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Boundary Values, Discontinuities, and Dispersion Integrals

A dispersion integral is Cauchy’s formula after a contour has been deformed onto the two banks of a cut. The deformation converts global analytic information into boundary data, but only after three questions have been answered: where is the function analytic, what happens on the large contour, and in what sense do the boundary values exist? If the large contour does not vanish, subtraction data remain.

Required background. Holomorphic Functions and Cauchy Theory supplies Cauchy’s formula, winding numbers, and the contour deformation used in the representation.

Helpful background. Branches, Sheets, Analytic Continuation, and Monodromy supplies cut and bank conventions; Test-Function Spaces, Distributions, Support, and Convergence supplies weak boundary values when pointwise limits fail.

This page develops the mathematics for one complex variable. The page does not derive amplitude domains, unitarity, positivity, sum rules, or high-energy bounds; those require additional QFT hypotheses.

Let

Ω=C[s0,),\Omega = \mathbb C\setminus[s_0,\infty),

and let FF be holomorphic on Ω\Omega. When the limits exist, define

F±(s)=limϵ0+F(s±iϵ),s>s0,F_\pm(s) = \lim_{\epsilon\to0^+}F(s\pm i\epsilon), \qquad s>s_0,

and fix the convention

DiscF(s)=F+(s)F(s).\operatorname{Disc}F(s) = F_+(s)-F_-(s).

These limits may exist pointwise, in an LpL^p sense, or as distributions. The mode is part of the theorem. A pointwise formula should not be inferred from distributional existence alone. The principal-value and (x±i0)1(x\pm i0)^{-1} distribution identities are stated in Dyatlov 2022, § 5.2.3 and Exercise 5.4(c), PDF.

If FF obeys Schwarz reflection,

F(z)=F(z),F(z^*)=F(z)^*,

then

F(s)=F+(s),DiscF(s)=2iImF+(s).F_-(s)=F_+(s)^*, \qquad \operatorname{Disc}F(s) = 2i\,\operatorname{Im}F_+(s).

Without reflection, the discontinuity and the imaginary part are different data. Some sources also reverse the sign of Disc\operatorname{Disc}, so this definition must travel with every formula.

Fix zΩz\in\Omega. Use finite positively oriented keyhole contours inside Ω\Omega: two banks approach the cut from heights ±ϵ\pm\epsilon, a small arc avoids the endpoint s0s_0, and an outer part ΓRoutΩ\Gamma_R^{\mathrm{out}}\subset\Omega closes the contour at scale RR. Choose each finite contour to have winding number 11 about zz. Assume for the moment that:

  1. FF has no poles in Ω\Omega;

  2. the boundary values F±(s)F_\pm(s) exist for almost every s>s0s>s_0, and for some ϵ0>0\epsilon_0>0 there is a function gzL1((s0,))g_z\in L^1((s_0,\infty)) such that

    F(s±iϵ)s±iϵzgz(s),0<ϵ<ϵ0;\left| \frac{F(s\pm i\epsilon)} {s\pm i\epsilon-z} \right| \leq g_z(s), \qquad 0<\epsilon<\epsilon_0;
  3. the outer parts have length O(R)O(R), stay a distance comparable to RR from zz, and obey

    supζΓRoutF(ζ)0(R);\sup_{\zeta\in\Gamma_R^{\mathrm{out}}} |F(\zeta)| \longrightarrow0 \qquad (R\to\infty);
  4. the small endpoint arc contributes zero as its radius tends to zero.

The second condition permits dominated convergence from the finite banks to the boundary integrals. The third makes the outer contribution vanish by the estimation lemma. Cauchy’s formula on each finite keyhole contour gives the cut representation; compare Conway 1978, Chapters IV and IX:

F(z)=12πiF(ζ)ζzdζ.F(z) = \frac{1}{2\pi i} \oint \frac{F(\zeta)}{\zeta-z} \,\mathrm d\zeta.

On the upper bank, the positively oriented boundary of the slit runs from s0s_0 toward ++\infty; on the lower bank it runs back toward s0s_0. After the outer and endpoint circles vanish, the two bank integrals combine:

F(z)=12πis0F+(s)F(s)szds.F(z) = \frac{1}{2\pi i} \int_{s_0}^{\infty} \frac{ F_+(s')-F_-(s') } {s'-z} \,\mathrm ds'.

