= Cauchy principal value
{c}
{title2=$\operatorname{PV}\int$}
{wiki}
= Principal value
{synonym}
The <Cauchy principal value> assigns an <integral> a value by a specified symmetric <limit>. Its simplest example is $1/x$ on $[-1,1]$: delete $(-\varepsilon,\varepsilon)$, integrate what remains, and obtain zero for every $\varepsilon>0$. 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.
\Image[/cauchy-principal-value-local-cutoffs.png]
{title=Equal local cutoffs cancel the signed areas of 1/x; unequal cutoffs leave a logarithmic remainder, even when both shrink to zero}
Precisely, let $a<c<b$ be finite and suppose that the two truncated <integrals> below exist for every sufficiently small $\varepsilon>0$. The local <Cauchy principal value> at $c$ is
$$
\operatorname{PV}\int_a^b f(x)\,dx
=\lim_{\varepsilon\downarrow0}\left(\int_a^{c-\varepsilon}f(x)\,dx+\int_{c+\varepsilon}^b f(x)\,dx\right),
$$
provided the combined <limit> is finite. The same $\varepsilon$ 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.https://dlmf.nist.gov/1.4#E24[NIST DLMF, equation 1.4.24: local principal value]{ref}
For $1/x$, the <natural logarithm> gives
$$
\int_{-1}^{-\varepsilon}\frac{dx}{x}=\log\varepsilon\longrightarrow-\infty,
\qquad
\int_{\varepsilon}^{1}\frac{dx}{x}=-\log\varepsilon\longrightarrow+\infty.
$$
Their sum is zero at every positive cutoff, so $\operatorname{PV}\int_{-1}^1dx/x=0$. 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 $\varepsilon_-$ and $\varepsilon_+$, then
$$
\int_{-1}^{-\varepsilon_-}\frac{dx}{x}+\int_{\varepsilon_+}^{1}\frac{dx}{x}
=\log\frac{\varepsilon_-}{\varepsilon_+}.
$$
Thus a ratio tending to $r>0$ produces $\log r$: choosing $\varepsilon_-=2\varepsilon_+$ gives $\log2$, and choosing $\varepsilon_-=\varepsilon_+^2$ 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 $A,B>0$,
$$
\operatorname{PV}\int_{-A}^{B}\frac{dx}{x}=\log\frac BA;
\qquad
\operatorname{PV}\int_{-1}^{2}\frac{dx}{x}=\log2.
$$
More generally, for a function $h$ with continuous <derivative> on $[a,b]$, subtract its value at the possible <simple pole>:
$$
\operatorname{PV}\int_a^b\frac{h(x)}{x-c}\,dx
=h(c)\log\frac{b-c}{c-a}
+\int_a^b\frac{h(x)-h(c)}{x-c}\,dx.
$$
The quotient in the last <integral> extends continuously to $h'(c)$ 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,
$$
\int_{-1}^{-\varepsilon}\frac{dx}{|x|}+\int_{\varepsilon}^{1}\frac{dx}{|x|}
=2\log(1/\varepsilon)\longrightarrow+\infty.
$$
Taking the <absolute value> destroys the cancellation. Even the <Cauchy principal value> need not exist: $1/x^2$ gives $2(1/\varepsilon-1)\to+\infty$ 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, define
$$
\operatorname{PV}\int_{-\infty}^{\infty}f(x)\,dx
=\lim_{R\to\infty}\int_{-R}^R f(x)\,dx,
$$
when the <limit> is finite. The ordinary <improper integral> instead requires convergence of the two tails separately.https://dlmf.nist.gov/1.4#E25[NIST DLMF, equation 1.4.25: principal value at infinity]{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 give
$$
\int_{-A}^B\frac{x}{1+x^2}\,dx
=\frac12\log\frac{1+B^2}{1+A^2}.
$$
Letting $B/A\to r>0$ 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> $\varphi$ is prescribed by
$$
\left\langle\operatorname{PV}\frac1x,\varphi\right\rangle
=\lim_{\varepsilon\downarrow0}\int_{|x|>\varepsilon}\frac{\varphi(x)}x\,dx
=\int_0^\infty\frac{\varphi(x)-\varphi(-x)}x\,dx.
$$
The <test function> is smooth with <compact support>, so the last numerator removes the singular constant contribution: its quotient tends to $2\varphi'(0)$ 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>.https://www.math.ucdavis.edu/~hunter/book/ch11.pdf[Hunter and Nachtergaele, Applied Analysis, Chapter 11, Example 11.7, p. 293: principal-value reciprocal distribution]{ref} Using an asymmetric local cutoff ratio $r>0$ would add $(\log r)\varphi(0)$ 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 $\widehat f(k)=\int_{\mathbb R}f(x)e^{-ikx}\,dx$, pairing $x$ and $-x$ gives, for $k\ne0$ and $0<\varepsilon<R$,
$$
\int_{\varepsilon<|x|<R}\frac{e^{-ikx}}x\,dx
=-2i\int_{\varepsilon}^{R}\frac{\sin(kx)}x\,dx
\longrightarrow-i\pi\operatorname{sgn}(k).
$$
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 $\operatorname{sgn}$ 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.
\Image[/cauchy-principal-value-fourier-cancellation.png]
{title=Fourier cancellation for 1/x: the odd cosine contribution disappears, the sine contribution approaches minus i times pi, and the Hilbert transform turns a cosine into a sine}
The <Hilbert transform> uses exactly this <principal-value reciprocal distribution> as a <convolution kernel>. In the convention already used here,
$$
Hf(y)=\frac1\pi\operatorname{PV}\int_{\mathbb R}\frac{f(x)}{y-x}\,dx,
\qquad
\widehat{Hf}(k)=-i\operatorname{sgn}(k)\widehat f(k)
$$
for <Schwartz functions> $f$, with the second identity interpreted through <tempered distributions>. Hence the singular <integral> has a simple description as a <Fourier multiplier>.https://ocw.mit.edu/courses/res-18-015-topics-in-fourier-analysis-spring-2024/mitres_18_015_s24_lec21.pdf[Stroock, MIT Topics in Fourier Analysis, Lecture 21, p. 1: principal-value convolution kernel and its Fourier multiplier; signs adapted to the displayed convention]{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.https://dlmf.nist.gov/1.14#v[NIST DLMF, equations 1.14.41 and 1.14.46: Hilbert transform and Fourier sign conventions]{ref}
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