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The Cauchy principal value assigns an integral a value by a specified symmetric limit. Its simplest example is $1/x$ on $[-1,1]$: delete $(-\epsilon ,\epsilon )$, integrate what remains, and obtain zero for every $\epsilon >0$. The positive signed area under a graph cancels the negative one, although each grows without bound as the gap closes. This is a useful cancellation rule, but it does not make the ordinary improper integral exist.

Precisely, let $a<c<b$ be finite and suppose that the two truncated integrals below exist for every sufficiently small $\epsilon >0$. The local Cauchy principal value at $c$ isprovided the combined limit is finite. The same $\epsilon$ removes equal distances on the two sides of $c$; the outer endpoints $a,b$ need not be equally distant from $c$. By contrast, the ordinary improper integral requires both one-sided limits to be finite separately before adding them. Whenever those separate limits exist, the Cauchy principal value agrees with their sum.^[ref]

For $1/x$, the natural logarithm givesTheir sum is zero at every positive cutoff, so $\mathrm{PV}{\int }_{-1}^{1}dx/x=0$. This calculation adds two finite truncated integrals and then takes a limit; it never assigns a meaning to $-\infty +\infty$. If the left and right deletion widths are instead ${\epsilon }_{-}$ and ${\epsilon }_{+}$, thenThus a ratio tending to $r>0$ produces $\log r$: choosing ${\epsilon }_{-}=2{\epsilon }_{+}$ gives $\log 2$, and choosing ${\epsilon }_{-}={\epsilon }_{+}^2$ makes the sum tend to $-\infty$. Both widths can tend to zero without selecting a unique answer. The local Cauchy principal value selects the ratio $1$.

Unequal outer endpoints leave a finite contribution even with symmetric local deletion. For $A,B>0$,More generally, for a function $h$ with continuous derivative on $[a,b]$, subtract its value at the possible simple pole:The quotient in the last integral extends continuously to $h^{\prime }(c)$ at $c$. This formula explains both the cancellation and a practical way to compute it: integrate a continuous remainder instead of subtracting two large, nearly equal divergent contributions.

A finite Cauchy principal value is weaker than absolute convergence. Indeed,Taking the absolute value destroys the cancellation. Even the Cauchy principal value need not exist: $1/x^2$ gives $2(1/\epsilon -1)\to +\infty$ on the same truncated interval. Absolute integrability makes the cutoff prescription unnecessary; an ordinary improper integral can also converge without absolute convergence, as the Dirichlet integral illustrates. The three notions should therefore be checked separately.

There is an independent Cauchy principal value at infinity. For a function integrable on every finite interval, definewhen the limit is finite. The ordinary improper integral instead requires convergence of the two tails separately.^[ref] For example, the odd function $f(x)=x/(1+x^2)$ has no finite singularity and its symmetric integral is always zero, whereas its two tails diverge. Unequal truncations giveLetting $B/A\to r>0$ as $A,B\to \infty$ yields $\log r$. Symmetry at infinity is therefore a separate choice from symmetry near a finite pole. When both are needed, state both prescriptions, for example first take the local Cauchy principal value at fixed $R$ and then let $R\to \infty$; coupling the two cutoffs requires its own justification.

The local cancellation also defines the principal-value reciprocal distribution. Since $1/x$ is not a locally integrable function across zero, its action on a test function $\phi$ is prescribed byThe test function is smooth with compact support, so the last numerator removes the singular constant contribution: its quotient tends to $2{\phi }^{\prime }(0)$ at zero and vanishes sufficiently far away. This defines a distribution, not a finite pointwise value of $1/x$ at zero. It extends to Schwartz functions as a tempered distribution, and is also the distributional derivative of the logarithmic modulus.^[ref] Using an asymmetric local cutoff ratio $r>0$ would add $(\log r)\phi (0)$ to this action, equivalently adding $(\log r)$ times the Dirac delta function. Thus the cutoff choice matters even after passing to distributions.

Why keep this prescription? A Fourier transform naturally encounters such a kernel. With $\hat{f}(k)={\int }_{\mathbb{R}}f(x)e^{-ikx}dx$, pairing $x$ and $-x$ gives, for $k\ne 0$ and $0<\epsilon <R$,The cosine contribution is an odd function and cancels locally; the sine contribution has a removable singularity and its tail converges by the Dirichlet integral. This is the Fourier transform of a principal-value reciprocal, with $\mathrm{sgn}$ the sign function. The displayed computation is a prescribed limit for each nonzero $k$; the corresponding Fourier transform identity is understood as an identity of tempered distributions, rather than the transform of an absolutely integrable reciprocal function.

