nLab
Cauchy principal value

Context

Integration theory

Functional analysis

Contents

Idea

The Cauchy principal value of a function which is integrable on the complement of one point is, if it exists, the limit of the integrals of the function over subsets in the complement of this point as these integration domains tend to that point symmetrically from all sides.

One also subsumes the case that the “point” is “at infinity”, hence that the function is integrable over every bounded domain. In this case the Cauchy principal value is the limit, if it exists, of the integrals of the function over bounded domains, as their bounds tend symmetrically to infinity.

The operation of sending a compactly supported smooth function (bump function) to Cauchy principal value of its pointwise product with a function ff that may be singular at the origin defines a distribution, usually denoted PV(f)PV(f).

When the Cauchy principal value exists but the full integral does not (hence when the full integral “diverges”) one may think of the Cauchy principal value as “exracting a finite value from a diverging quantity”. This is similar to the intuition of the early days of renormalization in perturbative quantum field theory (Schwinger-Tomonaga-Feynman-Dyson), but one has to be careful not to carry this analogy too far.

One point where the Cauchy principal value really does play a key role in perturbative quantum field theory is in the computation of Green functions (propagators) for the Klein-Gordon operator and the Dirac operator. See remark below and see at Feynman propagator for more on this.

Definition

As an integral

Definition

(Cauchy principal value of an integral over the real line)

Let f:f \colon \mathbb{R} \to \mathbb{R} be a function on the real line such that for every positive real number ϵ\epsilon its restriction to (ϵ,ϵ)\mathbb{R}\setminus (-\epsilon, \epsilon) is integrable. Then the Cauchy principal value of ff is, if it exists, the limit

PV(f)limϵ0(ϵ,ϵ)f(x)dx. PV(f) \coloneqq \underset{\epsilon \to 0}{\lim} \underset{\mathbb{R} \setminus (-\epsilon, \epsilon)}{\int} f(x) \, d x \,.

As a distribution

Definition

(Cauchy principal value as distribution on the real line)

Let f:f \colon \mathbb{R} \to \mathbb{R} be a function on the real line such that for all bump functions bC cp ()b \in C^\infty_{cp}(\mathbb{R}) the Cauchy principal value of the pointwise product function fbf b exists, in the sense of def. . Then this assignment

PV(f):bPV(fb) PV(f) \;\colon\; b \mapsto PV(f b)

defines a distribution PV(f)𝒟()PV(f) \in \mathcal{D}'(\mathbb{R}).

Examples

The principal value of 1/x1/x

Example

Let f:f \colon \mathbb{R} \to \mathbb{R} be an integrable function which is symmetric, in that f(x)=f(x)f(-x) = f(x) for all xx \in \mathbb{R}. Then the principal value integral (def. ) of xf(x)xx \mapsto \frac{f(x)}{x} exists and is zero:

limϵ0(ϵ,ϵ)f(x)xdx=0 \underset{\epsilon \to 0}{\lim} \underset{\mathbb{R}\setminus (-\epsilon, \epsilon)}{\int} \frac{f(x)}{x} d x \; = \; 0

This is because, by the symmetry of ff and the skew-symmetry of x1/xx \mapsto 1/x, the the two contributions to the integral are equal up to a sign:

ϵf(x)xdx= ϵ f(x)xdx. \int_{-\infty}^{-\epsilon} \frac{f(x)}{x} d x \;=\; - \int_{\epsilon}^\infty \frac{f(x)}{x} d x \,.
Example

The principal value distribution PV(1x)PV\left( \frac{1}{x}\right) (def. ) solves the distributional equation

(1)xPV(1x)=1AAA𝒟( 1). x PV\left(\frac{1}{x}\right) = 1 \phantom{AAA} \in \mathcal{D}'(\mathbb{R}^1) \,.

