The Cauchy Principal Value

When an improper integral diverges, it is sometimes still possible to attach a finite number to it by cutting out the trouble spot symmetrically and letting the cut shrink to nothing. The number produced this way is the Cauchy principal value. It is a value we assign to the symbol, not the value of the integral.

Item	Statement
Two notations for the same thing	$P . V . \int$ and the barred integral $- \int$
Doubly infinite interval	$P . V . \int_{- \infty}^{\infty} f = lim_{R \to \infty} \int_{- R}^{R} f$
Interior singularity at $c$	$P . V . \int_{a}^{b} f = lim_{ε \to 0^{+}} [\int_{a}^{c - ε} f + \int_{c + ε}^{b} f]$
What makes it different	one $ε$ (or one $R$ ) on both sides at once, so $+$ and $-$ can cancel
Ordinary convergence	needs each side of the trouble spot to converge separately
The one-way implication	convergence $\Rightarrow$ the principal value exists and agrees; the converse is false
Odd $f$ , symmetric window	$P . V . \int_{- a}^{a} f = 0$ and $P . V . \int_{- \infty}^{\infty} f = 0$
Model example	$P . V . \int_{- 1}^{1} \frac{d x}{x} = 0$ , yet $\int_{- 1}^{1} \frac{d x}{x}$ diverges
The interval need not be symmetric	$P . V . \int_{- 1}^{2} \frac{d x}{x} = \ln 2$
When even the principal value fails	$P . V . \int_{- 1}^{1} \frac{d x}{x^{2}}$ does not exist: both sides run to $+ \infty$
Linearity	linear whenever the pieces exist, but $P . V . \int (f + g)$ can exist when neither piece does
Classic trap	applying the Fundamental Theorem straight across an interior singularity
Named for	Augustin-Louis Cauchy (1789 to 1857)
Where it is used	contour integration, Fourier analysis, signal processing, quantum mechanics

Two Notations

When an improper integral diverges, it is sometimes possible to assign it a finite value by a symmetric limiting process. The result is called the Cauchy principal value. It is written in two equivalent ways:

P . V . \int or - \int

The second notation places a horizontal dash through the integral sign, and it appears frequently in complex analysis and distribution theory. The barred integral sign and $P . V . \int$ mean exactly the same thing, and we use both forms interchangeably below.

Both are read aloud the same way: "the principal value of the integral of $f$ ." Neither is a new kind of integral. Each is a shorthand for a particular limit, spelled out in the next section.

The Definition

There are two situations in which an integral can fail to exist and a symmetric limit can rescue it: the interval runs off to infinity in both directions, which is a Type I improper integral, or the integrand blows up at a point strictly inside the interval, which is a Type II improper integral. The definition therefore has two parts, one for each case.

Doubly infinite integrals. If $f$ is continuous on $ℝ$ , the Cauchy principal value is \bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\operatorname{P.V.}\!\int_{-\infty}^{\infty}f(x)\,dx=\lim_{R\to\infty}\int_{-R}^{R}f(x)\,dx}
provided this limit exists. Here $R > 0$ is the half-length of the window $[- R, R]$ , and the same $R$ is used at both ends. The barred form ${- \int}_{- \infty}^{\infty} f (x) d x$ denotes the same quantity.

Interior singularity at $c$ . If $f$ is continuous on $[a, b] ∖ {c}$ and has an infinite discontinuity at a point $c$ with $a < c < b$ , the Cauchy principal value is \bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\operatorname{P.V.}\!\int_{a}^{b}f(x)\,dx=\lim_{\varepsilon\to0^{+}}\left[\int_{a}^{c-\varepsilon}f(x)\,dx+\int_{c+\varepsilon}^{b}f(x)\,dx\right]}
provided this limit exists. Here $ε > 0$ is the half-width of the little interval $(c - ε, c + ε)$ that is deleted around the singularity, and the same $ε$ is used on both sides. The barred form is ${- \int}_{a}^{b} f (x) d x$ .

The whole content of the definition is in the words "the same". In part (a) the window grows outward at the same rate in both directions. In part (b) the gap closes at the same rate from both sides. Because the two sides are tied together, a large positive contribution on one side can be cancelled by an equally large negative contribution on the other before either one is allowed to become infinite.

