERCISES
10.1.1 Derivetherecurrencerelations
(cid:2)(z+1)=z(cid:2)(z)
fromtheEulerintegral[Eq.(10.5)]
(cid:2)
∞
(cid:2)(z)= e−ttz−1dt.
0
10.1.2 In a power series solution for the Legendre functions of the second
kind,weencountertheexpression
(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.
10.1.3 Showthat
(s−n)! (−1)n−s(2n−2s)!
= ,
(2s−2n)! (n−s)!
wheresandnareintegerswiths<n.Thisresultcanbeusedtoavoid
negativefactorialssuchasintheseriesrepresentationsofthespher-
ical Neumann functions and the Legendre functions of the second
kind.
10.1.4 Showthat(cid:2)(z)maybewritten
(cid:2)
∞
(cid:2)(z)=2 e−t2t2z−1dz, (cid:10)(z)>0,
0
(cid:2) (cid:3) (cid:4) (cid:5)(cid:6)
1 1 z−1
(cid:2)(z)= ln dt, (cid:10)(z)>0.
t
0
10.1.5 In a Maxwellian distribution the fraction of particles with speed
betweenvandv+dvis
(cid:4) (cid:5) (cid:4) (cid:5)
dN m 3/2 mv2
=4π exp − v2dv,
N 2πkT 2kT
where N isthetotalnumberofparti(cid:8)cles.Theaverageorexpectation
valueofvnisdefinedas(cid:15)vn(cid:16)= N−1 vndN.Showthat
(cid:4) (cid:5) (cid:4) (cid:5)(cid:27)
2kT n/2 n+1 1
(cid:15)vn(cid:16)= ! !.
m 2 2