Lemma del grande cerchio e di Jordan

Lemma

ſ: S → ℂ

S = {z ∈ ℂ: Im(z) ≥ 0}

ſ continue.

lim _{|z| → ∞} ſ(z) = 0

γ_R: [0, π] → ℂ, t → R_e^it

allora lim _{R → +∞} ∫_γR ſ(z) dz = 0

Esempio

∫_-∞^{+∞ 1}/_{x⁴ + x² + 1} dx = ?

ſ(z) = ¹/_{z⁴ + z² + 1}

z⁴ + z² + 1 = 0

z² = ^{-1 ± √1-4}/₂ = -¹/₂ ± ^√3/₂ i

z² = -¹/₂ - ^√3/₂ i = cos(-²/₃ π) + i sin(-²/₃ π)

z₀ = cos(^π/₃) + i sin(^π/₃) = ¹/₂ - ^√3/₂ i

z² = -¹/₂ + ^√3/₂ i = cos(^π/₃) + i sin(^π/₃) = -¹/₂ + ^√3/₂ i

z₁ = cos(^π/₃) + i sin(^π/₃) = ¹/₂ + ^√3/₂ i

z₂ = cos(^4π/₃) + i sin(^4π/₃) = ¹/₂ + ^√3/₂ i

Lemma (del Cerchio Grande)

f: S ⟶ ℂ

S = {z ∈ ℂ: Im(z) ≥ 0}

f continua .

lim_{|z| ⟶ +∞} f(z) = 0

γ_R: [0,π] ⟶ ℂ . t ⟶ Re^it

allora lim_{R ⟶ +∞} ∫_γR f(z) dz = 0

Esempio

∫_-∞^+∞ 1/(x⁴ + x² +1) dx = ?

f(z) = 1/(z⁴ + z² + 1)

z⁴ + z² + 1 = 0

z² = -1 ± i√1-4/2 = -1/2 ± √3 i/2

z² = -1/2 - √3/2 i = cos(-2π/3) + i sin(-2π/3)

z₀ = cos(-2π/3) + i sin(-2π/3) = cos(π/3) + i sin(π/3)

z₂ = cos(2π/3) + i sin(2π/3) = 1/2 - √3/2 i

z² = -1/2 + √3/2 i = cos(2π/3) + i sin(2π/3)

z₁ = cos(2π/3) + i sin(2π/3) = -1/2 + √3/2 i

z³: cos(4π/3) + i sin(4π/3) = 1/2 + √3/2 i

z₃: cos(4π/3) + i sin(4π/3) = 1/2 + √3/2 i

∫∫ β(z) dz = -∫₀^δ₀ β(z) dz + ∫_{γR β(z) dz = 2πi (Res(β, z1) + Res(β, z2))}

lim_R→+∞ ∫_{-1 → +R} g(x) dx + lim_R→+∞ ∫_γR β(z) dz = 2π i (Res(β, z₁) + Res(β, z₂))

lim_|z|→+∞ z g(z) = lim_|z|→+∞ ²/_{z⁴ + z² + 1} = 0

Quindi per il lemma il lim_R→+∞ ∫_γR β(z) dz = 0,

Quindi ∫_{-1 → +∞} g(x) dx = 2π i (Res(β, z₁) + Res(β, z₂))

β(z) = _g1(z)/^g₂(z)

g₁(z) = 1

g₂(z) = z⁴ + z² + 1

g'₁(z) = 1

g'₂(z) = 4z³ + 2z = 2z(z² + z + 1)

Res(β, z₁) = _g1(z)/^g'₂(z₁) = ₁/^{(-1 + √3 i) (-1 - √3 i + 1)} = ₁/^{3 - √3 i} = ₁/⁴ + √₃/₁₂ i

Res(β, z₂) = ₁/^{3 + √3 i} = ₁/⁴ - √₃/₁₂ i

∫_{-∞ → +∞} g(x) dx = ^√₃/₃ π

Lemma (di Jordan)

g: S -> 𝕌 S = {z ∈ 𝕌: Im(z) ≥ 0} g cont lim |z| -> +∞ g(z) = 0 𝕌R: [0,1] -> 𝕌 t -> R^ei𝘊t

Esempio

∫_-∞^+∞ (cos(3x) / (x² + 2)) dx = ? g(z) = e^i3z / (z² + 2) Re{g(x)} = cos(3x) / (x² + 2) g(z) = 1 / (z² + 2) lim |z| -> +∞ 1 / (z² + 2) = 0 g è continue z² + 2 = 0 => z² = -2 => zz = √2i, z₁ = -√2i

∫ g(z) dz = ∫_-R^R g(x) dx + ∫_𝕌R g(z) dz = 2πi Re{g(z₀)} lim_R->+∞ ∫_-R^R f(x) dx = lim_R->+∞ ∫_k g(z) dz = 2πi Re{Re(g(z₀))}

∫_-∞^+∞ e^I3x / (x² + 2) dx = 2πi Re{f(z₀)}