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

Statement

Let $λ \neq 0$ and let $R$ be a rational function such that $z R (z) \to 0$ as $∣ z ∣ \to \infty$ , with at most simple poles on the real axis. Then the oscillatory integral

$\int_{- \infty}^{\infty} e^{i λ x} R (x) d x$

is evaluated by closing in the upper half-plane when $λ > 0$ and in the lower half-plane when $λ < 0$ . More precisely:

if $λ > 0$ and $R$ has no real poles, then $\int_{- \infty}^{\infty} e^{i λ x} R (x) d x = 2 π i \sum_{ℑ a > 0} Res (e^{i λ z} R (z), a);$
if $λ < 0$ and $R$ has no real poles, then $\int_{- \infty}^{\infty} e^{i λ x} R (x) d x = - 2 π i \sum_{ℑ a < 0} Res (e^{i λ z} R (z), a);$
if the real poles $a_{j}$ are simple, then $PV ⁣ \int_{- \infty}^{\infty} e^{i λ x} R (x) d x = 2 π i \sum_{ℑ a > 0} Res (e^{i λ z} R (z), a) + i π \sum_{j} Res (e^{i λ z} R (z), a_{j})$ when $λ > 0$ , while for $λ < 0$ it equals $- 2 π i \sum_{ℑ a < 0} Res (e^{i λ z} R (z), a) - i π \sum_{j} Res (e^{i λ z} R (z), a_{j}) .$

Facts & Assumptions

Given: A nonzero real $λ$ and a rational function $R$ with $z R (z) \to 0$ at infinity and at most simple poles on the real axis.

[L1]

For $λ > 0$ , Jordan's lemma kills the upper large semicircle for $e^{i λ z} R (z)$ , and after replacing $z$ by $\overset{ˉ}{z}$ it kills the lower large semicircle for $λ < 0$ as well (Jordan's lemma for rational functions of one complex variable).

[L2]

The residue theorem evaluates the closed contour integral by the enclosed residues (The residue theorem for a null-homologous cycle).

[L3]

Indentation arcs around simple real poles contribute the signed half-residue terms (An indented arc around a simple singularity contributes the expected residue fraction).

[L4]

The real-axis indentation formulas compute principal values, not automatic improper convergence (This page keeps Cauchy principal values distinct from genuine improper convergence).

Proof

technique · cases

1.1

Assume first that $λ > 0$ and that $R$ has no real pole. Apply [L2] to [assume-case positive, L1, L2] the contour formed by $[- T, T]$ and the upper semicircle. The arc term tends to $0$ by, so the real-line integral is the sum of the residues of $e^{i λ z} R (z)$ in the upper half-plane.

1.2

If $λ < 0$ and $R$ has no real pole, close instead by the lower semicircle. The same computation gives a minus sign because the positively oriented contour now traverses the real segment from $T$ back to $- T$ , so the real integral equals $- 2 π i$ times the sum of the residues in the lower half-plane.

assume-case negativeL1L2

1.3

If $R$ has simple real poles and $λ > 0$ , indent them above the axis. Each indentation excludes its pole and contributes $- i π$ times its residue by [L3]; the residue theorem therefore gives the first displayed principal-value formula. If $λ < 0$ , use lower indentations and the lower semicircle. Each indentation contributes $+ i π$ times its residue, while the clockwise outer contour contributes $- 2 π i$ times the lower-half-plane residue sum, giving the second formula.

assume-case realpolesL1L2L3L4

2.1

Steps 1.1, 1.2, and 1.3 prove all cases listed in the statement.

step 1.1step 1.2step 1.3cases-exhaustive∎

Depends on

Used by

The integral of cos x / (1 + x²) over the real line is pi / e Example

Dependency tree · two levels

18 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)
R. Howell and J. Mathews, Complex Analysis, Ch. 8 §8.4 (standard reference, not scraped)