Page 386 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 15] GAMMA AND BETA FUNCTIONS 377

This factorial representation for positive integers suggests the possibility that
k! k x
ðx þ 1Þ¼ x! ¼ lim x 6¼ 1; 2; k ð6Þ
k!1 ðx þ 1Þ ... ðx þ kÞ
Gauss verified this identification back in the nineteenth century.
x
k! k
.Itis called Gauss’s
This infinite product is symbolized by ðx; kÞ, i.e., ðx; kÞ¼
ðx þ 1Þ ðx þ kÞ
function and through this symbolism,
ðx þ 1Þ¼ lim ðx; kÞ ð7Þ
k!1
1
The expression for (which has some advantage in developing the derivative of ðxÞ) results as
ðxÞ
follows. Put (6a)in the form
k x 1 1 1
lim x 6¼ ; ; ... ;
2 3 k
k!1 ð1 þ xÞð1 þ x=2Þ ... ð1 þ x=kÞ
Next, introduce

1 1 1

k ¼ 1 þ þ þ ln k
2 3 k
Then

¼ lim
k
k!1
is Euler’s constant. This constant has been calculated to many places, a few of which are

0:57721566 ... .
x
By letting k ¼ e x ln k ¼ e x½
k þ1þ1=2þ þ1=kŠ , the representation (6) can be further modified so that
e x e x=2 e x=k
x Y
x x ln k x
1

x
ðx þ 1Þ¼ e lim ¼ e e e = 1 þ
k!1 1 þ x 1 þ x=2 1 þ x=k k
k¼1
1
Y x 1 2 3 k x
k k!ðk þ xÞ¼ lim
¼ x ¼ lim ðx; kÞ ð8Þ
¼1 k!1 ðx þ 1Þðx þ 2Þ ðx þ kÞ k!1
Since ðx þ 1Þ¼ x ðxÞ;
1
x 1 þ x 1 þ x=2 1 þ x=k
x Y x=k
1
¼ xe lim ¼ xe ð1 þ x=kÞ e ð9Þ
k!1 e x e x=2 e x=k
¼1
ðxÞ
Another result of special interest emanates from a comparison of ðxÞ ð1 xÞ with the ‘‘famous’’
formula

x 1 1 1 Y 2
1
¼ lim
sin x 2 2 2 ¼ f1 ðx=kÞ g ð10Þ
k!1 1 x 1 ðx=2Þ
ð1 x=kÞ ¼1
(See Differential and Integral Calculus,byR. Courant (translated by E. J. McShane), Blackie & Son
Limited.)
1
ð1 xÞ is obtained from ð yÞ¼ ð y þ 1Þ by letting y ¼ x, i.e.,
y
1
or
ð xÞ¼ ð1 xÞ ð1 xÞ¼ x ð xÞ
x