How to Compute a Principal Value

Locate the trouble. Either the interval is $(- \infty, \infty)$ , or the integrand has an infinite discontinuity at some interior point $c$ . If the trouble is only at an endpoint, there is nothing symmetric to do and the principal value is just the ordinary improper integral.
Cut symmetrically. Replace $(- \infty, \infty)$ by $[- R, R]$ , or delete $(c - ε, c + ε)$ from $[a, b]$ . Do not use two different letters for the two sides.
Integrate exactly, in terms of $R$ or $ε$ . Use an antiderivative on each remaining piece. Remember $\int \frac{d x}{x} = \ln | x | + C$ : the absolute value matters on the left of the origin.
Add the pieces and simplify before taking the limit. This is the step where the cancellation happens. Terms such as $\ln ε$ and $- \ln ε$ , or $R^{2} / 2$ and $- R^{2} / 2$ , must be combined first; taking limits term by term would produce $\infty - \infty$ , which is meaningless.
Take the limit $ε \to 0^{+}$ or $R \to \infty$ . If it exists, that number is the principal value. If it does not exist, the principal value does not exist either.
Say what you have found. A principal value is not by itself a statement that the integral converges. Check convergence separately if the question asks for it.

Why the Cancellation Works

The picture below is the whole idea in one image. The function $y = 1 / x$ is odd, and the deleted gap $(- ε, ε)$ is centred on the singularity, so whatever area the curve piles up on the right of the gap is matched exactly by an equal amount of negative area on the left.

The gap $(- ε, ε)$ is removed symmetrically. The negative area on the left equals $\ln ε$ , which is a large negative number when $ε$ is small, and the positive area on the right equals $- \ln ε$ , the same size with the opposite sign. They cancel for every $ε > 0$ , so the principal value is $0$ even though the ordinary integral diverges. The computation is carried out in full in the first worked example below.

Three consequences are worth stating separately.

Symmetry is the whole difference. In part (b) of the definition the same $ε$ is removed on both sides of $c$ , so positive and negative contributions are allowed to cancel. The ordinary improper integral removes $δ$ on the left and $η$ on the right and insists that each piece settle down on its own, with no help from the other.
Odd functions give zero. If $f$ is odd, meaning $f (- x) = - f (x)$ , then $P . V . \int_{- a}^{a} f (x) d x = 0$ for every $a > 0$ , and $P . V . \int_{- \infty}^{\infty} f (x) d x = 0$ whenever the limit exists. Each piece of area to the right of the origin is matched by its mirror image below the axis on the left. This is proved in the exercises.
Zero is not a claim of convergence. The value $0$ above says nothing about whether $\int_{- 1}^{1} d x / x$ converges. It does not.

Principal Value Versus Ordinary Convergence

Recall from Type II improper integrals that when $f$ has an infinite discontinuity at an interior point $c$ of $[a, b]$ , the ordinary improper integral is defined by splitting at $c$ and requiring both halves to converge on their own:

\int_{a}^{b} f (x) d x = lim_{δ \to 0^{+}} \int_{a}^{c - δ} f (x) d x + lim_{η \to 0^{+}} \int_{c + η}^{b} f (x) d x .

Two separate letters, $δ$ and $η$ , and two separate limits. If either one fails, the whole integral diverges, no matter what the other one does.

For the ordinary integral the left cut $δ$ and the right cut $η$ are unrelated, and each half must converge by itself.

For the principal value there is a single letter $ε$ , so the two cuts are locked together and shrink at the same rate. That single change is what allows cancellation.

The central fact of this section, in one sentence: convergence is strictly stronger than the existence of a principal value.

If $\int_{a}^{b} f (x) d x$ converges in the ordinary sense, then $P . V . \int_{a}^{b} f (x) d x$ exists and the two are equal. Taking $δ = η = ε$ is one particular way of letting the two cuts close, so a limit that already exists for all ways of closing them certainly exists for this one.
The converse is false. The principal value can exist when the integral diverges, and $P . V . \int_{- 1}^{1} d x / x = 0$ is the standard example.
Therefore a principal value is an assigned value, not the value of the integral. Writing $P . V . \int_{- 1}^{1} d x / x = 0$ is a statement about a particular symmetric limit. Writing $\int_{- 1}^{1} d x / x = 0$ is simply false, because that integral does not converge.
Never quote a principal value as though it settled convergence, and never drop the symbol $P . V .$ from an answer that depends on it.