Since the delta distribution δ𝒟( 1)\delta \in \mathcal{D}'(\mathbb{R}^1) solves the equation

xδ(x)=0AAA𝒟(𝕣 1) x \delta(x) = 0 \phantom{AAA} \in \mathcal{D}'(\mathbb{r}^1)

we have that more generally every linear combination of the form

(2)F(x)PV(1/x)+cδ(x)AAA𝒟( 1) F(x) \coloneqq PV(1/x) + c \delta(x) \phantom{AAA} \in \mathcal{D}'(\mathbb{R}^1)

for cc \in \mathbb{C}, is a distributional solution to xF(x)=1x F(x) = 1.

The wave front set of all these solutions is

WF(PV(1/x)+cδ(x))={(0,k)|k *{0}}. WF\left( PV(1/x) + c \delta(x) \right) \;=\; \left\{ (0,k) \;\vert\; k \in \mathbb{R}^\ast \setminus \{0\} \right\} \,.
Proof

The first statement is immediate from the definition: For bC c ( 1)b \in C^\infty_c(\mathbb{R}^1) any bump function we have that

xPV(1x),b limϵ0 1(ϵ,ϵ)xxb(x)dx =b(x)dx =1,b \begin{aligned} \left\langle x PV\left(\frac{1}{x}\right), b \right\rangle & \coloneqq \underset{\epsilon \to 0}{\lim} \underset{\mathbb{R}^1 \setminus (-\epsilon, \epsilon)}{\int} \frac{x}{x}b(x) \, d x \\ & = \int b(x) d x \\ & = \langle 1,b\rangle \end{aligned}

Regarding the second statement: It is clear that the wave front set is concentrated at the origin. By symmetry of the distribution around the origin, it must contain both directions.

Proposition

In fact (2) is the most general distributional solution to (1).

This follows by the characterization of extension of distributions to a point, see there at this prop. (Hörmander 90, thm. 3.2.4)

Definition

(integration against inverse variable with imaginary offset)

Write

1x+i0 ±𝒟() \tfrac{1}{x + i0^\pm} \;\in\; \mathcal{D}'(\mathbb{R})

for the distribution which is the limit in 𝒟()\mathcal{D}'(\mathbb{R}) of the non-singular distributions which are given by the smooth functions x1x±iϵx \mapsto \tfrac{1}{x \pm i \epsilon} as the positive real number ϵ\epsilon tends to zero:

1x+i0 ±limϵ(0,)ϵ01x±iϵ \frac{1}{ x + i 0^\pm } \;\coloneqq\; \underset{ { \epsilon \in (0,\infty) } \atop { \epsilon \to 0 } }{\lim} \tfrac{1}{x \pm i \epsilon}

hence the distribution which sends bC ( 1)b \in C^\infty(\mathbb{R}^1) to

bb(x)x±iϵdx. b \mapsto \underset{\mathbb{R}}{\int} \frac{b(x)}{x \pm i \epsilon} \, d x \,.
Proposition

(Cauchy principal value equals integration with imaginary offset plus delta distribution)

The Cauchy principal value distribution PV(1x)𝒟()PV\left( \tfrac{1}{x}\right) \in \mathcal{D}'(\mathbb{R}) (def. ) is equal to the sum of the integration over 1/x1/x with imaginary offset (def. ) and a delta distribution.

PV(1x)=1x+i0 ±±iπδ. PV\left(\frac{1}{x}\right) \;=\; \frac{1}{x + i 0^\pm} \pm i \pi \delta \,.

