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An improper integral whose absolute-value integral is finite. The triangle inequality for an integral bounds each truncated tail by the corresponding tail of $|f|$, so the Cauchy criterion for an improper integral gives ordinary convergence. Absolute convergence is stronger than a Cauchy principal value, which may exist only because two divergent sides cancel. The continuous sine-product integrand $\sin (ax)\sin (bx)/x^2$ has a removable value $ab$ at zero and an integrable $1/x^2$ tail; its auxiliary complex contour integrand can nevertheless require a principal value.

If a Borel summation contour meets a simple pole, an upper or lower indentation fixes the exponentially small contribution. For the transform $(1-t^2{)}^{-1}$ on the positive ray, the relevant pole is $t=1$. A lower indentation gives the Cauchy principal value minus $i\pi e^{-z}/2$. Finite geometric expansion gives an exact remainder integral, which enables optimal truncation and error-function smoothing of a Stokes multiplier.

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]

For $0<\alpha <1$ and $-1<x<1$, the endpoint powers are integrable and the interior pole is taken as a Cauchy principal value. The analytic function $w(z)=(z-1{)}^{\alpha -1}/(z+1{)}^{\alpha }$, positive for $z>1$ and decaying like $1/z$, has boundary values $-e^{\pm i\pi \alpha }(1-t{)}^{\alpha -1}/(1+t{)}^{\alpha }$. Its jump gives a Cauchy reconstruction, while the mean of its boundary values and the Sokhotski–Plemelj formula give the displayed identity. At $\alpha =1/2$ the principal value is zero.

For $y_{tt}-y_{xx}=F(t)\delta (x-x_0)$ with zero endpoint displacement, a Fourier sine series gives $q_n^{\prime \prime }+k_n^2{q}_n=2F(t)\sin (k_n{x}_0)/L$. With the $e^{-i\omega t}$ Fourier transform convention, the causal inverse of this oscillator uses the displayed denominator. Its inverse transform is $H(t)\sin (k_{n}t)/k_n$, giving $q_n(t)=2\sin (k_n{x}_0)/(L{k}_n){\int }_{-\infty }^{t}F(s)\sin [k_n(t-s)]ds$ when the past integral exists. The plain denominator $k_n^2-{\omega }^2$ requires a pole prescription: its Cauchy principal value gives the time-symmetric kernel $\sin (k_n|t|)/(2k_n)$ instead. Homogeneous sine and cosine modes are additional freedom unless temporal data or causality are specified.

Let $X$ have the Standard Cauchy distribution and let $Y$, independently, have the uniform distribution on $[-h,h]$, where $h>0$. The convolution of probability densities gives the probability density of $Z=X+Y$ asThis is an even function. Its derivative isso its unique mode is zero. Its tails satisfy $p_h(z)\sim 1/(\pi z^2)$ as $|z|\to \infty$: multiplying the integral by $z^2$ makes the integrand converge uniformly to $1$ on the bounded interval $[-h,h]$. Consequently both positive and negative parts of the first moment diverge, so its expected value does not exist. The symmetric Cauchy principal value of that moment is zero, which is a different assertion. Adding a bounded independent displacement smooths the probability density without removing the divergent first moment of the Cauchy distribution.

Suppose $q(t)=A/(t-c)+r(t)$, where $r$ is continuous, the pole $c$ is fixed, and differentiable endpoints obey $a(x)<c<b(x)$ locally. ThenApply the chain rule and the fundamental theorem of calculus to getThe same boundary expression as in the Leibniz integral rule thus applies to this explicitly specified Cauchy principal value. If $A\ne 0$, the two ordinary one-sided improper integrals diverge, so their sum is not an ordinary integral. A finite boundary derivative alone does not make that divergent integral meaningful.

Integrating a circular ring's Newtonian gravitational potential over azimuth gives $d\Phi =-4Gs\Sigma (s)K(k)ds/(r+s)$, with modulus $k=2\sqrt{rs}/(r+s)$. Differentiating using the complete elliptic integral of the first kind derivative gives the displayed circular speed kernel, where $E$ is the complete elliptic integral of the second kind. For an interior point of a smooth disk use a Cauchy principal value; beyond the outer radius the integral is ordinary. A discontinuous nonzero surface-density edge in an exactly zero-thickness disk has a logarithmically singular in-plane force at that edge.

