212 CHAPTER 9. INTEGRATION

We know that sin(x)∈ [0,1] for x∈ [0,π/2] and so sin2m+1 (x)≥ sin2m+2 (x) . Thereforefrom 9.10,

m+1m

=2m+2

2m≥ 2m+2

2m

∫ π/20 sin2m+2 (x)dx∫ π/20 sin2m+1 (x)dx

=

∫ π/20 sin2m (x)dx∫ π/2

0 sin2m+1 (x)dx

=

(2m−1)(2m−3)···12m(2m−2)···2(2m)(2m−2)···2

(2m+1)(2m−1)···3

π

2=

π

2(2m+1)

(2m−1)2 (2m−3)2 · · ·122m (m!)2 ≥ 1

It follows from the squeezing theorem that

limm→∞

12m+1

22m (m!)2

(2m−1)2 (2m−3)2 · · ·1=

π

2

This exceedingly interesting formula is Wallis’ formula.Now multiply both the top and the bottom of the expression on the left by the product

(2m)2 (2(m−1))2 · · ·22 which is 22m (m!)2 . This is another version of the Wallis formula.

π

2= lim

m→∞

22m

2m+122m (m!)2 (m!)2

((2m)!)2

It follows that √π

2= lim

m→∞

22m√

2m+1(m!)2

(2m)!= lim

m→∞

22m√

2m(m!)2

(2m)!(9.11)

Now with this result, it is possible to find c in Stirling’s formula. Recall

limm→∞

m!mm+(1/2)e−mc

= 1 = limm→∞

mm+(1/2)e−mcm!

In particular, replacing m with 2m and using 9.11,

√π

2= lim

m→∞

→√

π2

22m√

2m(m!)2

(2m)!

→1(2m)!

(2m)2m+(1/2) e−2mc

→1

mm+(1/2)e−mcm!

2

= limm→∞

22m√

2m1

(2m)2m+(1/2) e−2mc

(mm+(1/2)e−mc

1

)2

= c limm→∞

22m m2m+1

(2m)2m+1 =c2

so c =√

2π . This proves Stirling’s formula.

Theorem 9.8.4 limm→∞m!

mm+(1/2)e−m =√

2π .

9.9 Fubini’s Theorem an IntroductionFubini’s theorem has become the name of a theorem which involves interchanging the orderof integration in iterated integrals. You may have seen it mentioned in a beginning calculuscourse. It is actually an incredibly deep result, much more so than what will be indicatedhere. Here I will only consider enough to allow what will be done in this book. It turns out