There is a familiar trap hiding here. A student who does not notice the singularity at $x = 0$ may apply the Fundamental Theorem of Calculus straight across the interval and write

\int_{- 1}^{1} \frac{d x}{x} = [\ln | x |]_{- 1}^{1} = \ln 1 - \ln 1 = 0.

The answer happens to agree with the principal value, but the reasoning is invalid: the Fundamental Theorem requires the integrand to be continuous on the closed interval, and $1 / x$ is not even defined at $x = 0$ . The same careless step applied to $\int_{- 1}^{1} d x / x^{2}$ gives

[- \frac{1}{x}]_{- 1}^{1} = - 1 - 1 = - 2,

a negative number for the integral of a strictly positive function, which is impossible. That integral diverges, and as the last exercise shows its principal value does not exist either.

Worked Examples

The two examples below are the two halves of the definition: an interior singularity first, then a doubly infinite interval. In each one, compute the ordinary integral and the principal value separately and compare them.

Compute $P . V . \int_{- 1}^{1} \frac{d x}{x}$ , and compare with the ordinary improper integral.

Solution

Ordinary integral. The integrand $1 / x$ has an infinite discontinuity at $x = 0$ , which lies inside $(- 1, 1)$ , so the integral must be split there and each half examined separately. We check the right half:

\begin{aligned} \int_{0}^{1}\frac{dx}{x} &= \lim_{\varepsilon\to0^{+}}\Bigl[\ln x\Bigr]_{\varepsilon}^{1}\\ &= \lim_{\varepsilon\to0^{+}}(0-\ln\varepsilon)\\ &= +\infty. \end{aligned}

Since $\int_{0}^{1} d x / x$ diverges, the ordinary improper integral $\int_{- 1}^{1} d x / x$ diverges. One divergent half is enough to sink the whole thing.

Cauchy principal value. Now delete a symmetric gap of half-width $ε$ about the singularity:

P . V . \int_{- 1}^{1} \frac{d x}{x} = lim_{ε \to 0^{+}} [\int_{- 1}^{- ε} \frac{d x}{x} + \int_{ε}^{1} \frac{d x}{x}] .

Evaluate each piece exactly, keeping $ε$ as a symbol. On the left piece $x$ is negative, so the antiderivative is $\ln | x |$ :

\begin{aligned} \int_{-1}^{-\varepsilon}\frac{dx}{x} &= \Bigl[\ln|x|\Bigr]_{-1}^{-\varepsilon}\\ &= \ln\varepsilon-\ln 1\\ &= \ln\varepsilon,\\ \int_{\varepsilon}^{1}\frac{dx}{x} &= \Bigl[\ln x\Bigr]_{\varepsilon}^{1}\\ &= 0-\ln\varepsilon\\ &= -\ln\varepsilon. \end{aligned}

Their sum is $\ln ε + (- \ln ε) = 0$ for every $ε > 0$ , so the quantity whose limit we are taking is the constant $0$ . Taking the limit,

P . V . \int_{- 1}^{1} \frac{d x}{x} = 0.

The ordinary integral diverges, but the principal value exists and equals $0$ . The cancellation occurs because $1 / x$ is an odd function and the interval $[- 1, 1]$ is symmetric about $0$ . Notice that the two pieces were added before the limit was taken; taking limits first would have given the meaningless expression $- \infty + \infty$ .

Compute $P . V . \int_{- \infty}^{\infty} \frac{x}{1 + x^{2}} d x$ , and determine whether the ordinary integral converges.

Over the symmetric window $[- R, R]$ the shaded areas are equal in size and opposite in sign, so they cancel for every $R > 0$ .