Thus

F(z)=12πis0DiscF(s)szds.\boxed{ F(z) = \frac{1}{2\pi i} \int_{s_0}^{\infty} \frac{\operatorname{Disc}F(s')} {s'-z} \,\mathrm ds' }.

The box is not a formula for every function with a cut. It is the conclusion of the four assumptions above. In particular, knowledge of a discontinuity does not erase isolated poles or a nonvanishing contribution from infinity.

The boundary-value mechanism can be checked in the opposite direction. Let DD be smooth with compact support in (s0,)(s_0,\infty) and define

CD(z)=12πis0D(s)szds.\mathcal C_D(z) = \frac{1}{2\pi i} \int_{s_0}^{\infty} \frac{D(s')}{s'-z} \,\mathrm ds'.

The distributional identities

1xi0=p.v.1x±iπδ(x)\frac{1}{x\mp i0} = \operatorname{p.v.}\frac1x \pm i\pi\delta(x)

give

(CD)+(s)=12πip.v.s0D(s)ssds+12D(s),(CD)(s)=12πip.v.s0D(s)ssds12D(s).\begin{aligned} (\mathcal C_D)_+(s) &= \frac{1}{2\pi i} \operatorname{p.v.} \int_{s_0}^{\infty} \frac{D(s')}{s'-s} \,\mathrm ds' + \frac12D(s),\\ (\mathcal C_D)_-(s) &= \frac{1}{2\pi i} \operatorname{p.v.} \int_{s_0}^{\infty} \frac{D(s')}{s'-s} \,\mathrm ds' - \frac12D(s). \end{aligned}

Therefore

DiscCD(s)=D(s).\operatorname{Disc}\mathcal C_D(s) = D(s).

For rougher densities, analogous statements require the appropriate LpL^p, almost-everywhere, or distributional theorem. The smooth compactly supported case already fixes all signs.

Growth determines the number of subtractions

Section titled “Growth determines the number of subtractions”

Suppose FF does not decay. Choose a subtraction point zΩz_*\in\Omega and an integer N1N\geq1. Remove the Taylor polynomial

PN1(z)=k=0N1F(k)(z)k!(zz)kP_{N-1}(z) = \sum_{k=0}^{N-1} \frac{F^{(k)}(z_*)}{k!} (z-z_*)^k

and define

GN(z)=F(z)PN1(z)(zz)N.G_N(z) = \frac{F(z)-P_{N-1}(z)} {(z-z_*)^N}.

The apparent singularity of GNG_N at zz_* is removable. The polynomial has no discontinuity, so

DiscGN(s)=DiscF(s)(sz)N.\operatorname{Disc}G_N(s) = \frac{\operatorname{Disc}F(s)} {(s-z_*)^N}.

If GNG_N satisfies every hypothesis of the unsubtracted theorem—including existence and domination of its bank limits, the winding condition, the outer-contour estimate, and the endpoint limit—then

F(z)=k=0N1F(k)(z)k!(zz)k+(zz)N2πis0DiscF(s)(sz)N(sz)ds.\boxed{ \begin{aligned} F(z) ={}& \sum_{k=0}^{N-1} \frac{F^{(k)}(z_*)}{k!} (z-z_*)^k\\ &+ \frac{(z-z_*)^N}{2\pi i} \int_{s_0}^{\infty} \frac{\operatorname{Disc}F(s')} {(s'-z_*)^N(s'-z)} \,\mathrm ds'. \end{aligned} }

This is an NN-times-subtracted dispersion representation. If F(z)=O(zp)F(z)=O(|z|^p) on the relevant large circles, choosing an integer N>pN>p is a simple sufficient condition for GNG_N to vanish there. The weakest legal choice can depend on angular uniformity and logarithmic factors, so the large-circle estimate should still be shown.

The coefficients of PN1P_{N-1} are subtraction data. The discontinuity does not determine them. They must come from normalization conditions, independent information, or a separate theorem.

