Residue calculus on rectangles
This files gathers definitions and basic properties about rectangles.
The border of a rectangle is the union of its four sides.
DefinitionRectangleBorder#
A Rectangle's border, given corners \(z\) and \(w\) is the union of the four sides.
Used by 4
DefinitionRectangleIntegral#
A RectangleIntegral of a function \(f\) is one over a rectangle determined by \(z\) and \(w\) in \(\mathbb{C}\). We will sometimes denote it by \(\int_{z}^{w} f\). (There is also a primed version, which is \(1/(2\pi i)\) times the original.)
Used by 16
DefinitionUpperUIntegral#
An UpperUIntegral of a function \(f\) comes from \(\sigma+i\infty\) down to \(\sigma+iT\), over to \(\sigma'+iT\), and back up to \(\sigma'+i\infty\).
Used by 2
DefinitionLowerUIntegral#
A LowerUIntegral of a function \(f\) comes from \(\sigma-i\infty\) up to \(\sigma-iT\), over to \(\sigma'-iT\), and back down to \(\sigma'-i\infty\).
Used by 2
It is very convenient to define integrals along vertical lines in the complex plane, as follows.
DefinitionVerticalIntegral#
Let \(f\) be a function from \(\mathbb{C}\) to \(\mathbb{C}\), and let \(\sigma\) be a real number. Then we define
We also have a version with a factor of \(1/(2\pi i)\).
LemmaDiffVertRect-eq-UpperLowerUs#
The difference of two vertical integrals and a rectangle is the difference of an upper and a lower U integrals.
Proof
Follows directly from the definitions.
Lean
Formalized as DiffVertRect_eq_UpperLowerUs.
Sources
Uses 4
TheoremexistsDifferentiableOn-of-bddAbove#
If \(f\) is differentiable on a set \(s\) except at \(c\in s\), and \(f\) is bounded above on \(s\setminus\{c\}\), then there exists a differentiable function \(g\) on \(s\) such that \(f\) and \(g\) agree on \(s\setminus\{c\}\).
Proof
This is the Riemann Removable Singularity Theorem, slightly rephrased from what's in Mathlib. (We don't care what the function \(g\) is, just that it's holomorphic.)
Lean
Formalized as existsDifferentiableOn_of_bddAbove.
Sources
Used by 5
TheoremHolomorphicOn.vanishesOnRectangle#
If \(f\) is holomorphic on a rectangle \(z\) and \(w\), then the integral of \(f\) over the rectangle with corners \(z\) and \(w\) is \(0\).
Proof
This is in a Mathlib PR.
Lean
Formalized as HolomorphicOn.vanishesOnRectangle.
Sources
Uses 1 · Used by 5
The next lemma allows to zoom a big rectangle down to a small square, centered at a pole.
LemmaRectanglePullToNhdOfPole#
If \(f\) is holomorphic on a rectangle \(z\) and \(w\) except at a point \(p\), then the integral of \(f\) over the rectangle with corners \(z\) and \(w\) is the same as the integral of \(f\) over a small square centered at \(p\).
Proof
Chop the big rectangle with two vertical cuts and two horizontal cuts into smaller rectangles, the middle one being the desired square. The integral over each of the outer rectangles vanishes, since \(f\) is holomorphic there. (The constant \(c\) being “small enough” here just means that the inner square is strictly contained in the big rectangle.)
Lean
Formalized as RectanglePullToNhdOfPole.
Sources
Uses 3
LemmaResidueTheoremAtOrigin#
The rectangle (square) integral of \(f(s) = 1/s\) with corners \(-1-i\) and \(1+i\) is equal to \(2\pi i\).
Proof
This is a special case of the more general result above.
Lean
Formalized as ResidueTheoremAtOrigin.
Sources
Uses 1
LemmaResidueTheoremOnRectangleWithSimplePole#
Suppose that \(f\) is a holomorphic function on a rectangle, except for a simple pole at \(p\). By the latter, we mean that there is a function \(g\) holomorphic on the rectangle such that, \(f = g + A/(s-p)\) for some \(A\in\mathbb{C}\). Then the integral of \(f\) over the rectangle is \(A\).
Proof
Replace \(f\) with \(g + A/(s-p)\) in the integral. The integral of \(g\) vanishes by Lemma HolomorphicOn.vanishesOnRectangle. To evaluate the integral of \(1/(s-p)\), pull everything to a square about the origin using Lemma RectanglePullToNhdOfPole, and rescale by \(c\); what remains is handled by Lemma ResidueTheoremAtOrigin.
Lean
Formalized as ResidueTheoremOnRectangleWithSimplePole.
Sources
Uses 3 · Used by 3
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