Advanced book on Mathematics Olympiad

540 Real Analysis

In=

⎧

⎪⎪⎨

⎪⎪⎩

1 · 3 · 5 ···( 2 k− 1 )

2 · 4 · 6 ···( 2 k)

·

π

2

, ifn= 2 k,

2 · 4 · 6 ···( 2 k)

1 · 3 · 5 ···( 2 k+ 1 )

, ifn= 2 k+ 1.

To prove the Wallis formula, we use the obvious inequality sin^2 n+^1 x<sin^2 nx<

sin^2 n−^1 x,x∈( 0 ,π 2 )to deduce thatI 2 n+ 1 <I 2 n<I 2 n− 1 ,n≥1. This translates into

2 · 4 · 6 ···( 2 n)

1 · 3 · 5 ···( 2 n+ 1 )

<

1 · 3 · 5 ···( 2 n− 1 )

2 · 4 · 6 ···( 2 n)

·

π

2

<

2 · 4 · 6 ···( 2 n− 2 )

1 · 3 · 5 ···( 2 n− 1 )

,

which is equivalent to

[

2 · 4 · 6 ···( 2 n)

1 · 3 · 5 ···( 2 n− 1 )

] 2

·

2

2 n+ 1

<π <

[

2 · 4 · 6 ···( 2 n)

1 · 3 · 5 ···( 2 n− 1 )

] 2

·

2

2 n

.

We obtain the double inequality

π<

[

2 · 4 · 6 ···( 2 n)

1 · 3 · 5 ···( 2 n− 1 )

] 2

·

1

n

<π·

2 n+ 1

2 n

.

Passing to the limit and using the squeezing principle, we obtain the Wallis formula.

467.Denote the integral from the statement byIn,n≥0. We have

In=

∫ 0

−π

sinnx

( 1 + 2 x)sinx

dx+

∫π

0

sinnx

( 1 + 2 x)sinx

dx.

In the first integral changexto−xto further obtain

In=

∫π

0

sinnx

( 1 + 2 −x)sinx

dx+

∫π

0

sinnx

( 1 + 2 x)sinx

dx

=

∫π

0

2 xsinnx

( 1 + 2 x)sinx

dx+

∫π

0

sinnx

( 1 + 2 x)sinx

dx

=

∫π

0

( 1 + 2 x)sinnx

( 1 + 2 x)sinx

dx=

∫π

0

sinnx

sinx

dx.

And these integrals can be computed recursively. Indeed, forn≥0 we have

In+ 2 −In=

∫π

0

sin(n+ 2 )x−sinnx

sinx

dx= 2

∫π

0

cos(n− 1 )xdx= 0 ,

a very simple recurrence. Hence forneven,In=I 0 =0, and fornodd,In=I 1 =π.

(3rd International Mathematics Competition for University Students, 1996)