For a continuous even function $f$ on $\mathbb{R}$, the Cauchy principal valuedefines a distribution of order of a distribution at most one. For test functions supported in $[-R,R]$, the absolute value is at most $2R{\max }_{[0,R]}|f|\Vert {\varphi }^{\prime }{\Vert }_{\infty }$, since the numerator is bounded by $2x\Vert {\varphi }^{\prime }{\Vert }_{\infty }$. When $\varphi (0)=0$, the expression equals the absolutely convergent ordinary integral $\int f\varphi /x$.

It is a regular distribution exactly when ${\int }_0^r|f(x)|/xdx<\infty$ for some $r>0$. Necessity follows by testing away from zero: any representing locally integrable function must equal $f/x$ almost everywhere off zero. Sufficiency follows from absolute convergence. Continuity with $f(0)=0$ alone does not suffice: $f(x)=1/\log (1/|x|)$ near zero gives the divergent integral ${\int }_1^{\infty }ds/s$.

Target: L1-normalized dilation of a distribution. Parent: Dilation of a distribution. Intended file: distribution-theory.bigb.

The finite Hilbert transform is a singular integral operator whose integration is restricted to a finite interval. The displayed convention uses the denominator $t-x$; reversing that denominator reverses the sign. The Cauchy principal value deletes equal intervals on the two sides of the interior point $x$. Endpoint behavior matters: the integrable function $1/\sqrt{x(1-x)}$ belongs to the null space for this interval. This nontrivial null space is one reason that an inversion formula needs a normalization or a stated endpoint class.

With the $e^{-ikx}$ convention, exponential damping on the positive half-line gives $1/(a+ik)$. Its even part tends to $\pi \delta (k)$ and its odd part to $-i\mathrm{PV}(1/k)$. Thus the Heaviside step function transform requires a Cauchy principal value, not an ordinary reciprocal function at zero.

For the radial induced-charge screening kernel, integrate the sine transform twice by parts, with $g(0)=g(\infty )=0$ and vanishing remaining endpoint terms. This gives ${\rho }_{\mathrm{ind}}=Ze\int g^{\prime \prime }(q)\sin (qr)dq/r^3$. A Cauchy principal value singularity at $Q>0$ contributes $C\pi \cos (Qr)$ to that integral: set $s=q-Q$ and use ${\int }_{-\infty }^{\infty }\sin (sr)ds/s=\pi$ for $r>0$. Smoothly localizing around $Q$ justifies the large-$r$ extension of the limits. In a three-dimensional isotropic Fermi gas, $Q=2k_F$.

For $z=2n+i\mu \sqrt{n}+O(1)$, the factor $t^{2n}{e}^{z(1-t)}$ has saddle scale $t=1+u/\sqrt{n}$ and becomes $e^{-u^2-i\mu u}$. If the amplitude has a simple pole there, its leading local integral contains $e^{-u^2-i\mu u}/u$. Differentiating its Cauchy principal value with respect to $\mu$ reduces it to a Gaussian integral, producing an error function. The nonsingular amplitude $(1+t{)}^{-1}$ gives $\frac{1}{2}\sqrt{\pi /n}{e}^{-{\mu }^2/4}$.

Let $\phi$ be continuous and suppose that the displayed improper integral converges separately at both infinite endpoints for some $T>0$. Absolute convergence is not required. Then its heat-kernel solution exists, is smooth and solves the heat equation for $0<t<T$, and tends to $\phi (x)$ as $t\downarrow 0$ at each fixed $x$.

To prove the smoothing claim, take a primitive $H$ of $\phi (\xi )e^{-{\xi }^2/(4T)}$. The assumed convergence makes $H$ bounded. For $q(x,\xi ,t)=K_t(x-\xi )e^{{\xi }^2/(4T)}$, integration by parts givesThe endpoint term vanishes. The coefficient of ${\xi }^2$ in the exponent of $q$ is $-(T-t)/(4tT)<0$. Its derivatives therefore have Gaussian decay on every compact subset of $\mathbb{R}\times (0,T)$, justifying differentiation under the integral sign in the absolutely convergent right-hand expression. Since $q_t=q_{xx}$, it satisfies the heat equation.