Solution

Ordinary integral. The interval is doubly infinite, so we split at a convenient point, say $0$ , and require both halves to converge. For the right half, substitute $u = 1 + x^{2}$ , or simply recognise the antiderivative $\frac{1}{2} \ln (1 + x^{2})$ :

\begin{aligned} \int_{0}^{\infty}\frac{x}{1+x^{2}}\,dx &= \lim_{b\to\infty}\Bigl[\tfrac{1}{2}\ln(1+x^{2})\Bigr]_{0}^{b}\\ &= \lim_{b\to\infty}\tfrac{1}{2}\ln(1+b^{2})\\ &= \infty. \end{aligned}

The integral $\int_{0}^{\infty}$ diverges, so the ordinary improper integral over $(- \infty, \infty)$ diverges. The integrand behaves like $1 / x$ for large $x$ , which is exactly the borderline case that fails the $p$ -test for the $p$ -integrals.

Cauchy principal value. Since $f (x) = x / (1 + x^{2})$ is an odd function, because

f (- x) = \frac{- x}{1 + (- x)^{2}} = - \frac{x}{1 + x^{2}} = - f (x),

the graph on $[- R, 0]$ is the graph on $[0, R]$ rotated through half a turn about the origin. The area below the axis on the left therefore matches the area above the axis on the right, and

\begin{aligned} \int_{-R}^{R}\frac{x}{1+x^{2}}\,dx &= \Bigl[\tfrac{1}{2}\ln(1+x^{2})\Bigr]_{-R}^{R}\\ &= \tfrac{1}{2}\ln(1+R^{2})-\tfrac{1}{2}\ln(1+R^{2})\\ &= 0 \end{aligned}

for every $R > 0$ . Taking the limit of the constant $0$ ,

P . V . \int_{- \infty}^{\infty} \frac{x}{1 + x^{2}} d x = lim_{R \to \infty} 0 = 0.

Once again the ordinary integral diverges while the principal value exists. The quantity $\frac{1}{2} \ln (1 + R^{2})$ marches off to infinity, but it is subtracted from itself at every stage, so we never see it.

The principal value of any odd function's integral over a symmetric interval $[- a, a]$ or over $(- \infty, \infty)$ is zero, whenever the limit exists. This does not mean the improper integral converges; as the example above shows, it may well diverge.
In physics and engineering a plain $\int$ is sometimes used loosely to mean the Cauchy principal value. When precision matters, write $P . V . \int$ or $- \int$ explicitly, to signal that only the symmetric limit is intended.

Where Principal Values Are Used

The Cauchy principal value is named after Augustin-Louis Cauchy (1789 to 1857), who introduced it in the context of complex analysis and singular integrals. Far from being a curiosity, it is the standard way of making sense of integrals that appear throughout applied mathematics.

Contour integration. Real integrals whose integrand has a pole on the path of integration are evaluated by indenting the contour with a small semicircle around the pole. The straight parts of the path are exactly the two pieces in part (b) of the definition, so what the residue calculation produces is a principal value.
Fourier analysis and the Hilbert transform. The Hilbert transform of a signal $u$ is defined by $P . V . \int_{- \infty}^{\infty} \frac{u (t)}{x - t} d t$ , up to a constant. The integrand has a $1 / (x - t)$ singularity right in the middle of the interval, and without the principal value the formula would be meaningless.
Signal processing and control. The Kramers and Kronig relations, which link the real and imaginary parts of a frequency response, are principal-value integrals. They are what lets an engineer recover the phase of a causal filter from its gain.
Quantum mechanics. Propagators and scattering amplitudes contain poles on the real axis. The identity attributed to Sokhotski and Plemelj, $lim_{ϵ \to 0^{+}} \frac{1}{x \pm i ϵ} = P . V . \frac{1}{x} \mp i π δ (x)$ , splits such an expression into a principal value plus a delta function, and both halves have physical meaning.
Aerodynamics and fracture mechanics. Thin-airfoil theory and crack-tip stress analysis both lead to singular integral equations whose kernels are principal-value integrals.

In every one of these settings the symbol $P . V .$ is doing real work: it records the extra convention, namely symmetric cancellation, that was needed before the expression meant anything at all.

Exercises

In each problem, cut symmetrically, integrate exactly in terms of $ε$ or $R$ , add the pieces, and only then take the limit. Where a problem asks about the ordinary improper integral as well, test each side of the trouble spot on its own.

Compute $P . V . \int_{- \infty}^{\infty} x d x$ . Does the ordinary improper integral converge?

Answer

$P . V . = 0$ ; the ordinary improper integral diverges.

