1) The document discusses calculating line integrals of rational functions around a unit circle using the residue theorem. 2) It provides an example of calculating the definite integral of cos2n(θ) from 0 to π by rewriting it as a line integral and applying the residue theorem. 3) The solution is that the integral equals π/(22n(2nn)), which is obtained by evaluating the residue of the rational function inside the integral.

1. Residue integration Hanpen Robot 2015/02/12
★Formula 01
∫ 𝑓(𝑒 𝑗𝜃
2𝜋
0
)𝑑𝜃 = ∫ 𝑓(𝑧)
𝑑𝑧
𝑗𝑧𝐶
, (𝐶: 𝑧(𝑡) = 𝑒 𝑗𝜃
, 0 ≤ 𝜃 < 2𝜋)
∴ ∫ 𝑓(𝑒 𝑗𝜃
2𝜋
0
)𝑑𝜃 = ∑ Res [
𝑓(𝑧)
𝑗𝑧
, 𝛼𝑖]
Where 𝑓(𝑧) is rational function.
★Proof: Obvious by definition of a line integral.
★Problem 01: Calculate the definite integral of the following:
∫ cos2𝑛(𝜃)
𝜋
0
𝑑𝜃, (𝑛 = 1,2,3, … )
★Answer: Using Formula 01,
∫ cos2𝑛(𝜃)
𝜋
0
𝑑𝜃 =
1
2
∫ cos2𝑛(𝜃)
2𝜋
0
𝑑𝜃 =
1
2
∫ (
𝑧 + 𝑧−1
2
)
2𝑛
𝑑𝑧
𝑗𝑧𝐶
=
1
𝑗22𝑛+1
∫ (𝑧 +
1
𝑧
)
2𝑛
𝑑𝑧
𝑧𝐶
=
1
𝑗22𝑛+1
∫ (
𝑧2
+ 1
𝑧
)
2𝑛
𝑑𝑧
𝑧𝐶
=
1
𝑗22𝑛+1
∫
(𝑧2
+ 1)2𝑛
𝑧2𝑛+1
𝑑𝑧
𝐶
=
1
𝑗22𝑛+1
∫
1
𝑧2𝑛+1
∑ (
2𝑛
𝑘
) 𝑧2(2𝑛−𝑘)
2𝑛
𝑘=0
𝑑𝑧
𝐶
=
1
𝑗22𝑛+1
∫
1
𝑧2𝑛+1
∑ (
2𝑛
𝑘
) 𝑧4𝑛−2𝑘
2𝑛
𝑘=0
𝑑𝑧
𝐶
=
1
𝑗22𝑛+1
∫ ∑ (
2𝑛
𝑘
) 𝑧4𝑛−2𝑘−2𝑛−1
2𝑛
𝑘=0
𝑑𝑧
𝐶
=
1
𝑗22𝑛+1
∫ ∑ (
2𝑛
𝑘
) 𝑧2𝑛−2𝑘−1
2𝑛
𝑘=0
𝑑𝑧
𝐶
=
1
𝑗22𝑛+1
∫ (
2𝑛
𝑛
) 𝑧−1
𝑑𝑧
𝐶
=
1
𝑗22𝑛+1
(
2𝑛
𝑛
) ∫
𝑑𝑧
𝑧𝐶
=
1
𝑗22𝑛+1
(
2𝑛
𝑛
) 2𝜋𝑗 =
𝜋
22𝑛
(
2𝑛
𝑛
)
∴ ∫ cos2𝑛(𝜃)
𝜋
0
𝑑𝜃 =
𝜋
22𝑛
(
2𝑛
𝑛
)