Begin2.DVI

Derivatives of ln Γ( z)

Using the Weierstrass definition of the Gamma function

1

Γ(z)=ze

γz

∏∞

n=1

[(

1 +

1

z

)

e−z/n

]

(12.135)

take the natural logarithm of both sides and show

ln Γ(z) = −ln z−γz −

∑∞

k=1

[

ln

(

1 +

z

k

)

−

z

k

]

(12.136)

Take the derivative of each term in this equation to show

d

dz

ln Γ(z) = Γ

′(z)

Γ(z)

=−^1

z

−γ−

∑∞

k=1

(

1

z+k

−^1

k

)

(12.137)

Differentiate equation (12.137) to obtain

d^2

dz^2 ln Γ(z) =

1

z^2 +

∑∞

k=1

1

(z+k)^2 =

1

z^2 +

1

(z+ 1)^2 +

1

(z+ 2)^2 +··· (12.138)

Continuing differentiating the function lnΓ(z)and use the fact that

d

dz

(z+k)−^2 =(−2)(z+k)−^3

d^2

dz^2

(z+k)−^2 =(−2)(−3)(z+k)−^4

d^3

dz^3 (z+k)

− (^2) =(−2)(−3)(−4)(z+k)− 5

..

.

..

.

and demonstrate that

dn

dznln Γ(z) =(−1)

n(n−1)!

∑∞

k=0

1

(z+k)n

dn+1

dn+1

ln Γ(z) =(−1)n+1n!

∑∞

k=0

1

(z+k)n+1

(12.139)

Make note of the following definitions. The function ζ(n, z) =

∑∞

k=0

1

(z+k)n

,where

any term where (z+k) = 0 is understood to be excluded from the summation process,