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182 9. Elliptic and Hypergeometric Functions

(5.8) Hypergeometric Differential Equation. The hypergeometric function F sat-
isfies the following differential equation called the hypergeometric differential equa-
tion:

d 2 d
z(1 − z) F + [c − (a + b + 1)z] F − abF = 0.
dz 2 dz
2

1/2 n
For the case of F(1/2, 1/2, 1; z) = z this function satisfies
0≤n n
d 2 d 1
z(1 − z) F + (1 − 2z) F − F = 0.
dz 2 dz 4
This hypergeometric function can be used to evaluate the following elliptic inte-
gral.
(5.9) Theorem. For a complex number λ with |λ| < 1

(
π/2 −1/2 1 1
2
2 1 − λ sin θ dθ = π F , , 1; λ .
0 2 2
Proof. Using the binomial series, we have
2 −1/2 − 1 n 2n
1 − λ sin θ = 2 (−λ) sin θ.
n
0≤n
By (5.5) and integrating term by term, we obtain

!
1 1
(
π/2 −1/2 − −
n n
2
2 1 − λ sin θ dθ = 2 (−1) λ π(−1) n 2
0 n n
0≤n
1 2
− n 1 1
= π 2 λ = π F , , 1; λ .
n 2 2
0≤n
This proves the theorem.
Exercises
1. Show that
(s)
(1 − s) = π/ sin πs for 0 < Re(s)< 1, and prove that
(s) prolongs
to a mermorphic function on C using this functional equation.
2. Show that the following formulas hold for |z| < 1.
(a) zF(1, 1, 2;−z) = log(1 + z).
2
−
(b) zF(1/2, 1, 3/2;−z ) = tan 1(z).
2 −1
(c) zF(1/2, 1/2, 3/2; z ) = sin (z).
3. Show that (d/dz)F(a, b, c; z) = (ab/c)F(a + 1, b + 1, c + 1; z).
4. Show that T n (z) = F(n, −n, 1/2; (1 − z)/2) is a polynomial of degree n.