Mathematical Methods for Physics and Engineering : A Comprehensive Guide

18.10 HYPERGEOMETRIC FUNCTIONS

F(a, b, b;x)=(1−x)−a, F(^12 ,^12 ,^32 ;x^2 )=x−^1 sin−^1 x,

F(1, 1 ,2;−x)=x−^1 ln(1 +x), F(^12 , 1 ,^32 ;−x^2 )=x−^1 tan−^1 x,

lim

m→∞

F(1,m,1;x/m)=ex, F(^12 , 1 ,^32 ;x^2 )=^12 x−^1 ln[(1 +x)/(1−x)],

F(^12 ,−^12 ,^12 ;sin^2 x)=cosx, F(m+1,−m,1; (1−x)/2) =Pm(x),

F(^12 ,p,p;sin^2 x)=secx, F(m,−m,^12 ;(1−x)/2) =Tm(x),

wheremis an integer,Pm(x)isthemth Legendre polynomial andTm(x)isthe

mth Chebyshev polynomial of the first kind. Some of these results are proved in

exercise 18.11.

Show thatF(m,−m,^12 ;(1−x)/2) =Tm(x).

Let us prove this result by transforming the hypergeometric equation. The form of the
result suggests that we should make the substitutionx=(1−z)/2 into (18.136), in which
cased/dx=− 2 d/dz. Thus, lettingu(z)=y(x) and settinga=m,b=−mandc=1/2,
(18.136) becomes

(1−z)

2

(1 +z)

2

(−2)^2

d^2 u

dz^2

+

[

1

2 −(m−m+1)

1 −z

2

]

(−2)

du

dz

−(m)(−m)u=0.

On simplifying, we obtain

(1−z^2 )

d^2 u

dz^2

−z

du

dz

+m^2 u=0,

which has the form of Chebyshev’s equation, (18.54). This equation hasu(z)=Tm(z)as
its power series solution, and soF(m,−m,^12 ;(1−z)/2) andTm(z) are equal to within a
normalisation factor. On comparing the expressions (18.141) and (18.56) atx=0,i.e.at
z= 1, we see that they both have value 1. Hence, the normalisations already agree and
we obtain the required result.

Integral representation

One of the most useful representations for the hypergeometric functions is in

terms of an integral, which may be derived using the properties of the gamma

and beta functions discussed in section 18.12. The integral representation reads

F(a, b, c;x)=

Γ(c)

Γ(b)Γ(c−b)

∫ 1

0

tb−^1 (1−t)c−b−^1 (1−tx)−adt,

(18.144)

and requiresc>b>0 for the integral to converge.

Prove the result (18.144).

From the series expansion (18.142), we have

F(a, b, c;x)=

Γ(c)

Γ(a)Γ(b)

∑∞

n=0

Γ(a+n)Γ(b+n)

Γ(c+n)

xn

n!

=

Γ(c)

Γ(a)Γ(b)Γ(c−b)

∑∞

n=0

Γ(a+n)B(b+n, c−b)

xn

n!

,