The Hilbert transform uses exactly this principal-value reciprocal distribution as a convolution kernel. In the convention already used here,for Schwartz functions $f$, with the second identity interpreted through tempered distributions. Hence the singular integral has a simple description as a Fourier multiplier.^[ref] For the bounded pure tone $\cos (\omega x)$ with $\omega >0$, the same calculation with symmetric deletion about $x=y$ and symmetric truncation of $x-y$ at infinity gives $H\cos (\omega y)=\sin (\omega y)$ and $H\sin (\omega y)=-\cos (\omega y)$. These tones lie outside the Schwartz space; their prescribed Cauchy principal values explain the quarter-cycle shift in the figure. Both figures are calculated mathematical illustrations. Reversing the Hilbert denominator reverses these signs; NIST uses that opposite convention.^[ref]

The locally integrable function $\log |x|$ has distributional derivative equal to the principal-value reciprocal distribution. Symmetric integration by parts makes the boundary term $\log \epsilon [\phi (\epsilon )-\phi (-\epsilon )]$ tend to zero.

The distributional derivative of $\log |x|$ is the principal-value reciprocal distribution. Its transform is $-i\pi \mathrm{sgn}k$. Dividing the derivative-transform relation by $ik$ gives the displayed finite-part inverse-absolute-value distribution, with an undetermined delta term fixed by the finite-part convention.

The real-space expression $1/x$ means the principal-value reciprocal distribution. Apply even exponential damping. Oddness givesDifferentiate the integral with respect to $k$; its derivative is ${\int }_0^{\infty }{e}^{-ax}\cos (kx)dx=a/(a^2+k^2)$. Its value at $k=0$ is zero, so the integral equals $\arctan (k/a)$. Taking $a\downarrow 0$ givesThe limit is a tempered distribution limit: the regularized transforms are uniformly bounded and converge away from zero, allowing dominated convergence against a Schwartz function. This Fourier transform of a principal-value reciprocal argument also establishes the needed Dirichlet integral ${\int }_0^{\infty }\sin u/udu=\pi /2$ in its Abel interpretation.

We place the definitions and the unheaded preliminary requests here before addressing the first labelled property. The Schwartz space consists of smooth functions with finite seminormsfor every pair of multi-indices. Convergence means convergence in each seminorm. The tempered distribution space ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ is its continuous dual space, with weak convergence tested against every Schwartz function. A continuous functional satisfies a bound by finitely many of these seminorms, equivalently by one sufficiently large weighted derivative seminorm.

Use the Fourier transform conventionDifferentiation under the integral and integration by parts express ${\xi }^{\alpha }{\partial }_{\xi }^{\beta }\hat{f}$ as a constant of modulus one times the Fourier transform of ${\partial }_x^{\alpha }(x^{\beta }f)$. Its supremum is bounded by the $L^1$ norm of that function. For $s>n$, this norm is at most a constant times finitely many Schwartz seminorms, using the integrable weight $(1+|x|{)}^{-s}$. Thus $\mathcal{F}:\mathcal{S}\to \mathcal{S}$ is continuous.

For completeness, Fourier inversion follows by inserting $e^{-\epsilon |\xi {|}^2}$ in the inverse integral and using Fubini's theorem. The result is convolution with the Gaussian approximate identityIt tends to $f$, while integrability of $\hat{f}$ allows the damping factor to be removed by dominated convergence theorem. Consequently ${\mathcal{F}}^{2}f=(2\pi {)}^{n}f(-\cdot )$. Reflection preserves every Schwartz seminorm, so the inverse transform is continuous as well. This proves the Fourier transform isomorphism of the Schwartz space.

Define the Fourier transform of a tempered distribution by transposition,The Schwartz-space continuity just proved makes this a tempered distribution. Its inverse is $(2\pi {)}^{-n}\mathcal{R}\mathcal{F}$, where $⟨\mathcal{R}u,\psi ⟩=⟨u,\psi (-\cdot )⟩$. These maps are continuous for weak convergence, since each pairing is a pairing with a fixed transformed test. They are also continuous for the strong dual topology, because the Schwartz-space maps take bounded sets to bounded sets.

