# Jordan's Lemma — Closing the Contour at Infinity

*The estimate that lets the large semicircular arc contribute nothing to a K–K or Fourier-type contour integral, even when the integrand itself does not decay.*

## The problem it solves

To apply Cauchy's theorem to an integral along the real axis, we close
the contour through a large semicircle of radius R and take R → ∞. For
this to give a useful identity, the semicircle's contribution must
**vanish** in the limit. A naive application of the ML inequality
(Chapter 5) asks: does `|f(z)| · πR → 0` as R → ∞? That requires
|f(z)| to fall faster than 1/R on the arc — too strong for many
physically interesting integrands.

The K–K integrand `χ(ω′)/(ω′ − ω)` does not have this decay in general.
What we need is an **improved** estimate that exploits an oscillatory
factor of the form `e^{iaz}` with `a > 0` when the contour is in the
upper half-plane: the exponential's own decay off the real axis
dramatically accelerates the death of the arc contribution. That is
exactly what Jordan's lemma provides.

## Statement

> **Jordan's lemma.** Let `C_R` be the upper semicircle
> `{z = R e^{iθ} : 0 ≤ θ ≤ π}`, traversed counter-clockwise. Suppose f
> is continuous on `C_R` for all sufficiently large R, and
>
> ```
> M_R  :=  max_{0 ≤ θ ≤ π}  | f(R e^{iθ}) |   →  0    as  R → ∞
> ```
>
> Then for any constant `a > 0`,
>
> ```
> | ∫_{C_R}  e^{iaz}  f(z)  dz |   →   0    as  R → ∞
> ```
>
> More precisely, the integral is bounded by `M_R · π/a`, so any decay
> of M_R (even logarithmic) is enough.

For `a < 0`, the analogous statement holds on the **lower** semicircle;
for `a = 0`, there is no oscillatory factor and Jordan gives nothing
beyond ML.

Corresponding variants exist for semicircles on the left or right (with
`e^{az}` oscillation pointing leftward or rightward). They all have the
same proof with reshuffled inequalities.

## Proof

Parametrise `z = R e^{iθ} = R cos θ + iR sin θ`, `dz = iR e^{iθ} dθ`.
Then

```
|e^{iaz}|  =  |e^{ia R cos θ}| · |e^{−a R sin θ}|  =  e^{−a R sin θ}
```

because the first factor has unit modulus. So

```
| ∫_{C_R} e^{iaz} f(z) dz |
   ≤  ∫_0^{π}  e^{−a R sin θ}  |f(R e^{iθ})|  · R  dθ
   ≤  M_R R  ∫_0^{π}  e^{−a R sin θ}  dθ
```

The integrand is symmetric about θ = π/2, so

```
∫_0^{π}  e^{−a R sin θ}  dθ   =  2 ∫_0^{π/2}  e^{−a R sin θ}  dθ
```

Now use the elementary **Jordan inequality**

```
sin θ   ≥   2θ/π       for  0 ≤ θ ≤ π/2
```

(the chord from (0, 0) to (π/2, 1) lies below `sin θ` on [0, π/2]). The
Jordan inequality is key: it replaces `sin θ` by a linear function so
that `e^{−a R sin θ}` is replaced by a plain decaying exponential that
integrates exactly. With this,

```
2 ∫_0^{π/2}  e^{−a R sin θ}  dθ
   ≤  2 ∫_0^{π/2}  e^{−2 a R θ / π}  dθ
   =  2 · (π/(2 a R))  [ 1 − e^{−a R} ]
   <  π / (a R)
```

Substituting back:

```
| ∫_{C_R} e^{iaz} f(z) dz |  ≤  M_R R · π/(a R)  =  π M_R / a
```

The factor of R from `dz` cancelled against the 1/R inside the
arclength-weighted exponential estimate. Since `M_R → 0`, the entire
bound tends to zero. QED.

