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The integral of $d z / (z - p)$ along a contour is the increment of a continuous logarithm

Statement

Let $γ : [a, b] \to C$ be a complex contour, let $p \in C$ with $p \notin γ^{*}$ , and let $λ$ be a continuous logarithm of $γ - p$ along $γ$ (Continuous logarithms and continuous arguments along a contour). Then

$\int_{γ} \frac{d z}{z - p} = λ (b) - λ (a) .$

The contour need not be closed, and the right-hand side is the same for every continuous logarithm of $γ - p$ along $γ$ .

Facts & Assumptions

Given: A complex contour $γ : [a, b] \to C$ , a point $p \notin γ^{*}$ , and a continuous logarithm $λ$ of $γ - p$ along $γ$ .

[L1]

A continuous logarithm of $γ - p$ along $γ$ is a continuous $λ : [a, b] \to C$ with $\exp (λ (t)) = γ (t) - p$ for every $t$ (Continuous logarithms and continuous arguments along a contour).

[L2]

For a complex contour $γ$ and $p \notin γ^{*}$ there is a continuous logarithm of $γ - p$ along $γ$ , and any two of them differ by a constant lying in $2 π i Z$ (Every contour missing a point admits a continuous logarithm, unique up to a constant in $2 π i Z$ ).

[L3]

For a complex contour $γ$ and $p \notin γ^{*}$ , the distance $d = \inf {∣ w - p ∣ : w \in γ^{*}}$ is positive and some partition $a = t_{0} < \dots < t_{r} = b$ satisfies $γ ([t_{i}, t_{i + 1}]) \subseteq D (γ (t_{i}), d)$ and $p \notin D (γ (t_{i}), d)$ for every $i < r$ (A contour missing a point subdivides into arcs lying in discs that miss it).

[L4]

If $D (c, ρ)$ is an open disc with $ρ > 0$ and $p \notin D (c, ρ)$ , there is a holomorphic $L$ on $D (c, ρ)$ with $\exp (L (z)) = z - p$ there, and every such $L$ satisfies $L^{'} (z) = 1 / (z - p)$ (A disc missing $p$ carries a holomorphic logarithm of $z - p$ ).

[L5]

If $F$ is a primitive of a continuous $f$ on an open set containing the trace of a rectifiable contour $γ : [a, b] \to C$ and $F^{'} = f$ is continuous, then $\int_{γ} f (z) d z = F (γ (b)) - F (γ (a))$ (The line integral of a continuous function admitting a primitive is that primitive's endpoint increment along every rectifiable path, A primitive of a complex function on an open set).

[L6]

If $ϕ : [c, d] \to [a, b]$ is a strictly increasing continuous bijection and $f$ is continuous on the trace of the rectifiable $γ$ , then $\int_{γ \circ ϕ} f d z = \int_{γ} f d z$ (Complex and absolute line integrals are invariant under increasing continuous reparametrization).

[L7]

For composable rectifiable contours $α, β$ , $\int_{α * β} f d z = \int_{α} f d z + \int_{β} f d z$ (Complex line integrals change sign under reversal and add under concatenation); concatenation of $α, β : [0, 1] \to C$ with $α (1) = β (0)$ is $(α * β) (s) = α (2 s)$ for $s \leq \frac{1}{2}$ and $β (2 s - 1)$ for $s \geq \frac{1}{2}$ (Rectifiable complex contours, reversal, concatenation, closedness, and orientation, Reversal, concatenation, closed paths, and oriented piecewise-C1 reparametrizations).

[L8]

For a rectifiable $γ$ and $f$ continuous on its trace, $\int_{γ} f d z$ exists (Continuous integrands have complex and absolute line integrals along every rectifiable path).

[L9]

Arc length is additive across a split of the parameter interval, and $γ$ is rectifiable exactly when both restrictions are (Arc length is additive across every subdivision point and decreases under restriction).

[L10]

$\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

[L11]

The continuous image of a connected subset is connected (A continuous image of a connected space is connected, and connectedness is a topological property), and a connected subset of $R$ is order-convex (The connected subspaces of $R$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $R$ ").

[L12]

If a property holds at $0$ and passes from $n$ to $n + 1$ , it holds for every natural number (The principle of mathematical induction).

[L14]

A composite of continuous maps is continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous), and a function complex differentiable at a point is continuous there (Complex differentiability at a point implies continuity there).

[L15]

For $z = a + b i$ with $a, b$ real, $Im z = b$ and $∣ z ∣ = \sqrt{a^{2} + b^{2}}$ (Real and imaginary parts, complex conjugation, and modulus).

[L16]

The integers form an ordered commutative ring, and their canonical image in $R$ is discrete; hence if $m < n$ then $m + \frac{1}{2}$ lies strictly between them and is not an integer (The integers form a commutative ring, The integers form a totally ordered ring, Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ).

