Recall Euler’s infinite product formula for sin θ(see Example 4-38)
sin(πx ) = πx
(
1 −x
2
12
)(
1 −x
2
22
)(
1 −x
2
32
)
··· (12.111)
and compare this infinite product with the one occurring in equation (12.110) to
show
Γ(x)Γ(1 −x) =
π
sin(πx ) (12.112)
which is known as Euler’s reflection formula for the Gamma function. Make note of
the fact that if the value x= 1/ 2 is substituted into equation (12.112) one obtains
Γ(^1
2
)Γ(^1
2
) = π or Γ(^1
2
) =
√
π (12.114)
which agrees with our previous result.
Using the previous result nΓ(n) = Γ(1 + n), the equation (12.112) is sometimes
written in the form
Γ(1 + n)Γ(1 −n) = nπ
sin nπ
(12.114)
The Zeta function related to the Gamma function
The Gamma function Γ(z)and the Riemann Zeta function ζ(z)are related. Recall
that one definition of the Zeta function is (see Example 4-38)
ζ(z) =
1
1 z+
1
2 z+
1
3 z+
1
4 z+··· =
∑∞
n=1
1
nz, z >^1 (12.115)
and the integral form for representing the Gamma function is given by
Γ(z) =
∫∞
0
tz−^1 e−tdt, z > 0 (12.116)
In equation (12.116) make the change of variable t=rx and show
Γ(z) =
∫∞
0
(rx )z−^1 e−rx r dx =rz
∫∞
0
xz−^1 e−rx dx (12.117)
One can express equation (12.117) in the form
1
rz
=^1
Γ(z)
∫∞
0
xz−^1 e−rx dx (12.118)
A summation of equation (12.118) over integer values for rproduces the result
ζ(z) =
∑∞
r=1
1
rz
=^1
Γ(z)
∑∞
r=1
∫∞
0
xz−^1 e−rx dx (12.119)
