1550251515-Classical_Complex_Analysis__Gonzalez_

638 Chapters

= z ~ (a+ n)(a)n(b + n)(b)n Zn

n=O LI n!(c)n

= z(O + a)(O + b)w

which shows that w = F( a, b, ~; z) satisfies

[0(0 + c - 1) - z(O + a)(O + b)]w = 0. (8.22-5)
But Ow == zw' and 0( () - 1 )w = z^2 w^11 • Hence (8.22-5) can be written as
z^2 w^11 + czw' -z[z^2 w" -zw' +(a+ b)zw' + abw] = 0
or
(z -z^2 )w^11 + [c - (a+ b + l)z]w' - abw = 0
A second independent solution of this differential equation can be ob-
tained in terms of a hypergeometric function with different parameters,
provided that c is not a positive integer. In fact, letting w = z^1 -c(, we get
w' = (1 -c)z-c(, + z^1 -c('
w" = - c(l - c)z-c-l( + 2(1 - c)z-c(' + z^1 -c(^11

and substituting in (8.22-4), we find that

(z - z^2 )(^11 + [c1 - (a1 +bi+ l)z](' - aib1( = 0 (8.22-6)

where ai = a - c + 1, bi = b - c + 1, c1 = 2 - c.
Equation (8.22-6) has the solution ( = F( a 1 , b 1 , c 1 ; z), and hence (8.22-4)
has as a second solution

w = zl-c F(a - c + 1, b-c + 1, 2 - c; z)

whenever c-=/= 1, 2, 3, ... , the general solution now being any linear combi-

nation of the two particular solutions just found. If c is a positive integer,

then the second solution contains in general a logarithmic term. See [30] or

[33], Vol. II, 646-652. For instance, if c = 1, the second solution is given by

w=F(a,b,l;z)Logz+ (:a+ :b +2:c)F(a,b,c;z) lc=l

(6) From Theorem 8.50 (3) and formula (8.20-33) we have

(b)n r(b+n)r(c) r(c) r(b+n)r(c-b)

(c)n r(c+n)r(b) r(b)r(c-b) r(c+n)

r(c)

r(b)r(c - b) B(b + n, c - b)

= r(c) fl tb+n-l(l - ty-b-1 dt (8.22-7)

r(b)r(c - b) lo