[L17]

Nonvanishing quotients of functions complex differentiable at a point are complex differentiable there (Linearity, product, reciprocal, and quotient rules for complex derivatives).

Proof

technique · direct

1.1givenL8L14L17

The function $z \mapsto 1 / (z - p)$ is defined and continuous on $γ^{*}$ by [L14] and [L17], since $p \notin γ^{*}$ , so the integral $\int_{γ} d z / (z - p)$ exists by [L8].

1.2L2

By [L2] any two continuous logarithms of $γ - p$ along $γ$ differ by a constant, so the increment $λ (b) - λ (a)$ is the same for all of them.

1.3givenL3L4L9

Assume $a < b$ . By [L3] fix $d > 0$ and a partition $a = t_{0} < \dots < t_{r} = b$ with $γ ([t_{i}, t_{i + 1}]) \subseteq D_{i} : = D (γ (t_{i}), d)$ and $p \notin D_{i}$ for $i < r$ , and by [L4] fix a holomorphic $L_{i}$ on $D_{i}$ with $\exp (L_{i} (z)) = z - p$ and $L_{i}^{'} (z) = 1 / (z - p)$ there. By [L9] each restriction $γ_{i} : = γ ∣_{[t_{i}, t_{i + 1}]}$ is rectifiable.

2.1givenstep 1.3L1L10L11L14L15L16

Fix $i < r$ . For $t \in [t_{i}, t_{i + 1}]$ both $\exp (λ (t))$ and $\exp (L_{i} (γ (t)))$ equal $γ (t) - p$ , so $λ (t) - L_{i} (γ (t)) \in 2 π i Z$ by [L10]; that difference is continuous by [L14], its scaled imaginary part is a continuous integer-valued real function by [L15], and [L11] with [L16] forces it to be constant on the interval. Hence $λ (t_{i + 1}) - λ (t_{i}) = L_{i} (γ (t_{i + 1})) - L_{i} (γ (t_{i}))$ .

2.2step 1.3L4L5L14L17

Fix $i < r$ . The trace of $γ_{i}$ lies in the open disc $D_{i}$ , on which $L_{i}$ is a primitive of the continuous function $1 / (z - p)$ , so [L5] gives $\int_{γ_{i}} d z / (z - p) = L_{i} (γ (t_{i + 1})) - L_{i} (γ (t_{i}))$ .

2.3step 1.1step 1.3L6L7L12

For $a \leq u < v < w \leq b$ the increasing affine reparametrisations $α (s) = γ (u + s (v - u))$ and $β (s) = γ (v + s (w - v))$ of $[0, 1]$ satisfy $α (1) = β (0)$ and $α * β = γ ∣_{[u, w]} \circ ϕ$ for the strictly increasing continuous bijection $ϕ : [0, 1] \to [u, w]$ that is affine on $[0, \frac{1}{2}]$ and on $[\frac{1}{2}, 1]$ with $ϕ (\frac{1}{2}) = v$ , so [L6] and [L7] split the integral at $v$ ; applying this at $t_{1}$ , then to $γ ∣_{[t_{1}, b]}$ at $t_{2}$ , and so on, an induction on the number of partition points ([L12]) gives $\int_{γ} d z / (z - p) = \sum_{i < r} \int_{γ_{i}} d z / (z - p)$ .

3.1step 2.1step 2.2step 2.3algebra

Substituting step 2.2 into step 2.3 and then step 2.1, the integral equals $\sum_{i < r} (λ (t_{i + 1}) - λ (t_{i}))$ . Expanding this finite sum, every intermediate value $λ (t_{i})$ with $0 < i < r$ appears once with sign $+$ and once with sign $-$ , so the sum telescopes to $λ (t_{r}) - λ (t_{0}) = λ (b) - λ (a)$ .

4.1step 1.2L4L5∎

If instead $a = b$ , choose $ρ > 0$ with $p \notin D (γ (a), ρ)$ ; [L4] gives a holomorphic $L$ on that disc with $L^{'} (z) = 1 / (z - p)$ . The trace of the constant contour $γ$ lies in that disc, so [L5] gives $\int_{γ} d z / (z - p) = L (γ (a)) - L (γ (a)) = 0$ , while $λ (b) - λ (a) = 0$ . Thus the identity also holds when $a = b$ ; and by step 1.2 the value asserted is independent of which continuous logarithm is used.

The integral of $d z / (z - p)$ along a contour is the increment of a continuous logarithm

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The integral of dz/(z−p)dz/(z−p) along a contour is the increment of a continuous logarithm

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The integral of $d z / (z - p)$ along a contour is the increment of a continuous logarithm