In particular, by prop. this means that 1x+i0 ±\tfrac{1}{x + i 0^\pm} solves the distributional equation

x1x+i0 ±=1AA𝒟( 1). x \frac{1}{x + i 0^\pm} \;=\; 1 \phantom{AA} \in \mathcal{D}'(\mathbb{R}^1) \,.
Proof

Using that

1x±iϵ =xiϵ(x+iϵ)(xiϵ) =xiϵ(x 2+ϵ 2) \begin{aligned} \frac{1}{x \pm i \epsilon} & = \frac{ x \mp i \epsilon }{ (x + i \epsilon)(x - i \epsilon) } \\ & = \frac{ x \mp i \epsilon }{(x^2 + \epsilon^2)} \end{aligned}

we have for every bump function bC cp ( 1)b \in C^\infty_{cp}(\mathbb{R}^1)

limϵ0 1b(x)x±iϵdx =limϵ0 1x 2x 2+ϵ 2b(x)xdx(A)iπlimϵ0 11πϵx 2+ϵ 2b(x)dx(B) \begin{aligned} \underset{\epsilon \to 0}{\lim} \underset{\mathbb{R}^1}{\int} \frac{b(x)}{x \pm i \epsilon} d x & \;=\; \underset{ (A) }{ \underbrace{ \underset{\epsilon \to 0}{\lim} \underset{\mathbb{R}^1}{\int} \frac{x^2}{x^2 + \epsilon^2} \frac{b(x)}{x} d x } } \mp i \pi \underset{(B)}{ \underbrace{ \underset{\epsilon \to 0}{\lim} \underset{\mathbb{R}^1}{\int} \frac{1}{\pi} \frac{\epsilon}{x^2 + \epsilon^2} b(x) \, d x }} \end{aligned}

Since

x 2x 2+ϵ 2 |x|<ϵϵ0 |x|>ϵϵ0 0 1 \array{ && \frac{x^2}{x^2 + \epsilon^2} \\ & {}^{\mathllap{ { {\vert x \vert} \lt \epsilon } \atop { \epsilon \to 0 } }}\swarrow && \searrow^{\mathrlap{ {{\vert x\vert} \gt \epsilon} \atop { \epsilon \to 0 } }} \\ 0 && && 1 }

it is plausible that (A)=PV(b(x)x)(A) = PV\left( \frac{b(x)}{x} \right), and similarly that (B)=b(0)(B) = b(0). In detail:

(A) =limϵ0 1xx 2+ϵ 2b(x)dx =limϵ0 1ddx(12ln(x 2+ϵ 2))b(x)dx =12limϵ0 1ln(x 2+ϵ 2)dbdx(x)dx =12 1ln(x 2)dbdx(x)dx = 1ln(|x|)dbdx(x)dx =limϵ0 1(ϵ,ϵ)ln(|x|)dbdx(x)dx =limϵ0 1(ϵ,ϵ)1xb(x)dx =PV(b(x)x) \begin{aligned} (A) & = \underset{\epsilon \to 0}{\lim} \underset{\mathbb{R}^1}{\int} \frac{x}{x^2 + \epsilon^2} b(x) d x \\ & = \underset{\epsilon \to 0}{\lim} \underset{\mathbb{R}^1}{\int} \frac{d}{d x} \left( \tfrac{1}{2} \ln(x^2 + \epsilon^2) \right) b(x) d x \\ & = -\tfrac{1}{2} \underset{\epsilon \to 0}{\lim} \underset{\mathbb{R}^1}{\int} \ln(x^2 + \epsilon^2) \frac{d b}{d x}(x) d x \\ & = -\tfrac{1}{2} \underset{\mathbb{R}^1}{\int} \ln(x^2) \frac{d b}{d x}(x) d x \\ & = - \underset{\mathbb{R}^1}{\int} \ln({\vert x \vert}) \frac{d b}{d x}(x) d x \\ & = - \underset{\epsilon \to 0}{\lim} \underset{\mathbb{R}^1\setminus (-\epsilon, \epsilon)}{\int} \ln( {\vert x \vert} ) \frac{d b}{d x}(x) d x \\ & = \underset{\epsilon \to 0}{\lim} \underset{\mathbb{R}^1\setminus (-\epsilon, \epsilon)}{\int} \frac{1}{x} b(x) d x \\ & = PV\left( \frac{b(x)}{x} \right) \end{aligned}

