1000 Solved Problems in Modern Physics

1.3 Solutions 43

Thus the gamma function is an extension of the factorial function to numbers

which are not integers.

1.24 B(m,n)=

∫ 1

0

xm−^1 (1−x)n−^1 dx (1)

With the substitutionx=sin 2Φ(1) becomes

B(m,n)= 2

∫π/ 2

0

(sinΦ)^2 m−^1 (cosΦ)^2 n−^1 dΦ (2)

NowΓ(n)= 2

∫∞

0 y

2 n− (^1) e−y^2 dy
Γ(m)= 2

∫∞

0

y^2 m−^1 e−x

2

dx

∴Γ(m)Γ(n)= 4

∫∞

0

∫∞

0

x^2 m−^1 y^2 n−^1 exp−(x^2 +y^2 )dxdy (3)

The double integral may be evaluated as a surface integral in the first

quadrant of thexy-plane. Introducing the polar coordinatesx=rcosθand

y=rsinθ, the surface element ds=rdrdθ, (3) becomes

Γ(m)Γ(n)= 4

∫π/ 2

0

∫∞

0

r^2 m−^1 (cosθ)^2 m−^1 (sinθ)^2 n−^1 e−r

2

rdrdθ

Γ(m)Γ(n)= 2

∫π/ 2

0

(cosθ)^2 m−^1 (sinθ)^2 n−^1 dθ. 2

∫∞

0

r2(m+n)−^1 e−r

2

dr (4)

In (4), the first integral is identified asB(m,n) and the second one as

Γ(m+n). It follows that

B(m,n)=

Γ(m)Γ(n)

Γ(m+n)

1.25 One form of Beta function is

2

∫π/ 2

0

(cosθ)^2 m−^1 (sinθ)^2 n−^1 dθ=B(m,n)=

Γ(m)Γ(n)

Γ(m+n)

(m> 0 ,n>0)

(1)

Letting 2m− 1 =r, that ism=r+ 21 and 2n− 1 =0, that isn= 1 /2, (1)

becomes

∫π/ 2

0

(cosθ)rdθ=

1

2

Γ

(r+ 1

2

)

Γ

( 1

2

)

Γ

(r

2 +^1

) (2)