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Jordan's lemma for rational functions of one complex variable

Statement

Let $λ > 0$ , let $R$ be a rational function, and let $γ_{T}^{+} (t) = T e^{i t}$ for $0 \leq t \leq π$ . Assume that $γ_{T}^{+}$ meets no pole of $R$ for all sufficiently large $T$ , and that

$\sup_{0 \leq t \leq π} ∣ γ_{T}^{+} (t) R (γ_{T}^{+} (t)) ∣ ⟶ 0 (T \to \infty) .$

Then

$\int_{γ_{T}^{+}} e^{i λ z} R (z) d z ⟶ 0 (T \to \infty) .$

Facts & Assumptions

Given: A real number $λ > 0$ , a rational function $R$ , and the upper semicircles $γ_{T}^{+} (t) = T e^{i t}$ .

[L1]

If a twice differentiable function on an interval has nonnegative second derivative, then it is convex (A twice-differentiable function on an open interval is convex if and only if its second derivative is nonnegative).

Proof

technique · direct

1.1givenalgebra

For $z = T e^{i t}$ on the upper semicircle, $∣ e^{i λ z} ∣ = ∣ e^{i λ T (\cos t + i \sin t)} ∣ = e^{- λ T \sin t} .$

1.2L1

On $[0, π / 2]$ the function $g (t) = - \sin t$ has [L1, algebra] $g^{''} (t) = \sin t \geq 0$ , so makes $g$ convex there. A convex graph lies below the chord joining its endpoint values, hence $- \sin t \leq - 2 t / π$ and therefore $\sin t \geq \frac{2 t}{π} (0 \leq t \leq \frac{π}{2}) .$ By symmetry, also $\sin t \geq \frac{2 (π - t)}{π} (\frac{π}{2} \leq t \leq π) .$

2.1step 1.1given

Put $M_{T} : = \sup_{0 \leq t \leq π} ∣ γ_{T}^{+} (t) R (γ_{T}^{+} (t)) ∣ .$ Then along the arc $∣ e^{i λ z} R (z) d z ∣ \leq e^{- λ T \sin t} \frac{M_{T}}{T} T d t = M_{T} e^{- λ T \sin t} d t .$ So $∣ \int_{γ_{T}^{+}} e^{i λ z} R (z) d z ∣ \leq M_{T} \int_{0}^{π} e^{- λ T \sin t} d t .$

3.1step 2.1step 1.2algebra

Splitting the integral at $π / 2$ and using step 1.2 gives $\int_{0}^{π} e^{- λ T \sin t} d t \leq 2 \int_{0}^{π / 2} e^{- 2 λ T t / π} d t = \frac{π}{λ T} (1 - e^{- λ T}) \leq \frac{π}{λ T} .$ Hence $∣ \int_{γ_{T}^{+}} e^{i λ z} R (z) d z ∣ \leq \frac{π M_{T}}{λ T} .$

4.1step 3.1∎

Since $M_{T} \to 0$ , the bound in step 3.1 tends to $0$ . Therefore the arc integral tends to $0$ .

Depends on

Used by

Rational Fourier integrals are evaluated by residues and Jordan's lemma Theorem

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

E. Stein and R. Shakarchi, Complex Analysis, Ch. 3 §2.1 (standard reference, not scraped)