and

(B) =1πlimϵ0 1ϵx 2+ϵ 2b(x)dx =1πlimϵ0 1(ddxarctan(xϵ))b(x)dx =1πlimϵ0 1arctan(xϵ)dbdx(x)dx =12 1sgn(x)dbdx(x)dx =b(0) \begin{aligned} (B) & = \tfrac{1}{\pi} \underset{\epsilon \to 0}{\lim} \underset{\mathbb{R}^1}{\int} \frac{\epsilon}{x^2 + \epsilon^2} b(x) \, d x \\ & = \tfrac{1}{\pi} \underset{\epsilon \to 0}{\lim} \underset{\mathbb{R}^1}{\int} \left( \frac{d}{d x} \arctan\left( \frac{x}{\epsilon} \right) \right) b(x) \, d x \\ & = - \tfrac{1}{\pi} \underset{\epsilon \to 0}{\lim} \underset{\mathbb{R}^1}{\int} \arctan\left( \frac{x}{\epsilon} \right) \frac{d b}{d x}(x) \, d x \\ & = - \frac{1}{2} \underset{\mathbb{R}^1}{\int} sgn(x) \frac{d b}{d x}(x) \, d x \\ & = b(0) \end{aligned}

where we used that the derivative of the arctan function is ddxarctan(x)=1/(1+x 2)\frac{d}{ d x} \arctan(x) = 1/(1 + x^2) and that limϵ+arctan(x/ϵ)=π2sgn(x)\underset{\epsilon \to + \infty}{\lim} \arctan(x/\epsilon) = \tfrac{\pi}{2}sgn(x) is proportional to the sign function.

Example

(Fourier integral formula for step function)

The Heaviside distribution Θ𝒟()\Theta \in \mathcal{D}'(\mathbb{R}) is equivalently the following Cauchy principal value:

Θ(x) =12πi e iωxωi0 + limϵ0 +12πi e iωxωiϵdω, \begin{aligned} \Theta(x) & = \frac{1}{2\pi i} \int_{-\infty}^\infty \frac{e^{i \omega x}}{\omega - i 0^+} \\ & \coloneqq \underset{ \epsilon \to 0^+}{\lim} \frac{1}{2 \pi i} \int_{-\infty}^\infty \frac{e^{i \omega x}}{\omega - i \epsilon} d\omega \,, \end{aligned}

where the limit is taken over sequences of positive real numbers ϵ(,0)\epsilon \in (-\infty,0) tending to zero.

Proof

We may think of the integrand e iωxωiϵ\frac{e^{i \omega x}}{\omega - i \epsilon} uniquely extended to a holomorphic function on the complex plane and consider computing the given real line integral for fixed ϵ\epsilon as a contour integral in the complex plane.

If x(0,)x \in (0,\infty) is positive, then the exponent

iωx=Im(ω)x+iRe(ω)x i \omega x = - Im(\omega) x + i Re(\omega) x

has negative real part for positive imaginary part of ω\omega. This means that the line integral equals the complex contour integral over a contour C +C_+ \subset \mathbb{C} closing in the upper half plane. Since iϵi \epsilon has positive imaginary part by construction, this contour does encircle the pole of the integrand e iωxωiϵ\frac{e^{i \omega x}}{\omega - i \epsilon} at ω=iϵ\omega = i \epsilon. Hence by the Cauchy integral formula in the case x>0x \gt 0 one gets

limϵ0 +12πi e iωxωiϵdω =limϵ0 +12πi C +e iωxωiϵdω =limϵ0 +(e iωx| ω=iϵ) =limϵ0 +e ϵx =e 0=1. \begin{aligned} \underset{\epsilon \to 0^+}{\lim} \frac{1}{2 \pi i} \int_{-\infty}^\infty \frac{e^{i \omega x}}{\omega - i \epsilon} d\omega & = \underset{\epsilon \to 0^+}{\lim} \frac{1}{2 \pi i} \oint_{C_+} \frac{e^{i \omega x}}{\omega - i \epsilon} d \omega \\ & = \underset{\epsilon \to 0^+}{\lim} \left(e^{i \omega x}\vert_{\omega = i \epsilon}\right) \\ & = \underset{\epsilon \to 0^+}{\lim} e^{- \epsilon x} \\ & = e^0 = 1 \end{aligned} \,.