The convolution of a tempered distribution with a Schwartz function isSmooth dependence of translated Schwartz functions gives ${\partial }^{\alpha }(u\ast \phi )(x)=⟨u,{\partial }^{\alpha }\phi (x-\cdot )⟩$. The finite-seminorm estimate and $1+|y|\le (1+|x|)(1+|x-y|)$ show that each derivative has at most polynomial growth. In particular, $u\ast \phi$ is a smooth function defining a tempered distribution. It need not itself be a Schwartz function; for example $1\ast \phi =\int \phi$.

Writing $\check{\phi }(y)=\phi (-y)$, its distributional pairing is $⟨u\ast \phi ,\psi ⟩=⟨u,\check{\phi }\ast \psi ⟩$. The inner convolution is a Schwartz function, and this identity follows by integration in the Schwartz topology, justified by the weighted seminorm estimates. A direct Fubini's theorem calculation gives $\check{\phi }\ast \hat{\psi }=\mathcal{F}(\hat{\phi }\psi )$. HenceMultiplication is well defined because multiplication by $\hat{\phi }$ acts continuously on $\mathcal{S}$.

Now write the Hilbert transform as convolution with $K={\pi }^{-1}\mathrm{pv}(1/x)$, the principal-value reciprocal distribution. The given Heaviside function transform, together with ${\mathcal{F}}^{2}H=2\pi H(-\cdot )$, yieldsThe value of the sign function at zero is irrelevant to its regular distribution. Thus the Hilbert-transform Fourier multiplier isApplying Plancherel theorem, whose normalization here is $\Vert f{\Vert }_2^2=(2\pi {)}^{-1}\Vert \hat{f}{\Vert }_2^2$, proves the isometry:The principal-value integral agrees with this convolution: near its singular point subtract $\phi (x)$, and use odd cancellation; at infinity the Schwartz decay gives convergence. The Hilbert transform has domain $\mathcal{S}$, but generally does not take values in $\mathcal{S}$, as the tail in part (c) demonstrates.

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$. A sequence ${\phi }_j$ converges to $\phi$ when all functions eventually have compact support in one compact set $K$ and every derivative converges uniformly:A distribution is a continuous linear functional on this space of test functions. Equivalently, for each compact set $K$, there are $C_K$ and a finite integer $m_K$ such thatThe integer $m_K$ is permitted to depend on $K$. The usual weak convergence of distributions isThus the convergence convention on the distribution space is its weak dual topology.

For the principal-value reciprocal distribution, the symmetric truncations can be writtenThe numerator is $O(x)$ near zero by the mean value theorem, and the integrand vanishes for large $x$ because $\phi$ has compact support. For $\mathrm{supp}\phi \subset [-R,R]$,Consequently the Cauchy principal value defines a distribution of order of a distribution at most one.

The function $\log |x|$ has local integrability, since ${\int }_0^1|\log x|dx<\infty$, and therefore defines a distribution. For its distributional derivative, remove $(-\epsilon ,\epsilon )$ and use integration by parts on both remaining intervals:The boundary term is $O(\epsilon |\log \epsilon |)$ and tends to zero. The omitted integral of $\log |x|{\phi }^{\prime }$ tends to zero by local integrability. Hence the distributional derivative of the logarithmic modulus satisfiesSymmetric truncation is essential to this normalization of the principal-value reciprocal distribution.

The principal-value reciprocal distribution isThe limit exists because the constant part of the numerator cancels symmetrically at zero. Multiplying by $x$ gives $x\mathrm{pv}(1/x)=1$.

If $v$ is any other solution, $w=v-\mathrm{pv}(1/x)$ obeys $xw=0$. To identify this kernel of multiplication by a coordinate, choose a cutoff function $\eta$ equal to one near zero. Every test function has the form $\phi =\phi (0)\eta +x\psi$ with $\psi \in \mathcal{D}$. Thus $⟨w,\phi ⟩=\phi (0)⟨w,\eta ⟩$, proving $w=c{\delta }_0$. ThereforeFor real distributions the constants are real.

The reciprocal in this transform must mean the principal-value reciprocal distribution. Its inverse transform isusing the given Dirichlet integral and the sign of $t$. Also ${\mathcal{F}}^{-1}[\delta ](t)=1/(2\pi )$. The inverse of the proposed expression is consequently $(iA/2)\mathrm{sgn}t+B/(2\pi )$. Comparison with $H(t)=[1+\mathrm{sgn}t]/2$ for $t\ne 0$ givesThe value assigned at the single point $t=0$ does not affect the distribution. This is the Fourier transform of the Heaviside step function; treating $1/\omega$ as an ordinary integral would miss the required prescription.