Solution

Cauchy principal value. By definition:

\begin{aligned} \operatorname{P.V.}\!\int_{-\infty}^{\infty}x\,dx &= \lim_{R\to\infty}\int_{-R}^{R}x\,dx\\ &= \lim_{R\to\infty}\left[\frac{x^{2}}{2}\right]_{-R}^{R}\\ &= \lim_{R\to\infty}\left(\frac{R^{2}}{2}-\frac{R^{2}}{2}\right)\\ &= \lim_{R\to\infty}0=0. \end{aligned}

The Cauchy principal value equals $0$ .

Ordinary integral. The ordinary integral requires both halves to converge independently. For the right half:

\begin{aligned} \int_{0}^{\infty}x\,dx &= \lim_{b\to\infty}\left[\frac{x^{2}}{2}\right]_{0}^{b}\\ &= \lim_{b\to\infty}\frac{b^{2}}{2}\\ &= \infty. \end{aligned}

Since $\int_{0}^{\infty} x d x$ diverges, the ordinary improper integral $\int_{- \infty}^{\infty} x d x$ also diverges.

This example shows that the symmetric cancellation $R^{2} / 2 - R^{2} / 2 = 0$ can hide divergence. The Cauchy principal value uses the symmetric limit as a device to assign a finite number to a divergent integral, but that number is not the value of the integral in the ordinary sense.

Compute $P . V . \int_{- 2}^{2} \frac{d x}{x^{3}}$ . Does the ordinary improper integral converge?

Answer

$P . V . = 0$ ; the ordinary improper integral diverges.

Solution

Cauchy principal value. The singularity is at $x = 0$ , which is interior to $[- 2, 2]$ . By definition:

P . V . \int_{- 2}^{2} \frac{d x}{x^{3}} = lim_{ε \to 0^{+}} [\int_{- 2}^{- ε} \frac{d x}{x^{3}} + \int_{ε}^{2} \frac{d x}{x^{3}}] .

Using the antiderivative $\int x^{- 3} d x = - \frac{1}{2 x^{2}}$ :

\begin{aligned} \int_{-2}^{-\varepsilon}\frac{dx}{x^{3}} &= \left[-\frac{1}{2x^{2}}\right]_{-2}^{-\varepsilon}\\ &= -\frac{1}{2\varepsilon^{2}}+\frac{1}{8},\\ \int_{\varepsilon}^{2}\frac{dx}{x^{3}} &= \left[-\frac{1}{2x^{2}}\right]_{\varepsilon}^{2}\\ &= -\frac{1}{8}+\frac{1}{2\varepsilon^{2}}. \end{aligned}

Adding the two pieces:

(- \frac{1}{2 ε^{2}} + \frac{1}{8}) + (- \frac{1}{8} + \frac{1}{2 ε^{2}}) = 0.

This sum is $0$ for every $ε > 0$ , so:

P . V . \int_{- 2}^{2} \frac{d x}{x^{3}} = 0.

Ordinary integral. For the right half:

\begin{aligned} \int_{0}^{2}\frac{dx}{x^{3}} &= \lim_{\varepsilon\to0^{+}}\left[-\frac{1}{2x^{2}}\right]_{\varepsilon}^{2}\\ &= \lim_{\varepsilon\to0^{+}}\left(-\frac{1}{8}+\frac{1}{2\varepsilon^{2}}\right)\\ &= \infty. \end{aligned}

The ordinary integral diverges. The principal value is $0$ only because the symmetric cancellation of the $+ 1 / (2 ε^{2})$ and $- 1 / (2 ε^{2})$ terms hides the underlying divergence.

Let $f$ be an odd function, so $f (- x) = - f (x)$ for all $x \neq 0$ , continuous on $[- a, a]$ except at $x = 0$ , where $| f (x) | \to \infty$ . Show that $P . V . \int_{- a}^{a} f (x) d x = 0$ for every $a > 0$ . Why does this not imply that the ordinary integral $\int_{- a}^{a} f (x) d x$ converges?

Answer

The principal value is $0$ for every such $f$ , yet the ordinary integral may diverge, as $f (x) = 1 / x$ shows.

Solution

The hypothesis puts us in case (b) of the definition, with the singularity at the interior point $c = 0$ . By definition:

P . V . \int_{- a}^{a} f (x) d x = lim_{ε \to 0^{+}} [\int_{- a}^{- ε} f (x) d x + \int_{ε}^{a} f (x) d x] .