For the initial limit, apply the same construction to $\phi (\xi )-\phi (x)$ and split at $|\xi -x|=\delta$. Continuity controls the local part by its small oscillation times the unit kernel mass. On each remaining tail, $q$ is monotone for sufficiently small $t$: its maximum lies at $\xi =xT/(T-t)$, approaching $x$. If the relevant primitive has absolute bound $M$, integration by parts bounds that tail by $2M$ times the value of $q$ at its finite endpoint. For either endpoint $\xi =x\pm \delta$, that value tends to zero like $t^{-1/2}{e}^{-{\delta }^2/(4t)}$ times a fixed constant. Both tails vanish, proving the limit. Separate endpoint convergence matters here; merely assuming a symmetric Cauchy principal value does not provide the bounded primitive used in this proof.

In the upper half-plane, a line source of outward volume flux $Q$ at height $a$ above an impermeable wall has an equal image source below it. On the wall $u=Qx/[\pi (x^2+a^2)]$ and the normal velocity vanishes. The steady Bernoulli equation gives $p-p_{\infty }=-\rho Q^2{x}^2/[2{\pi }^2(x^2+a^2{)}^2]$. Sincethe ambient-relative force per unit length pulls the wall toward the source with the displayed magnitude. The torque about the source's projection onto the wall is zero in the symmetric Cauchy principal value sense. Individual one-sided moments diverge logarithmically. This planar result differs from the three-dimensional hydrodynamic attraction of a plane wall to a point source.

A line integral can be defined as an improper integral of its parametrized integrand when that integrand is undefined or unbounded at isolated parameter values. At an interior singular parameter, split the integral and require both one-sided limits to exist separately; their cancellation in a symmetric Cauchy principal value is a different convention. For a scalar arc length integral, use $f(\mathbf{r}(t))|{\mathbf{r}}^{\prime }(t)|$ in place of the vector-field dot product.

If the restricted integrand has a removable singularity, integrating its continuous extension gives the same value. This does not require the original vector field to extend continuously to a full neighbourhood of the curve. Likewise, a differential identity may hold only on the regular segments: a zero integrand there has improper integral zero, although its printed classical expression may not be defined at the omitted points.

Let $Z$ have standard normal distribution, let $m\ge 1$ be an integer and let $\alpha \in \mathbb{R}$. A finite expected value of $Z{e}^{\alpha Z^{2m}}$ requires absolute integrability, which reduces to finiteness ofFor $m=1$ it is finite exactly when $\alpha <1/2$. For every $m\ge 2$ it is finite exactly when $\alpha \le 0$: the higher power dominates the negative quadratic when $\alpha >0$, and for $\alpha \le 0$ the integrand is bounded by $z{e}^{-z^2/2}$. In the finite cases the signed expected value is zero by symmetry. Otherwise each signed part has infinite integral; a formal symmetric cancellation is only a Cauchy principal value.

Causality constrains the real and imaginary parts of a frequency-dependent linear response. For a dielectric with ${ϵ}_r(\omega )\to 1$ at large frequency, the $e^{-i\omega t}$ convention givesThe integral is a Cauchy principal value. Absorption and optical dispersion therefore cannot be specified independently.

For a thin axisymmetric astrophysical disk, angular integration of the Newtonian kernel eliminates odd Legendre polynomials. Define $I_n={\int }_0^R\Sigma (s)s^{n+1}ds$ and $J_n={\int }_R^{\infty }\Sigma (s)s^{-n}ds$. The even coefficient is ${\alpha }_n=\pi [n!/(2^n((n/2)!{)}^2){]}^2$. In differentiating the paired inner and outer contributions, their moving-boundary terms cancel. The resulting galaxy rotation curve includes an inward inner-ring contribution and an outward exterior-ring contribution. For smooth surface density, the force at the disk is interpreted with a symmetric Cauchy principal value or a vanishing-thickness regularization.

The Cauchy principal value integral $\mathrm{li}(x)=\mathrm{PV}{\int }_0^{x}dt/\log t$ is the logarithmic integral. The offset version $\mathrm{Li}(x)={\int }_2^{x}dt/\log t$ differs by a constant and is often used with the prime-counting function. It has $\mathrm{Li}(x)\sim x/\log x$.