Product of odd and even integers
One can now apply the previous results
Γ(n) = (n−1)Γ(n−1), Γ(1) = 1, Γ(^1
2
) =
√
π (12 .88)
to show that for na positive integer
Γ
(
2 n+ 1
2
)
=
(
2 n− 1
2
)(
2 n− 3
2
)(
2 n− 5
2
)
···
(
5
2
)(
3
2
)(
1
2
)√
π (12 .89)
This result demonstrates that an alternative representation for the product of the
odd integers is given by
1 · 3 · 5 ···(2 n−5)(2 n−3)(2n−1) =^2
n
√
π
Γ
(
2 n+ 1
2
)
(12 .90)
Using the equations (12.88) one can demonstrate
Γ
(
2 n+ 2
2
)
=
(
2 n
2
)(
2 n− 2
2
)(
2 n− 4
2
)
···
(
6
2
)(
4
2
)(
2
2
)
(12 .91)
which shows that the product of the even integers can be represented in the form
2 · 4 · 6 ···(2 n−4)(2 n−2)(2 n) = 2 nΓ(n+ 1) (12 .92)
Example 12-5.
Let Sn=
∫π/ 2
0
sin nx dx and integrate by parts using
U= sin n−^1 x
dU =(n−1) sin n−^2 xcos x dx
dV = sin x dx
V =−cos x
to obtain
Sn=−sin n−^1 xcos x
]π/ 2
0 +
∫π/ 2
0
(n−1) sin n−^2 xcos^2 x dx
Sn=(n−1)
∫π/ 2
0
sin n−^2 x(1 −sin^2 x)dx = (n−1)Sn− 2 −(n−1)Sn
which simplifies to the recurrence formula
Sn=
n− 1
n Sn−^2 (12 .93)
The above recurrence relation implies that
Sn=
n− 1
n Sn−^2 =
(
n− 1
n
)(
n− 3
n− 2
)
Sn− 4 =··· (12 .94)