If FF has poles away from the cut, the contour deformation crosses their Cauchy kernels and the representation gains rational pole terms. A function such as

F(z)=1m2zF(z)=\frac{1}{m^2-z}

with m2[s0,)m^2\notin[s_0,\infty) has zero cut discontinuity but is not zero. Omitting its pole would make any cut-only reconstruction false.

More generally, suppose FF has finitely many simple poles pjΩp_j\in\Omega with residues rjr_j, and retain the same boundary-value, bank-domination, outer-contour, endpoint, and winding hypotheses. Then the unsubtracted representation becomes

F(z)=jrjzpj+12πis0DiscF(s)szds.F(z) = \sum_j\frac{r_j}{z-p_j} + \frac{1}{2\pi i} \int_{s_0}^{\infty} \frac{\operatorname{Disc}F(s')}{s'-z} \,\mathrm ds'.

For a subtraction point zΩ{pj}z_*\in\Omega\setminus\{p_j\}, the additional pole contribution in the NN-times-subtracted formula is

jrj(zz)N(pjz)N(zpj).\sum_j \frac{ r_j(z-z_*)^N }{ (p_j-z_*)^N(z-p_j) }.

Pole contributions, cut contributions, and subtraction polynomials are logically distinct.

The branch analysis introduced

B(s)=01Log(m2sx(1x)μ2)dx,m>0,μ>0,B(s) = \int_0^1 \operatorname{Log} \left( \frac{m^2-sx(1-x)} {\mu^2} \right) \,\mathrm dx, \qquad m>0,\quad\mu>0,

on the plane cut along [4m2,)[4m^2,\infty). Its principal-logarithm convention fixes the sign

DiscB(s)=2πi14m2s,s>4m2,\operatorname{Disc}B(s) = -2\pi i \sqrt{1-\frac{4m^2}{s}}, \qquad s>4m^2,

the discontinuity approaches a constant at large ss. The unsubtracted integral would therefore diverge logarithmically. Because B(s)B(s) itself grows only logarithmically, one subtraction at s=0s=0 is sufficient:

B(s)B(0)=s2πi4m2DiscB(s)s(ss)ds=s4m214m2/ss(ss)ds.\begin{aligned} B(s)-B(0) &= \frac{s}{2\pi i} \int_{4m^2}^{\infty} \frac{\operatorname{Disc}B(s')} {s'(s'-s)} \,\mathrm ds'\\ &= -s \int_{4m^2}^{\infty} \frac{ \sqrt{1-4m^2/s'} } {s'(s'-s)} \,\mathrm ds'. \end{aligned}

The underlying scalar-bubble parameter integral is given in Schwartz 2014, § 16.1, pp. 302–303, while the subtracted dispersion treatment and threshold interpretation are compared in Zwicky 2016, §§ 2.1–2.3.

The formula holds for sC[4m2,)s\in\mathbb C\setminus[4m^2,\infty) and inherits the declared upper-minus-lower discontinuity convention. Its subtraction constant is

B(0)=logm2μ2.B(0)=\log\frac{m^2}{\mu^2}.

An independent derivative check is immediate from the parameter integral:

B(0)=1m201x(1x)dx=16m2.B'(0) = -\frac1{m^2} \int_0^1x(1-x)\,\mathrm dx = -\frac1{6m^2}.

The dispersive side gives

B(0)=4m214m2/s(s)2ds.B'(0) = - \int_{4m^2}^{\infty} \frac{ \sqrt{1-4m^2/s'} } {(s')^2} \,\mathrm ds'.

With

u=14m2s,s=4m21u2,u=\sqrt{1-\frac{4m^2}{s'}}, \qquad s'=\frac{4m^2}{1-u^2},

the last integral becomes

12m201u2du=16m2,-\frac1{2m^2} \int_0^1u^2\,\mathrm du = -\frac1{6m^2},

matching the direct calculation.

This example checks the complex-analysis machinery for a logarithmic one-loop function. It does not infer positivity from the sign shown here: overall amplitude conventions and physical spectral definitions have not been supplied. Scattering develops those added assumptions on Subtracted Dispersion Relations.

A drawn cut is not an analyticity proof. State the domain and account for every pole, endpoint, left-hand cut, and additional threshold before deforming a contour.