In the first integral, substitute $x = - t$ , so $d x = - d t$ . When $x = - a$ , $t = a$ ; when $x = - ε$ , $t = ε$ :

\begin{aligned} \int_{-a}^{-\varepsilon}f(x)\,dx &= \int_{a}^{\varepsilon}f(-t)\,(-dt)\\ &= \int_{a}^{\varepsilon}(-f(t))\,(-dt)\\ &= \int_{a}^{\varepsilon}f(t)\,dt\\ &= -\int_{\varepsilon}^{a}f(t)\,dt. \end{aligned}

We used the oddness $f (- t) = - f (t)$ , and then reversed the limits to absorb the minus sign. Substituting back:

\int_{- a}^{- ε} f (x) d x + \int_{ε}^{a} f (x) d x = - \int_{ε}^{a} f (t) d t + \int_{ε}^{a} f (x) d x = 0.

Since this sum is $0$ for every $ε > 0$ , taking the limit gives $P . V . \int_{- a}^{a} f (x) d x = 0$ .

Why the ordinary integral may still diverge. The ordinary integral requires each half to converge independently: $\int_{- a}^{0} f$ and $\int_{0}^{a} f$ must each be finite. Even if both diverge, one to $+ \infty$ and the other to $- \infty$ , the symmetric combination still cancels. For example, $f (x) = 1 / x$ is odd, and the divergence of $\int_{0}^{1} d x / x$ , established for the Type II improper integrals, shows that the ordinary integral $\int_{- 1}^{1} d x / x$ diverges. Yet by the first worked example above, $P . V . \int_{- 1}^{1} d x / x = 0$ .

Evaluate each principal value, and in each case say whether the ordinary improper integral converges.

$P . V . \int_{- 1}^{1} \frac{d x}{x^{3}}$
$P . V . \int_{- a}^{a} \frac{d x}{x}$ , where $a > 0$ is any constant

Answer

(a) $0$ ; the ordinary integral diverges. (b) $0$ for every $a > 0$ ; the ordinary integral diverges.

Solution

(a) The singularity is at $x = 0$ , interior to $[- 1, 1]$ . With the antiderivative $- 1 / (2 x^{2})$ :

\begin{aligned} \int_{-1}^{-\varepsilon}\frac{dx}{x^{3}} &= \left[-\frac{1}{2x^{2}}\right]_{-1}^{-\varepsilon}\\ &= -\frac{1}{2\varepsilon^{2}}+\frac{1}{2},\\ \int_{\varepsilon}^{1}\frac{dx}{x^{3}} &= \left[-\frac{1}{2x^{2}}\right]_{\varepsilon}^{1}\\ &= -\frac{1}{2}+\frac{1}{2\varepsilon^{2}}. \end{aligned}

The sum is $0$ for every $ε > 0$ , so

P . V . \int_{- 1}^{1} \frac{d x}{x^{3}} = lim_{ε \to 0^{+}} 0 = 0.

The ordinary integral diverges, since $\int_{0}^{1} d x / x^{3} = lim_{ε \to 0^{+}} (- \frac{1}{2} + \frac{1}{2 ε^{2}}) = \infty$ . Compare this with the interval $[- 2, 2]$ treated earlier: only the finite constants $\frac{1}{2}$ and $\frac{1}{8}$ changed, and they cancel too, so the answer is $0$ on any symmetric interval.

(b) Fix $a > 0$ and take $0 < ε < a$ . Then

\begin{aligned} \int_{-a}^{-\varepsilon}\frac{dx}{x} &= \Bigl[\ln|x|\Bigr]_{-a}^{-\varepsilon}\\ &= \ln\varepsilon-\ln a,\\ \int_{\varepsilon}^{a}\frac{dx}{x} &= \Bigl[\ln x\Bigr]_{\varepsilon}^{a}\\ &= \ln a-\ln\varepsilon. \end{aligned}

Adding, the $\ln ε$ terms cancel and so do the $\ln a$ terms:

(\ln ε - \ln a) + (\ln a - \ln ε) = 0.

