28.4. THE PROPER VALUE OF β (n) 1005

and so

Γ

(12

)= 2

∫∞

0e−u2

du =√

π

This proves the lemma.Next let n be a positive integer.

Theorem 28.4.2 α(n) = πn/2(Γ(n/2+1))−1 where Γ(s) is the gamma function

Γ(s) =∫

∞

0e−tts−1dt.

Proof: First let n = 1.

Γ(32) =

12

Γ

(12

)=

√π

2.

Thusπ

1/2(Γ(1/2+1))−1 =2√π

√π = 2 = α (1) .

and this shows the theorem is true if n = 1.Assume the theorem is true for n and let Bn+1 be the unit ball in Rn+1. Then by the

result in Rn,

mn+1(Bn+1) =∫ 1

−1α(n)(1− x2

n+1)n/2dxn+1

= 2α(n)∫ 1

0(1− t2)n/2dt.

Doing an integration by parts and using Lemma 28.4.1

= 2α(n)n∫ 1

0t2(1− t2)(n−2)/2dt

= 2α(n)n12

∫ 1

0u1/2(1−u)n/2−1du

= nα(n)∫ 1

0u3/2−1(1−u)n/2−1du

= nα(n)Γ(3/2)Γ(n/2)(Γ((n+3)/2))−1

= nπn/2(Γ(n/2+1))−1(Γ((n+3)/2))−1

Γ(3/2)Γ(n/2)

= nπn/2(Γ(n/2)(n/2))−1(Γ((n+1)/2+1))−1

Γ(3/2)Γ(n/2)

= 2πn/2

Γ(3/2)(Γ((n+1)/2+1))−1

= π(n+1)/2(Γ((n+1)/2+1))−1.

This proves the theorem.From now on, in the definition of Hausdorff measure, it will always be the case that

β (s) = α (s) . As shown above, this is the right thing to have β (s) to equal if s is a posi-tive integer because this yields the important result that Hausdorff measure is the same asLebesgue measure. Note the formula, πs/2(Γ(s/2+1))−1 makes sense for any s≥ 0.