Taylor series expansion for ln Γ( x+ 1)
Make reference to the equations (12.136), (12.137), (12.138), (12.139), and verify
that when zis replaced by (x+1) in these equations, one obtains the following values.
ln Γ(x+ 1)
x=0
= lnΓ(1) = 0
d
dx
ln Γ(x+ 1)
x=0
=−γ because (12.137) is a telescoping series
d^2
dx^2
ln Γ(x+ 1)
x=0
=^1
12
+^1
22
+^1
32
+··· =ζ(2)
..
.
..
.
dn
dx n
ln Γ(x+ 1)
x=0
=(−1)n(n−1)!ζ(n) n≥ 2
(12.141)
where
ζ(n) =
∑∞
k=1
1
kn
is the Riemann Zeta function. The above values of the derivatives of ln Γ(x+ 1),
evaluated at x= 0, produces the Taylor series expansion for ln Γ(x+ 1) as
ln Γ(x+ 1) = −γx +ζ(2)
x^2
2 −ζ(3)
x^3
3 +ζ(4)
x^4
4 +···+ (−1)
nζ(n)xn
n +··· (12.142)
which converges if xis less than unity.
Another product formula
Define the function
φ(z) =
nzΓ(z)Γ
(
z+^1 n
)
Γ
(
z+n^2
)
···Γ
(
z+(n−n1)
)
nΓ(nz ) (12.143)
and use the equation (12.106) to write
Γ
(
z+
r
n
)
= limm→∞
( (m−1)!mz+r/n
z+rn
)(
z+rn+ 1
)(
z+nr+ 2
)
···
(
z+rn+m− 1
) (12.144)
for r= 0, 1 , 2 ,.. ., n − 1 and
Γ(nz ) = limm→∞
(nm −1)!(nm )nz
(nz )(nz + 1)(nz + 2) ···(nz +nm −1) (12.145)
to express equation (12.143) in the form
φ(z) =
nnz
n∏− 1
r=0
mlim→∞ (m−1)!m
z+r/n
(
z+rn
)(
z+nr+ 1
)(
z+nr+ 2
)
···
(
z+rn+m− 1
)
nmlim→∞ (nm −1)!(nm )
nz
nz (nz + 1)(nz + 2) ···(nz +nm −1)
(12.146)
