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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]

Suppose $w$ is locally integrable andThen $\hat{w}(\lambda )+(c_{+}-c_{-})\log |\lambda |$ has finite limits at zero from both sides. With the convention $\hat{w}(\lambda )=\int e^{-i\lambda x}w(x)dx$, the positive-frequency limit minus the negative-frequency limit isSubtracting suitable reflected copies of $x^{-1}{1}_{(1,\infty )}(x)$ leaves an $L^1$ function, whose Fourier transform is continuous; the jump then follows from the Dirichlet integral.

Integrate $e^{iz}/z$ around the positively oriented indented upper-half-plane contour. The outer semicircle runs from $R$ to $-R$; the inner semicircle runs clockwise from $-r$ to $r$, excluding zero. The integrand is holomorphic inside this region, so Cauchy's integral theorem gives total integral zero.

The outer arc tends to zero by part (a) with $f(z)=1/z$ and $\lambda =1$. On the inner arc $z=r{e}^{i\theta }$, $\theta$ decreases from $\pi$ to zero, andThe two real segments combine asPassing to the limits gives $2iI-i\pi =0$, and hence the Dirichlet integralThis is an ordinary conditionally convergent improper integral: the integrand tends to one at zero, and integration by parts gives a tail bounded by $2/A$ beyond $A$. The clockwise indentation sign is essential.

There is an important convention issue in the printed statement. The ordinary Cauchy principal value exists only for $n=1$. Closing the Fourier contour in the lower half-plane, with a small indentation above the real-axis pole, givesEquivalently, the odd cosine contribution cancels and the Dirichlet integral ${\int }_0^{\infty }\sin x/xdx=\pi /2$ supplies the imaginary part.

For $n\ge 2$, the paired integrand near zero is $2\cos x/x^n$ if $n$ is even and $-2i\sin x/x^n$ if $n$ is odd. Its leading even power is nonintegrable: $x^{-n}$ in the first case and $x^{1-n}$ in the second. Thus the ordinary principal value diverges for every $n>1$.

If the intended convention is the Hadamard finite-part integral, define it consistently by distributional differentiation,Integration by parts in this distributional pairing, or regularized Fourier transformation, then givesThe distinction is essential: subtracting the divergent local Taylor terms supplies a finite part, but symmetric exclusion alone does not.

Integration by parts is legitimate after symmetrically excluding a neighborhood of zero; the boundary terms vanish because $\cos (nx)-\cos (mx)=O(x^2)$ at zero and is bounded at infinity. HenceThe standard Dirichlet integral, obtained by closing the contour for $e^{iaz}/z$ in the upper half-plane and indenting above its pole at zero, isTherefore

With the stated sign convention, write $t=y-x$ to obtainThe Cauchy principal value of the cosine part vanishes by oddness, while the Dirichlet integral givesThus the Hilbert transform has Fourier multiplier $-i\mathrm{sgn}(\omega )$. Taking the real part yields

For $\lambda >0$, subtracting the two formulas from part (iv) yieldsUsing the stated Dirichlet integral and taking the one-sided limit gives

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.

Let $B=|b|>0$. For $a>0$, integrate $e^{iaz}/(z^2+B^2)$ over the upper semicircle. Jordan's lemma removes the arc, and the residue theorem at $z=iB$ givesTaking the real part and using evenness in both $x$ and $a$ yieldsThe case $a=0$ follows directly from an inverse tangent primitive. The complex exponential is essential on the contour: the cosine alone has a growing exponential component.

For $a>0$, differentiation under the integral sign gives $-I_1^{\prime }(a)={\int }_0^{\infty }x\sin (ax)/(x^2+B^2)dx$. This derivative is a conditionally convergent improper integral; the Dirichlet test makes its tails uniform on compact subintervals of $a>0$, justifying the differentiation. Setting $a=1$ givesFinally let $J(a)={\int }_0^{\infty }\sin (ax)/[x(x^2+B^2)]dx$. Its differentiated integrand is absolutely integrable, so $J^{\prime }(a)=I_1(a)$ and $J(0)=0$. HenceIf $b=0$, $I_1$ and $I_3$ diverge at the origin, whereas $I_2={\int }_0^{\infty }\sin x/xdx=\pi /2$ is the corresponding Dirichlet integral.