7B (Complex Methods)

Take the unit circle counterclockwise and

n \geq 0

. The integrand is a Laurent polynomial with its only possible singularity at zero. Its expansion is

\frac{1}{z} (z - z^{- 1})^{2 n} = \sum_{j = 0}^{2 n} (- 1)^{j} (j 2 n) z^{2 n - 2 j - 1} .

The

z^{- 1}

term is the term

j = n

, so the residue is

(- 1)^{n} (n 2 n)

\oint_{∣ z ∣ = 1} (z - z^{- 1})^{2 n} \frac{d z}{z} = 2 πi (- 1)^{n} \frac{( 2 n )!}{( n ! ) ^{2}} .

b

Parametrize the same contour by

z = e^{i t}

0 \leq t \leq 2 π

. Then

d z / z = i d t

and

z - z^{- 1} = 2 i sin t

. Part a becomes

i (- 1)^{n} 2^{2 n} \int_{0}^{2 π} sin^{2 n} t d t = 2 πi (- 1)^{n} (n 2 n) .

Canceling the nonzero common factors gives

\int_{0}^{2 π} sin^{2 n} t d t = \frac{2 π}{2 ^{2 n}} (n 2 n) = \frac{π ( 2 n )!}{2 ^{2 n - 1} ( n ! ) ^{2}} .

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