The large circle does not vanish by convention. Bound it. If it survives, introduce enough subtractions or retain its contribution.

Subtraction constants are not reconstructed from the cut. They are independent polynomial data unless another result fixes them.

A pole is not a narrow cut. Keep its residue term separate. A cut-only formula misses stable isolated poles.

Boundary values may be distributions. Principal values and delta terms cannot be recovered by ordinary pointwise substitution at the cut.

Discontinuity is not automatically twice an imaginary part. Schwarz reflection is the missing hypothesis.

Convergence at the two ends is separate. Check threshold integrability near s0s_0 and ultraviolet convergence as ss'\to\infty.

QFT constraints are additional input. Causality, spectral conditions, unitarity, crossing, fixed-variable domains, and polynomial bounds are not consequences of Cauchy’s theorem alone.

  1. For

    CD(z)=12πiD(s)szds,\mathcal C_D(z) = \frac{1}{2\pi i} \int \frac{D(s')}{s'-z} \,\mathrm ds',

    verify which boundary value contains +D(s)/2+D(s)/2.

    Check

    At z=s+i0z=s+i0, the denominator is ssi0s'-s-i0, and

    1xi0=p.v.1x+iπδ(x).\frac1{x-i0} = \operatorname{p.v.}\frac1x +i\pi\delta(x).

    Multiplication by 1/(2πi)1/(2\pi i) gives +D(s)/2+D(s)/2. Thus the upper boundary is the plus case and (CD)+(CD)=D(\mathcal C_D)_+-(\mathcal C_D)_-=D.

  2. Explain why the cut discontinuity cannot reconstruct F(z)=1/(m2z)F(z)=1/(m^2-z).

    Check

    The function is meromorphic with an isolated pole and has no branch cut, so its cut discontinuity is zero. The contour deformation must retain the pole residue. Cut data alone would incorrectly return zero.

  3. Derive the once-subtracted kernel algebraically from

    1sz1sz.\frac1{s'-z} - \frac1{s'-z_*}.
    Check

    The difference is

    zz(sz)(sz).\frac{z-z_*} {(s'-z)(s'-z_*)}.

    Do not subtract two unsubtracted dispersion integrals: either one may diverge. Instead apply the unsubtracted theorem directly to

    G1(w)=F(w)F(z)wz.G_1(w) = \frac{F(w)-F(z_*)}{w-z_*}.

    Its discontinuity is DiscG1(s)=DiscF(s)/(sz)\operatorname{Disc}G_1(s') =\operatorname{Disc}F(s')/(s'-z_*). If G1G_1 satisfies the theorem’s endpoint, bank, and outer-contour hypotheses, then

    F(z)F(z)=zz2πis0DiscF(s)(sz)(sz)ds,F(z)-F(z_*) = \frac{z-z_*}{2\pi i} \int_{s_0}^{\infty} \frac{\operatorname{Disc}F(s')} {(s'-z)(s'-z_*)} \,\mathrm ds',

    after multiplying its representation by zzz-z_*. Equivalently, one may subtract the two finite-radius Cauchy formulas before taking the limit.

  • John B. Conway, Functions of One Complex Variable I, 2nd ed., Chapters IV and IX, Springer, 1978. Book record. This is the structural source for Cauchy representations, contour deformation, analytic continuation, and the hypotheses behind the cut contour.
  • Semyon Dyatlov, Lecture Notes for 18.155: Differential Analysis, MIT, 2022, § 5.2.3 and Exercise 5.4(c). Open PDF. This is the distributional source for principal values and the (x±i0)1(x\pm i0)^{-1} identities.
  • Matthew D. Schwartz, Quantum Field Theory and the Standard Model, §16.1, pp. 302–303, especially Eqs. (16.4)–(16.12), Cambridge University Press, 2014. Book record. This supplies the scalar-bubble Feynman-parameter logarithm used in the worked consistency check.
  • Roman Zwicky, “A Brief Introduction to Dispersion Relations and Analyticity”, §§ 2.1–2.3, 2016. This is the QFT-facing teaching source for cut contours, discontinuities, subtraction polynomials, two-point functions, and the equal-mass two-particle threshold.