# Cauchy Principal Value and the Sokhotski–Plemelj Formula

*How to make sense of integrals with a non-integrable pole on the real axis, and the distributional identity that is the single most important technical step in the K–K derivation.*

## The problem

The K–K relations involve integrals of the form

```
∫_{−∞}^{∞}  χ(ω′) / (ω′ − ω)  dω′
```

with ω a real number. The integrand has a **non-integrable singularity**
at ω′ = ω: near the pole, the integrand behaves like `1/(ω′ − ω)`, whose
ordinary improper integral over a neighbourhood of ω does not converge
(it is like `∫ dx/x` near 0, which diverges logarithmically from both
sides).

However, the singularity is **odd** about ω′ = ω: `1/(ω′ − ω)` is
positive to the right, negative to the left, with matching magnitudes.
If we excise a symmetric interval `(ω − ε, ω + ε)` around the pole, the
two divergences **cancel** before integration, leaving a finite limit as
ε → 0. This symmetric excision defines the Cauchy principal value.

## Definition

The **Cauchy principal value** of a singular integral with pole at x₀
is

```
P ∫_{−∞}^{∞}  f(x) / (x − x₀)  dx
   :=  lim_{ε → 0⁺}  [ ∫_{−∞}^{x₀ − ε}  f(x)/(x − x₀) dx
                       + ∫_{x₀ + ε}^{∞}  f(x)/(x − x₀) dx ]
```

provided the limit exists. More generally, for an integrand with a
simple pole at x₀,

```
P ∫_a^b g(x) dx := lim_{ε → 0⁺} [ ∫_a^{x₀ − ε} g(x) dx + ∫_{x₀ + ε}^b g(x) dx ]
```

The principal value is a linear functional on the space of smooth test
functions — in distribution-theory terminology, the distribution
denoted `P(1/x)` or `pv(1/x)`.

## Why the symmetric limit matters

The two one-sided integrals individually diverge as ε → 0; only their
**sum** with a **common** ε converges. Using different cut-offs on the
two sides,

```
lim_{ε → 0}  [ ∫_{−∞}^{x₀ − ε}  +  ∫_{x₀ + κε}^{∞} ]  f/(x − x₀) dx
   =  P ∫  f/(x − x₀) dx  −  f(x₀) · ln κ
```

for any positive κ. The P value is the canonical choice κ = 1, and the
symmetric cancellation is what generates a finite number. This is why
people often emphasise "symmetric limit" when defining it.

## Small worked evaluation: P ∫ dx/(x² − 1)

Write `1/(x² − 1) = [1/(x − 1) − 1/(x + 1)] / 2`. Each term has a
simple real pole, so

```
P ∫_{−2}^{2}  dx/(x² − 1)
  = (1/2) P ∫_{−2}^{2} dx/(x − 1)  −  (1/2) P ∫_{−2}^{2} dx/(x + 1)
```

Each P integral evaluates by symmetric cancellation:

```
P ∫_{−2}^{2} dx/(x − 1)
 = lim_{ε → 0} [ ∫_{−2}^{1 − ε} dx/(x − 1) + ∫_{1 + ε}^{2} dx/(x − 1) ]
 = lim_{ε → 0} [ ln|x − 1| ]_{−2}^{1 − ε}  +  [ ln|x − 1| ]_{1 + ε}^{2}
 = lim_{ε → 0} [ ln ε − ln 3  +  ln 1 − ln ε ]
 = −ln 3
```

Similarly `P ∫_{−2}^{2} dx/(x + 1) = ln 3` (pole at −1, symmetric
interval has right-end distance 3, left-end distance 1 — but redo the
arithmetic: `[ln|x+1|]_{−2}^{−1−ε} + [ln|x+1|]_{−1+ε}^{2}
= ln ε − ln 1 + ln 3 − ln ε = ln 3`). So

```
P ∫_{−2}^{2}  dx/(x² − 1)  =  (1/2)(−ln 3 − ln 3)  =  −ln 3
```

The logarithmic divergences on the two sides of each pole cancel inside
the principal value, leaving a finite answer.

## The Sokhotski–Plemelj formula

Now the key identity that links the principal value to contour
integration. Consider the function `1/(x − x₀ ∓ iε)` for small ε > 0 —
i.e., shift the pole **off** the real axis, into the lower half-plane
(upper sign in `∓`) or upper half-plane (lower sign).