## Why the Jordan inequality works: geometric picture

The curve `sin θ` on `[0, π/2]` is concave — it sits **above** its
chord. The chord connects `(0, 0)` to `(π/2, 1)` and has equation
`y = 2θ/π`. So `sin θ ≥ 2θ/π` for `θ ∈ [0, π/2]`, with equality only at
the endpoints. Exponentiating with a negative coefficient flips the
inequality to the direction we want: `e^{−a R sin θ} ≤ e^{−2 a R θ / π}`.

This is a purely geometric observation — no calculus needed beyond
concavity of sine — yet it is precisely the step that gives Jordan its
power over ML. Without it, one would have to work with
`∫ e^{−a R sin θ} dθ` directly, which has no elementary closed form.

## Reformulation for Fourier-type integrals

Most uses of Jordan's lemma look like this. We want to evaluate a real
integral

```
I  =  ∫_{−∞}^{∞}  e^{iax}  g(x)  dx    with  a > 0
```

by contour methods. Extend the integrand to complex z and close the
contour through the upper semicircle. Jordan's lemma says the arc
vanishes provided `|g(z)| → 0` uniformly on the arc as R → ∞ — note:
**g need not decay faster than 1/R**. Even g ~ 1/log R would suffice.
Then

```
∫_{−∞}^{∞}  e^{iax}  g(x)  dx   =   2πi Σ_{UHP poles} Res [e^{iaz} g(z)]
```

For `a < 0`, close in the **lower** half-plane and pick up −2πi times
the residues there.

**Example.** Evaluate `I = ∫_{−∞}^{∞} e^{iax} / (x² + b²) dx` for
`a, b > 0`. The function `g(z) = 1/(z² + b²)` decays as 1/R², so Jordan
(and even plain ML) kills the UHP arc. The only UHP pole is z = ib,
simple, with residue

```
Res(e^{iaz}/(z² + b²);  ib)  =  e^{ia · ib} / (2ib)  =  e^{−ab} / (2ib)
```

So `I = 2πi · e^{−ab}/(2ib) = (π/b) e^{−ab}`.

## Why this is the keystone for K–K

The K–K contour consists of the real axis (indented around ω′ = ω) and
the large UHP semicircle. Cauchy's theorem gives a relation between
their sum and (zero, since there are no enclosed poles). For the
relation to reduce to one involving only real-axis integrals, the
large semicircle must vanish as R → ∞. Three flavours of argument
appear in the literature, each a specialisation of Jordan:

**1. χ(ω) itself decays at infinity.** Many physical response
functions satisfy `|χ(ω)| → 0` as `|ω| → ∞` — e.g., dielectric
susceptibility, which goes to zero at high frequency because bound
charges cannot follow. Then `χ(ω′)/(ω′ − ω)` decays like (decay of
χ)/R on the arc. If χ decays faster than any power, ML alone
handles it. If χ decays only as 1/ω, the 1/(ω′ − ω) contributes
another 1/R, giving overall 1/R² · πR = π/R → 0. ML wins.

**2. χ does not decay, but the integrand carries an extra factor.**
If we multiply χ by an `e^{iaω′}` factor — the situation in the
**time-domain** / causality derivation, where the inverse Fourier
transform is

```
h(t) = (1/2π) ∫ χ(ω) e^{−iωt} dω
```

and we want to argue h(t) = 0 for t < 0 by closing the ω-contour in
the UHP — Jordan's lemma applies directly with `a = −t > 0` on the
UHP arc, and it tolerates χ bounded but not decaying.

**3. Subtracted dispersion relations.** When χ grows at infinity
(e.g., permittivity that diverges logarithmically, or an amplitude
that grows polynomially), one applies K–K to `(χ(ω) − χ(ω_sub))/(ω
− ω_sub)` or higher-order subtractions that force decay. Jordan then
applies to the subtracted integrand. Chapter 12 carries out this
construction.

In every case, Jordan's lemma is what converts "analytic in the UHP
plus mild asymptotic control" into "the real-axis integral equals the
residues enclosed". It is the last ingredient before the K–K theorem
itself.

## See also

- [05_contour_integration.md](05_contour_integration.md)
- [06_cauchy_theorems.md](06_cauchy_theorems.md)
- [07_residues.md](07_residues.md)
- [08_principal_value.md](08_principal_value.md)