Hence $P . V . \int_{- a}^{a} d x / x = 0$ for every $a > 0$ , which agrees with the general odd-function result. The ordinary integral diverges for every $a > 0$ , because $\int_{0}^{a} d x / x = lim_{ε \to 0^{+}} (\ln a - \ln ε) = + \infty$ .

Compute $P . V . \int_{- 1}^{2} \frac{d x}{x}$ . What does the answer show about the requirement that the interval be symmetric?

Answer

$\ln 2 \approx 0.693147$ . The interval need not be symmetric about the singularity; only the deleted gap must be.

Solution

The singularity is at $c = 0$ , which lies inside $[- 1, 2]$ even though the interval extends further to the right than to the left. The definition still applies: delete $(- ε, ε)$ and take $ε \to 0^{+}$ . Take $0 < ε < 1$ so that both remaining pieces are non-degenerate.

P . V . \int_{- 1}^{2} \frac{d x}{x} = lim_{ε \to 0^{+}} [\int_{- 1}^{- ε} \frac{d x}{x} + \int_{ε}^{2} \frac{d x}{x}] .

The two pieces are

\begin{aligned} \int_{-1}^{-\varepsilon}\frac{dx}{x} &= \Bigl[\ln|x|\Bigr]_{-1}^{-\varepsilon}\\ &= \ln\varepsilon-\ln 1\\ &= \ln\varepsilon,\\ \int_{\varepsilon}^{2}\frac{dx}{x} &= \Bigl[\ln x\Bigr]_{\varepsilon}^{2}\\ &= \ln 2-\ln\varepsilon. \end{aligned}

Adding them, the terms $\ln ε$ and $- \ln ε$ cancel and a constant survives:

\begin{aligned} \ln\varepsilon+(\ln 2-\ln\varepsilon) &= \ln 2. \end{aligned}

The sum equals $\ln 2$ for every admissible $ε$ , so

P . V . \int_{- 1}^{2} \frac{d x}{x} = lim_{ε \to 0^{+}} \ln 2 = \ln 2 \approx 0.693147 .

What matters is that the gap is symmetric about the singularity, not that the interval is. The unbalanced part of the interval, namely $[1, 2]$ , contributes the ordinary convergent integral $\int_{1}^{2} d x / x = \ln 2$ , while $[- 1, 1]$ contributes $0$ by the first worked example. Splitting the calculation that way gives the same answer:

P . V . \int_{- 1}^{2} \frac{d x}{x} = P . V . \int_{- 1}^{1} \frac{d x}{x} + \int_{1}^{2} \frac{d x}{x} = 0 + \ln 2 = \ln 2.

The ordinary improper integral still diverges, since $\int_{0}^{2} d x / x = + \infty$ .

Compute $P . V . \int_{- \infty}^{\infty} \frac{x d x}{x^{2} + 1}$ and $P . V . \int_{- \infty}^{\infty} \frac{x d x}{x^{2} + 4}$ . Do the two answers differ? What does that say about subtracting one from the other?

Answer

Both principal values equal $0$ , and both ordinary integrals diverge. Their difference, $\int_{- \infty}^{\infty} (\frac{x}{x^{2} + 1} - \frac{x}{x^{2} + 4}) d x$ , converges in the ordinary sense to $0$ .

Solution

First integral. An antiderivative of $x / (x^{2} + 1)$ is $\frac{1}{2} \ln (x^{2} + 1)$ , so

\begin{aligned} \int_{-R}^{R}\frac{x\,dx}{x^{2}+1} &= \left[\tfrac{1}{2}\ln(x^{2}+1)\right]_{-R}^{R}\\ &= \tfrac{1}{2}\ln(R^{2}+1)-\tfrac{1}{2}\ln(R^{2}+1)\\ &= 0, \end{aligned}

because $x^{2}$ takes the same value at $x = R$ and $x = - R$ . Hence the principal value is $lim_{R \to \infty} 0 = 0$ .

Second integral. Identical reasoning with the antiderivative $\frac{1}{2} \ln (x^{2} + 4)$ :

\begin{aligned} \int_{-R}^{R}\frac{x\,dx}{x^{2}+4} &= \left[\tfrac{1}{2}\ln(x^{2}+4)\right]_{-R}^{R}\\ &= 0, \end{aligned}

so this principal value is $0$ as well. Both integrands are odd, and the odd-function result predicts $0$ in both cases without any computation.

