Resolved: "Proving ∫02π(cos⁡(t)2n)=(2nn)2π22n using the following result ∫γ(z+1z)2ndzz=(2nn)2πi Prove..."

Seamus Kent

Answered question

2022-01-26

Proving

\int_{0}^{2 π} (\cos (t)^{2 n}) = (\binom{2 n}{n}) \frac{2 π}{2^{2 n}}

using the following result

\int_{γ} (z + \frac{1}{z})^{2 n} \frac{d z}{z} = (\binom{2 n}{n}) 2 π i

Prove

\int_{0}^{2 π} (\cos (t)^{2 n}) = (\binom{2 n}{n}) \frac{2 π}{2^{2 n}}

Deegan Mullen

Beginner2022-01-27Added 12 answers

Take

γ = {(x, y) | x^{2} + y^{2} = 1}

. Then

\int_{γ} (z + 1 / z)^{2 n} (1 / z) d z = \int_{0}^{2 π} (2 \cos t)^{2 n} (\cos t - i \sin t) (- \sin t + i \cos t) d t = i \int_{0}^{2 π} (2 \cos t)^{2 n} d t = (\binom{2 n}{n}) 2 π i

And the rest follows.

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