1000 Solved Problems in Modern Physics

44 1 Mathematical Physics

NowΓ(n)=

∫∞

0 x

n− (^1) e−xdx, putx=y (^2) ,dx= 2 ydy, so that
Γ(n)= 2

∫∞

0

y^2 n−^1 e−y

2

dy

Γ(1/2)= 2

∫∞

0

e−y

2

dy=

2

√

π

2

=

√

π

So that

∫ π 2

0

(cosθ)rdθ=

√

π

2

Γ

(r+ 1

2

)

Γ

(r

2 +^1

)

1.26 (a)B(m,n)=

∫ 1

0

xm−^1 (1−x)n−^1 dx (1)

Putx=

y

1 +y

(2)

B(m,n)=

∫∞

0

yn−^1 dy

(1+y)m+n

=

Γ(m)Γ(n)

Γ(m+n)

Lettingm= 1 −n;0<n< 1

∫∞

0

yn−^1

(1+y)

dy=

Γ(1−n)Γ(n)

Γ(1)

ButΓ(1)=1 and

∫∞

0

yn−^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)

NowΓ(n)=Γ(nn+1)

Γ(−in)=

Γ(1−in)

−in

∴|Γ(in)|^2 =

Γ(in)Γ(1−in)

−in(siniπn)

by (3)

Further sinh(πn)=isiniπn

∴|Γ(in)|^2 =

π

nsinh(πn)

1.3.4 Matrix Algebra

1.27 LetHbe the hermitian matrix with characteristic rootsλi. Then there exists a
non-zero vectorXisuch that