Funcion Gamma

International Journal of Pure and Applied Mathematics
————————————————————————–
Volume 61 No. 4 2010, 375-379

A GENERALIZATION OF KNAR’S FORMULA
Jake Logsdon
Department of Mathematics
North Carolina State University
2311, Stinson Dr., Raleigh, NC 27695, USA
e-mail: jtlogsdo@ncsu.edu
Abstract: We produce two derivations for an infinite double product functional equation for the Gammafunction, a generalization of Knar’s formula.
Series are derived for the Polygamma function using the double product, as
well as certain symmetric product identities.
AMS Subject Classification: 33B15
Key Words: Gamma function, Knar’s formula, functional equation

*
The Gamma function
+∞

Γ(z ) =

xz −1 e−x dx

0

has many identities – we cite the relevant literature (see [2], [1]) forits study
and development. Knar’s formula (see [2]), a seemingly little-known product
formula, is one of them:
∞

1
1
z
√Γ
+n.
(1)
π
22
n=1
√
Using the well-known value Γ(1/2) = π , letting z = 1 and taking a square
root throughout the given expression yields the highly symmetric equation
Γ(1 + z ) = 22z

∞

1

Γ1/2 ( 1 )
2
n=1

Γ1/2

1
1
+
2 2n

1
=.
2

(2)The purpose of this paper is to generalize (1) and derive results analogous to
Received:

February 27, 2010

c 2010 Academic Publications

376

J. Logsdon

the above. We will use the following formulas:
k!kz
,
Γ(z ) = lim
k →∞ (z + 1) · · · (z + k )

(3)

n−1

(2π )(n−1)/2 n1/2−nz Γ(nz ) =

Γ z+
k =0

k
n

(n ∈ Z+ ),

(4)

which are known as Euler’s limit formand the Gauss multiplication formula,
respectively, both shown and proven in [1].
Our first derivation of the titular identity comes from considering the finite
product
p
i=1

nΓ(n1−i z )
Γ(z )
Γ(z/n)
Γ(n1−p z )
np
= np ·
·
···
=
Γ(z ).
Γ(n−i z )
Γ(z/n) Γ(z/n2 )
Γ(z/np )
Γ(z/np )

(5)

From (3), np /Γ(z/np ) may be given by
np
=
Γ(z/np )

−p

k!kn z
lim
k →∞ z (n−p z+ 1) · · · (n−p z + k )

−1

;

and as p → ∞ we see easily that np /Γ(z/np ) → z . Substituting the above into
(5),
∞

i=1

nΓ(n1−i z )
= z Γ(z ) = Γ(1 + z ).
Γ(n−i z )

(6)

Now, the Gauss multiplication formula (4) can be used to give (8) by taking
n−1
out the k = 0 term to give the k=1 term; solving for Γ(nz )/Γ(z ), multiplying
−i z :
through by n and letting z → n
1−i√
n−1
nΓ(n1−i z )
nn z n
z
k
=
+i.
Γ
−i z )
(n−1)/2
Γ(n
nn
(2π )
k =1

Substituting this into (6) yields
√
n−1
∞
n
Γ
(2π )(n−1)/2 k=1
i=1

k
z
+i
nn

∞

n −n

= Γ(1 + z )

1−i z

.

(7)

i=1

But the product on the right-hand side of the above is just the exponential form
of a geometric series, i.e.,
∞

∞

n
i=1

−n1−i z

= exp −z log(n)

i=01
ni

= n−nz/(n−1)

A GENERALIZATION OF KNAR’S FORMULA

377

which gives
√

∞

i=1

n−1

n

(2π )(n−1)/2

k
z
+i
nn

Γ
k =1

= n−nz/(n−1) Γ(1 + z )

when substituted into (7). Note that by the Gauss multiplication formula
√
n−1
n
1
=
,
Γ(k/n)
(2π )(n−1)/2
k =1
which gives the desired result
∞ n−1
i=1 k =1

1
Γ
Γ(k/n)

k
z
+
n ni

=n−nz/(n−1) Γ(1 + z ) (n ∈ Z+ , z ∈ C\Z− ). (8)

A second derivation involves continually resubstituting directly from the
Gauss multiplication formula (4). As before, we take out the k = 0 term of the
formula to give the slightly modified version
n−1

(2π )(n−1)/2 n1/2−nz Γ(nz ) = Γ(z )

Γ z+
k =1

k
n

;

first we solve for Γ(z ) and, let z → z/n, and then solve back for Γ(z ) on the
left.This yields the two formulas
n−1
(1−n)/2 z −1/2

Γ(z ) = (2π )

n

Γ(z/n)

k
z
+
nn

Γ
k =1

,

(2π )(n−1)/2 n1/2−nz Γ(nz )
.
n−1
k =1 Γ(z + k/n)
Equating the right-hand sides of the prior equations and solving for Γ(nz ) we
find
Γ(z ) =

n−1

Γ(nz ) = (2π )2·(1−n)/2 nnz +z −2·1/2 Γ(z/n)

k
n

Γ z+
k =1

Γ

k
z
+
nn

;

we can see that if we...