Conversely, for x<0x \lt 0 the real part of the integrand decays as the negative imaginary part increases, and hence in this case the given line integral equals the contour integral for a contour C C_- \subset \mathbb{C} closing in the lower half plane. Since the integrand has no pole in the lower half plane, in this case the Cauchy integral formula says that this integral is zero.

Conversely, by the Fourier inversion theorem, the Fourier transform of the Heaviside distribution is the Cauchy principal value as in prop. :

Example

(relation to Fourier transform of Heaviside distribution / Schwinger parameterization)

Θ^(x) = 0 e ikxdk =i1x+i0 + \begin{aligned} \widehat \Theta(x) & = \int_0^\infty e^{i k x} \, dk \\ & = i \frac{1}{x + i 0^+} \end{aligned}

Here the second equality is also known as complex Schwinger parameterization.

Proof

As generalized functions consider the limit with a decaying component:

0 e ikxdk =limϵ0 + 0 e ikxϵkdk =limϵ0 +1ixϵ =i1x+i0 + \begin{aligned} \int_0^\infty e^{i k x} \, dk & = \underset{\epsilon \to 0^+}{\lim} \int_0^\infty e^{i k x - \epsilon k} \, dk \\ & = - \underset{\epsilon \to 0^+}{\lim} \frac{1}{ i x - \epsilon} \\ & = i \frac{1}{x + i 0^+} \end{aligned}

The principal value of 1/(q(x)+m 2)1/(q(x) + m^2)

Let q: nq \colon \mathbb{R}^{n} \to \mathbb{R} be a non-degenerate real quadratic form analytically continued to a real quadratic form

q: n. q \;\colon\; \mathbb{C}^n \longrightarrow \mathbb{C} \,.

Write Δ\Delta for the determinant of qq

Write q *q^\ast for the induced quadratic form on dual vector space. Notice that qq (and hence a *a^\ast) are assumed non-degenerate but need not necessarily be positive or negative definite.

Proposition

(Fourier transform of principal value of power of quadratic form)

Let mm \in \mathbb{R} be any real number, and κ\kappa \in \mathbb{C} any complex number. Then the Fourier transform of distributions of 1/(q+m 2+i0 +) κ1/(q + m^2 + i 0^+)^\kappa is

(1q+m 2+i0 +)^=2 1κ(2π) nm n/2κΓ(κ)ΔK n/2κ(mq *i0 +)(q *i0 +) n/2κ, \widehat { \left( \frac{1}{q + m^2 + i0^+} \right) } \;=\; \frac{ 2^{1- \kappa} (\sqrt{2\pi})^{n} m^{n/2-\kappa} } { \Gamma(\kappa) \sqrt{\Delta} } \frac{ K_{n/2 - \kappa}\left( m \sqrt{q^\ast - i 0^+} \right) } { \left(\sqrt{q^\ast - i0^+ }\right)^{n/2 - \kappa} } \,,

where

  1. Γ\Gamma deotes the Gamma function

  2. K νK_{\nu} denotes the modified Bessel function.

Notice that K ν(a)K_\nu(a) diverges for a0a \to 0 as a νa^{-\nu} (DLMF 10.30.2).

