10E (Analysis)

Solution

sin^{n - 1} x sin x

, for

n \geq 2

I_{n} = [- sin^{n - 1} x cos x]_{0}^{π /2} + (n - 1) \int_{0}^{π /2} sin^{n - 2} x cos^{2} x d x = (n - 1) (I_{n - 2} - I_{n}) .

Both boundary terms are zero. Therefore the Wallis integrals satisfy

n I_{n} = (n - 1) I_{n - 2} .

Starting with

I_{0} = π /2

and

I_{1} = 1

, repeated application gives

I_{2 n} I_{2 n + 1} = \frac{1 \cdot 3 \dots ( 2 n - 1 )}{2 \cdot 4 \dots 2 n} \frac{π}{2} = \frac{( 2 n )!}{2 ^{2 n} ( n ! ) ^{2}} \frac{π}{2}, = \frac{2 \cdot 4 \dots 2 n}{3 \cdot 5 \dots ( 2 n + 1 )} = \frac{2 ^{2 n} ( n ! ) ^{2}}{( 2 n + 1 )!} .

These formulas also hold at

n = 0

For

0 < x < π /2

0 < sin x < 1

, so

0 < sin^{n} x < sin^{n - 1} x

when

n \geq 1

. Integrating the strict inequalities proves

0 < I_{n} < I_{n - 1}

. In particular

I_{2 n + 1} / I_{2 n} < 1

. For

n \geq 1

, the recurrence and

I_{2 n - 1} > I_{2 n}

give

I_{2 n + 1} = \frac{2 n}{2 n + 1} I_{2 n - 1} > \frac{2 n}{2 n + 1} I_{2 n} .

Thus

\frac{2 n}{2 n + 1} < \frac{I _{2 n + 1}}{I _{2 n}} < 1.

n = 0

the same displayed bounds are

0 < 2/ π < 1

, also true directly from

I_{1} < I_{0}

I_{2 n + 1} / I_{2 n} \to 1

. Substituting the evaluated integrals shows

\frac{I _{2 n + 1}}{I _{2 n}} = \frac{1}{π} \frac{2 ^{4 n + 1} ( n ! ) ^{4}}{( 2 n + 1 ) (( 2 n )! ) ^{2}},

and hence

n \to \infty lim \frac{2 ^{4 n + 1} ( n ! ) ^{4}}{( 2 n + 1 ) (( 2 n )! ) ^{2}} = π .

This is twice the finite-product form of the Wallis product; the proof used only the integral recurrence and strict comparison of its integrands.

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