Cauchy Principal Values and Improper Integrals

Two Different Limits at an Interior Singularity

At an interior singularity, the integral is approached from both sides. An ordinary improper integral requires the two one-sided limits to exist separately. The Cauchy principal value instead removes equal lengths on the two sides and takes one combined limit. The distinction matters: symmetric cancellation can give a finite principal value even when neither one-sided integral converges.

Throughout, let \(A<c<B\) be real numbers, and let \(f\) be a real-valued function continuous on each closed subinterval that avoids \(c\). For measure-theoretic statements, we regard \(f\) as a measurable function on \([A,B]\) with Lebesgue measure \(m\), assigning any finite value at \(c\). That value does not affect any integral. Every truncated integral below is proper, and its Riemann and Lebesgue values agree.

Definition: The ordinary improper integral of \(f\) over \([A,B]\) at \(c\) exists as a finite value if both one-sided limits \(\displaystyle\lim_{\varepsilon\downarrow0}\int_A^{c-\varepsilon}f(x)\,dx\) and \(\displaystyle\lim_{\varepsilon\downarrow0}\int_{c+\varepsilon}^{B}f(x)\,dx\) exist as finite real numbers. Its value is the sum of those limits. The Cauchy principal value, when the limit exists as a finite real number, is \(\displaystyle\operatorname{PV}\int_A^B f(x)\,dx=\lim_{\varepsilon\downarrow0}\left(\int_A^{c-\varepsilon}f(x)\,dx+\int_{c+\varepsilon}^{B}f(x)\,dx\right)\), where \(0<\varepsilon<\min(c-A,B-c)\). The same distance \(\varepsilon\) is removed on both sides of \(c\).

The ordinary definition takes two limits before adding their values. The principal-value definition adds the truncated integrals first and then takes one limit. A finite value in the second definition does not establish convergence in the first.

When the Ordinary Integral Does Exist

Proposition: If the ordinary improper integral at \(c\) exists as a finite value, then the Cauchy principal value exists and equals it.

Proof. Suppose the left and right one-sided limits exist and equal \(L_-\) and \(L_+\), respectively. As \(\varepsilon\downarrow0\), the sum of the two truncated integrals tends to \(L_-+L_+\), because each summand tends to its stated finite limit. This is precisely the ordinary improper integral, and the same sum is the expression defining the principal value. Therefore the principal value exists and equals \(L_-+L_+\). \(\square\)

The converse is false. The next examples show how equal truncation can cancel logarithmic divergences that remain on each side separately.

Worked Example: The Pole \(1/x\) on a Symmetric Interval

On \([-1,1]\), with Lebesgue measure \(m\), define \(f(x)=1/x\) for \(x\ne0\). Compute the ordinary improper integral and the principal value at \(0\).

For \(0<\varepsilon<1\), an antiderivative on either side of zero is \(\ln|x|\). Thus the two truncated integrals are

\int_{-1}^{-\varepsilon}\frac{dx}{x} =\ln\varepsilon-\ln1=\ln\varepsilon, \qquad \int_{\varepsilon}^{1}\frac{dx}{x} =\ln1-\ln\varepsilon=-\ln\varepsilon.

As \(\varepsilon\downarrow0\), \(\ln\varepsilon\to-\infty\) and \(-\ln\varepsilon\to+\infty\). Neither one-sided limit is finite, so the ordinary improper integral does not exist. But their sum is \(0\) for every \(0<\varepsilon<1\), and therefore

\operatorname{PV}\int_{-1}^{1}\frac{dx}{x}=0.

Check: substituting \(\varepsilon=1/2\), the left integral is \(\ln(1/2)=-\ln2\), and the right integral is \(-\ln(1/2)=\ln2\); their sum is \(0\). This verifies the cancellation at a specific cutoff, but the separate limits still diverge.

This principal value is not a Lebesgue integral of \(1/x\) on \([-1,1]\). Indeed, the positive and negative parts each have infinite integral: for example, \(\int_\varepsilon^1 dx/x=-\ln\varepsilon\to+\infty\). A signed Lebesgue integral is not defined when both the positive-part and negative-part integrals are infinite. Principal value is a limit under a particular symmetric truncation, not a way to assign an ordinary Lebesgue integral.

Worked Example: Unequal Distances to the Endpoints

On \([-2,1]\), again take \(f(x)=1/x\) with singularity \(c=0\). Find its principal value, and compare it with the ordinary improper integral.

For \(0<\varepsilon<1\), the symmetric truncation gives

\int_{-2}^{-\varepsilon}\frac{dx}{x} =\ln\varepsilon-\ln2, \qquad \int_{\varepsilon}^{1}\frac{dx}{x} =-\ln\varepsilon.

Adding before taking the limit yields \(-\ln2\) for every such \(\varepsilon\), hence

\operatorname{PV}\int_{-2}^{1}\frac{dx}{x}=-\ln2.

The left one-sided integral tends to \(-\infty\), and the right one tends to \(+\infty\). Thus the ordinary improper integral does not exist. The nonzero principal value comes from the unequal distances from \(0\) to the endpoints, not from a finite one-sided integral.

Check: at \(\varepsilon=1/2\), the left integral is \(\ln(1/2)-\ln2=-2\ln2\), while the right integral is \(-\ln(1/2)=\ln2\). The sum is \(-\ln2\), as given by the formula.

