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 $f$ that may be singular at the origin defines a distribution, usually denoted $PV(f)$.

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

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

As a distribution

Definition

(Cauchy principal value as distribution on the real line)

Let $f \colon \mathbb{R} \to \mathbb{R}$ be a function on the real line such that for all bump functions $b \in C^\infty_{cp}(\mathbb{R})$ the Cauchy principal value of the pointwise product function $f b$ exists, in the sense of def. 1. Then this assignment

$PV(f) \;\colon\; b \mapsto PV(f b)$

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

Examples

The principal value of $1/x$

Example

Let $f \colon \mathbb{R} \to \mathbb{R}$ be an integrable function which is symmetric, in that $f(-x) = f(x)$ for all $x \in \mathbb{R}$. Then the principal value integral (def. 1) of $x \mapsto \frac{f(x)}{x}$ exists and is zero:

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

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

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

The principal value distribution $PV\left( \frac{1}{x}\right)$ (def. 2) solves the distributional equation

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

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

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

we have that more generally every linear combination of the form

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

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

Proof

This is immediate from the definition: For $b \in C^\infty_c(\mathbb{R}^1)$ any bump function we have that

\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}
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

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

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

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

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

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

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

The Cauchy principal value distribution $PV\left( \tfrac{1}{x}\right) \in \mathcal{D}'(\mathbb{R})$ (def. 2) is equal to the sum of the integration over $1/x$ with imaginary offset (def. 3) and a delta distribution.

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

In particular, by prop. 2 this means that $\tfrac{1}{x + i 0^\pm}$ solves the distributional equation

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

Using that

\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 $b \in C^\infty_{cp}(\mathbb{R}^1)$

\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

$\array{ && \frac{x^2}{x^2 + \epsilon^2} \\ & {}^{\mathllap{ { {\vert x \vert} \lt \epsilon } \atop { \epsilon \to 0 } }}\swarrow && \searrow^{\mathrlap{ {{\vert x\vert} \gt \epsilon} \atop { \epsilon \to 0 } }} \\ 0 && && 1 }$

it follows (…) that $(A) = PV\left( \frac{b(x)}{x} \right)$.

Similarly $(B) = b(0)$. (…)

Example

(Fourier integral formula for step function)

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

\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 $\epsilon \in (-\infty,0)$ tending to zero.

Proof

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

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

$i \omega x = - Im(\omega) x + i Re(\omega) x$

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

\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 $x \lt 0$ 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 $C_- \subset \mathbb{C}$ 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.

Remark

(Feynman propagator)

The Fourier form of the step function in prop. \ref{FourierIntegralFormula} gives rise to the standard expression for the advanced propagator, retarded propagator and Feynman propagator used in perturbative quantum field theory. See at Feynman propagator for more.

References

Named after Augustin Cauchy

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

References on homogeneous distributions

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