(Gel’fand-Shilov 66, III 2.8 (8) and (9), p 289)

Example

(Feynman propagator)

Let qη 1q \coloneqq \eta^{-1} be the dual Minkowski metric in dimension p+1p+1. Then

Δ F(x)1η 1(k,k)m 2+i0 +^ \Delta_F(x) \;\propto\; \widehat{ \frac{1}{ -\eta^{-1}(k,k) - m^2 + i0^+ } }

is the Feynman propagator for the Klein-Gordon equation on Minkowski spacetime. In this case prop. implies that its singular support is the light cone {x p,1|η(x,x)=0}\{x \in \mathbb{R}^{p,1} \;\vert\; \eta(x,x) = 0\}.

The Fourier transform of δ(q+m 2)\delta(q + m^2)

Let q: nq \colon \mathbb{R}^{n} \to \mathbb{R} be a non-degenerate real quadratic form analytically continued to a real quadratic form

q: n. q \;\colon\; \mathbb{C}^n \longrightarrow \mathbb{C} \,.

Write Δ\Delta for the determinant of qq. Write tt \in \mathbb{N} for the number of negative eigenvalues.

Write q *q^\ast for the induced quadratic form on dual vector space. Notice that qq (and hence a *a^\ast) are assumed non-degenerate but need not necessarily be positive or negative definite.

Proposition

(Fourier transform of delta distribution applied to mass shell)

Let mm \in \mathbb{R}, then the Fourier transform of distributions of the delta distribution δ\delta applied to the “mass shell” q+m 2q + m^2 is

δ(q+m 2)^=i|Δ|(e iπt/2K n/21(mq *+i0 +)(q *+i0 +) n/21e iπt/2K n/21(mq *i0 +)(q *i0 +) n/21), \widehat{ \delta(q + m^2) } \;=\; - \frac{i}{\sqrt{{\vert\Delta\vert}}} \left( e^{i \pi t /2 } \frac{ K_{n/2-1} \left( m \sqrt{ q^\ast + i0^+ } \right) }{ \left(\sqrt{q^\ast + i0^+}\right)^{n/2 - 1} } \;-\; e^{-i \pi t /2 } \frac{ K_{n/2-1} \left( m \sqrt{ q^\ast - i0^+ } \right) }{ \left(\sqrt{q^\ast - i0^+}\right)^{n/2 - 1} } \right) \,,

where K νK_\nu denotes the modified Bessel function of order ν\nu.

Notice that K ν(a)K_\nu(a) diverges for a0a \to 0 as a νa^{-\nu} (DLMF 10.30.2).

(Gel’fand-Shilov 66, III 2.11 (7), p 294)

Example

(causal propagator)

Let qη 1q \coloneqq \eta^{-1} be the dual Minkowski metric in dimension p+1p+1. Then

Δ S(x)δ(η(k,k)m 2)sgn(k 0)^ \Delta_S(x) \;\propto\; \widehat{ \delta(-\eta(k,k) - m^2) \, sgn(k_0) }

is the causal propagator for the Klein-Gordon equation on Minkowski spacetime. In this case prop. implies that its singular support is the light cone {x p,1|η(x,x)=0}\{x \in \mathbb{R}^{p,1} \;\vert\; \eta(x,x) = 0\}.

References

Named after Augustin Cauchy

  • Ram Kanwal, section 8.3 of Linear Integral Equations Birkhäuser 1997

Detailed discussion of relation to Bessel functions is in

  • I. M. Gel'fand, G. E. Shilov, Generalized functions, 1–5 , Acad. Press (1966–1968) transl. from И. М. Гельфанд, Г. Е. Шилов Обобщенные функции, вып. 1-3, М.:Физматгиз, 1958; 1: Обобщенные функции и действия над ними, 2: Пространства основных обобщенных функций, 3: Некоторые вопросы теории дифференциальных уравнений

References on homogeneous distributions

  • Lars Hörmander, The Analysis of Linear Partial Differential Operators I (Springer, 1990, 2nd ed.)

See also

Last revised on November 28, 2017 at 05:31:38. See the history of this page for a list of all contributions to it.