44
1 Mathematical Physics
Now Γ(n) =
∞
0 x
n−1 e
−x dx, put x = y
2 , dx = 2ydy, so that
Γ(n) = 2
∞
0
y
2n−1 e
−y
2 dy
Γ(1/2) = 2
∞
0
e
−y
2 dy =
2
√
π
2
=
√ π
So that
π
2
0
(cos θ )
r dθ =
√ π
2
Γ
r +1
2
Γ
r
2
+ 1
1.26 (a) B(m, n) =
1
0
x
m−1 (1 − x)
n−1 dx
(1)
Put x =
y
1 + y
(2)
B(m, n) =
∞
0
y
n−1 dy
(1 + y) m+n =
Γ (m)Γ (n)
Γ (m + n)
Letting m = 1 − n; 0 < n < 1
∞
0
y
n−1
(1 + y)
dy =
Γ (1 − n)Γ (n)
Γ (1)
But Γ (1) = 1 and
∞
0
y
n−1
(1 + y)
dy =
π
sin(nπ)
; 0 < n < 1
Γ (n)Γ (1 − n) =
π
sin(nπ)
(3)
(b) |Γ (in)|
2
= Γ (in)Γ (−in)
Now Γ (n) =
Γ (n+1)
n
Γ (−in) =
Γ (1 − in)
−in
∴ |Γ (in)|
2
=
Γ (in)Γ (1 − in)
−in(sin iπ n)
by (3)
Further sinh(π n) = i sin iπn
∴ |Γ (in)|
2
=
π
n sinh(πn)
1.3.4 Matrix Algebra
1.27 Let H be the hermitian matrix with characteristic roots λ i . Then there exists a
non-zero vector X i such that