Separating a Simple Pole from a Regular Part

The examples are instances of a useful calculation. A simple pole contributes a logarithm on each side. Equal truncation cancels the two divergent logarithms, while the remaining finite term records the endpoint distances.

Theorem: Let \(A<c<B\), let \(\alpha\in\mathbb{R}\), and let \(g:[A,B]\to\mathbb{R}\) be continuous. Define \(f(x)=\alpha/(x-c)+g(x)\) for \(x\ne c\). Then the principal value exists and \(\displaystyle\operatorname{PV}\int_A^B f(x)\,dx=\alpha\ln\!\left(\frac{B-c}{c-A}\right)+\int_A^B g(x)\,dx\). The ordinary improper integral exists as a finite value if and only if \(\alpha=0\); in that case its value is \(\int_A^B g(x)\,dx\).

Proof. Fix \(0<\varepsilon<\min(c-A,B-c)\). On each truncated interval, integrate the pole and the continuous function separately. The pole terms are

\int_A^{c-\varepsilon}\frac{\alpha}{x-c}\,dx =\alpha\bigl(\ln\varepsilon-\ln(c-A)\bigr), \qquad \int_{c+\varepsilon}^{B}\frac{\alpha}{x-c}\,dx =\alpha\bigl(\ln(B-c)-\ln\varepsilon\bigr).

Their sum is \(\alpha\ln((B-c)/(c-A))\), independent of \(\varepsilon\). Since \(g\) is continuous on the compact interval, it is bounded and integrable. The omitted intervals \([c-\varepsilon,c]\) and \([c,c+\varepsilon]\) have total length \(2\varepsilon\). If \(|g(x)|\le M\), the absolute value of the omitted integral is at most \(2M\varepsilon\), which tends to zero. Therefore the truncated integrals of \(g\) tend to \(\int_A^B g(x)\,dx\). Adding these limits proves the principal-value formula.

For the ordinary integral, consider the left one-sided pole integral. It equals \(\alpha(\ln\varepsilon-\ln(c-A))\), which has no finite limit when \(\alpha\ne0\). The one-sided integral of \(g\) has a finite limit because \(g\) is bounded, so it cannot cancel that divergence within the left limit. Thus the ordinary integral fails to exist when \(\alpha\ne0\). If \(\alpha=0\), then \(f=g\) is continuous, so both one-sided integrals converge and their sum is \(\int_A^B g(x)\,dx\). This proves both directions. \(\square\)

Worked Example: A Pole Plus a Continuous Term

On \([-1,2]\), let \(f(x)=1/x+x^2\) for \(x\ne0\), with Lebesgue measure \(m\). Compute its principal value and determine whether its ordinary improper integral exists.

Here the pole coefficient is \(\alpha=1\), the singularity is \(c=0\), and \(g(x)=x^2\) is continuous. The theorem gives a pole contribution \(\ln(2/1)=\ln2\). The regular contribution is

\int_{-1}^{2}x^2\,dx =\left[\frac{x^3}{3}\right]_{-1}^{2} =\frac{8}{3}-\left(-\frac{1}{3}\right)=3.

Therefore \(\operatorname{PV}\int_{-1}^{2}(1/x+x^2)\,dx=\ln2+3\). The ordinary integral does not exist: its left one-sided integral contains the term \(\ln\varepsilon\to-\infty\), while its \(x^2\) contribution has a finite limit.

Check: directly, the two truncated integrals add to \(\ln2+(1-\varepsilon^3)/3+(8-\varepsilon^3)/3=\ln2+3-2\varepsilon^3/3\). Taking \(\varepsilon\downarrow0\) gives \(\ln2+3\), agreeing with the theorem.

Why the Symmetry Requirement Matters

A principal value depends on how the singularity is truncated. For \(f(x)=1/x\) on \([-1,1]\), remove distance \(\varepsilon\) on the left but distance \(2\varepsilon\) on the right. For \(0<2\varepsilon<1\), the resulting sum is

\int_{-1}^{-\varepsilon}\frac{dx}{x} +\int_{2\varepsilon}^{1}\frac{dx}{x} =\ln\varepsilon-\ln(2\varepsilon) =-\ln2.

This differs from the symmetric principal value, which is \(0\). In general, changing the relative cutoffs can change the finite remainder, so the phrase “principal value” specifies equal distances from the singularity. It does not mean that divergent one-sided integrals may simply be added, and it does not make the ordinary improper integral exist.

Takeaway: At an interior singularity, first test the two one-sided limits separately to decide ordinary convergence. Only if the question asks for a principal value should you combine equal-distance truncations before taking the limit.

Check Your Understanding

For each question, distinguish the two one-sided limits from the symmetric truncated sum.

Compute the principal value of \(1/x\) on \([-3,2]\), and state whether the ordinary improper integral exists.
For \(f(x)=2/(x-1)\) on \([0,3]\), compute the principal value at \(1\) and determine whether the ordinary improper integral exists.
Explain why the principal value of \(1/x\) on \([-1,1]\) is not its Lebesgue integral.
For \(1/x\) on \([-1,1]\), compute the limit when the left cutoff is \(\varepsilon\) and the right cutoff is \(3\varepsilon\), and compare it with the symmetric principal value.
Suppose \(f(x)=\alpha/(x-c)+g(x)\) with \(g\) continuous and \(\alpha=0\). Which theorem above guarantees that the ordinary integral exists, and what is its value?

Cauchy Principal Values and Improper Integrals

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