> **Sokhotski–Plemelj formula.** For a test function f on ℝ and a real
> point x₀,
>
> ```
> lim_{ε → 0⁺}  ∫  f(x) / (x − x₀ ∓ iε)  dx
>    =  P ∫  f(x) / (x − x₀)  dx  ±  iπ f(x₀)
> ```
>
> Equivalently, as a distributional identity,
>
> ```
> 1 / (x − x₀ ∓ iε)  =  P(1/(x − x₀))  ±  iπ δ(x − x₀)     as ε → 0⁺
> ```

The two signs correspond to approaching the real axis from **below**
(pole pushed up, `−iε`, upper sign) or from **above** (pole pushed
down, `+iε`, lower sign). Crossing the real axis costs a jump of
`2πi δ(x − x₀)` — the classic Plemelj jump.

### Heuristic derivation

Split into real and imaginary parts:

```
1/(x − x₀ − iε)  =  (x − x₀)/[(x − x₀)² + ε²]  +  iε/[(x − x₀)² + ε²]
```

**Real part.** As ε → 0, the function `(x − x₀)/[(x − x₀)² + ε²]` is
odd about x₀, has zeros inside a neighbourhood of radius ε, and
converges (in the distributional sense, pairing against smooth test
functions) to `P(1/(x − x₀))`. The `ε²` in the denominator is exactly
the regularisation that makes the symmetric-cancellation P value
emerge.

**Imaginary part.** The function `ε/[(x − x₀)² + ε²]` is a standard
Lorentzian / nascent delta function with total integral π, concentrated
in a window of width ε around x₀. As ε → 0, it converges to
`π δ(x − x₀)`.

Adding: `1/(x − x₀ − iε) → P(1/(x − x₀)) + iπ δ(x − x₀)`. Flipping the
sign of ε flips the sign of the δ term.

### Contour interpretation

Sokhotski–Plemelj can equivalently be read as "indenting a real-axis
pole contributes half a residue". If the K–K contour indent goes **over**
the pole (upper semicircle, clockwise, of radius ε), then by direct
parametrisation

```
∫_{indent}  χ(ω′)/(ω′ − ω)  dω′  →  −iπ · Res = −iπ χ(ω)    as ε → 0
```

The negative sign is the clockwise traversal; a counterclockwise indent
below the pole would give `+iπ χ(ω)`. The difference between the two,
`2πi χ(ω)`, is the full residue of the encircled pole — consistent with
the residue theorem (Chapter 7). Either indentation, combined with
Cauchy's theorem on the closed contour and Jordan's lemma on the large
arc, yields the K–K relations.

## Application to K–K: the key step spelled out

With χ analytic in the UHP and decaying at infinity, integrate
`χ(ω′)/(ω′ − ω)` around the closed contour of Chapter 6: real axis
indented at ω, plus large UHP semicircle. No enclosed poles, so
Cauchy's theorem gives zero total. The large semicircle vanishes
(Chapter 9). The indent gives `−iπ χ(ω)`. The rest of the real axis
gives the principal value. Assembling:

```
P ∫_{−∞}^{∞}  χ(ω′)/(ω′ − ω) dω′   −  iπ χ(ω)  =  0
⇒  χ(ω)  =  (1 / iπ)  P ∫_{−∞}^{∞}  χ(ω′)/(ω′ − ω) dω′
          =  −(i/π)  P ∫_{−∞}^{∞}  χ(ω′)/(ω′ − ω) dω′
```

Writing χ = χ′ + iχ″ and splitting the equation into real and imaginary
parts gives

```
χ′(ω)  =  (1/π)  P ∫_{−∞}^{∞}  χ″(ω′)/(ω′ − ω) dω′
χ″(ω)  = −(1/π)  P ∫_{−∞}^{∞}  χ′(ω′)/(ω′ − ω) dω′
```

which is the K–K pair in its unfolded form.

Every step above except the indentation is from earlier chapters. The
Sokhotski–Plemelj identity is the single new technical ingredient that
turns Cauchy's formula into the dispersion relation.

## See also

- [06_cauchy_theorems.md](06_cauchy_theorems.md)
- [07_residues.md](07_residues.md)
- [09_jordan_lemma.md](09_jordan_lemma.md)