Neither ordinary integral converges. For large $x$ both integrands behave like $1 / x$ , and $\int_{1}^{\infty} d x / x$ diverges. Explicitly, $\int_{0}^{b} x d x / (x^{2} + 1) = \frac{1}{2} \ln (b^{2} + 1) \to \infty$ .

The difference. The two principal values are equal, so their difference is $0$ . This time, however, the difference is a genuinely convergent integral, because the leading $1 / x$ behaviour cancels:

\frac{x}{x^{2} + 1} - \frac{x}{x^{2} + 4} = \frac{x [(x^{2} + 4) - (x^{2} + 1)]}{(x^{2} + 1) (x^{2} + 4)} = \frac{3 x}{(x^{2} + 1) (x^{2} + 4)},

which decays like $3 / x^{3}$ . On $[0, \infty)$ , substituting $u = x^{2}$ and using partial fractions,

\begin{aligned} \int_{0}^{\infty}\frac{3x\,dx}{(x^{2}+1)(x^{2}+4)} &= \frac{3}{2}\int_{0}^{\infty}\frac{du}{(u+1)(u+4)}\\ &= \frac{1}{2}\int_{0}^{\infty}\left(\frac{1}{u+1}-\frac{1}{u+4}\right)du\\ &= \frac{1}{2}\left[\ln\frac{u+1}{u+4}\right]_{0}^{\infty}\\ &= \frac{1}{2}\left(0-\ln\tfrac{1}{4}\right)\\ &= \ln 2, \end{aligned}

a finite number. The integrand is odd, so the left half contributes $- \ln 2$ and the ordinary integral over $(- \infty, \infty)$ converges to $0$ . The moral: two expressions with the same principal value can differ by something that is a perfectly ordinary convergent integral, and the divergent parts, here $\frac{1}{2} \ln (R^{2} + 1)$ and $\frac{1}{2} \ln (R^{2} + 4)$ , cancel each other in the limit.

Compute $P . V . \int_{- \infty}^{\infty} \sin x d x$ . Does the ordinary improper integral converge?

Answer

$P . V . = 0$ ; the ordinary improper integral diverges.

Solution

Cauchy principal value. The integrand is continuous everywhere, so the only trouble is at the two infinite ends. Truncate symmetrically:

\begin{aligned} \int_{-R}^{R}\sin x\,dx &= \Bigl[-\cos x\Bigr]_{-R}^{R}\\ &= -\cos R+\cos(-R)\\ &= -\cos R+\cos R\\ &= 0, \end{aligned}

using $\cos (- R) = \cos R$ . This holds for every $R > 0$ , so

P . V . \int_{- \infty}^{\infty} \sin x d x = lim_{R \to \infty} 0 = 0.

Ordinary integral. Each half must converge separately. For the right half,

\begin{aligned} \int_{0}^{b}\sin x\,dx &= \Bigl[-\cos x\Bigr]_{0}^{b}\\ &= 1-\cos b. \end{aligned}

As $b \to \infty$ , $\cos b$ oscillates between $- 1$ and $1$ and approaches no limit, so $1 - \cos b$ oscillates between $0$ and $2$ and the half-line integral diverges by oscillation. Therefore the ordinary improper integral $\int_{- \infty}^{\infty} \sin x d x$ diverges.

This is the same phenomenon as in the earlier exercises, but with oscillation rather than growth: $\sin x$ is odd, so every hump above the axis on the right is matched by a hump below the axis on the left, and the symmetric truncation cancels them in pairs.

Compute $P . V . \int_{0}^{2} \frac{d x}{x - 1}$ .

Answer

$0$ . The ordinary improper integral diverges.

Solution

The integrand blows up at $c = 1$ , which is the midpoint of $[0, 2]$ . Delete the symmetric gap $(1 - ε, 1 + ε)$ with $0 < ε < 1$ :

P . V . \int_{0}^{2} \frac{d x}{x - 1} = lim_{ε \to 0^{+}} [\int_{0}^{1 - ε} \frac{d x}{x - 1} + \int_{1 + ε}^{2} \frac{d x}{x - 1}] .

An antiderivative is $\ln | x - 1 |$ , and the absolute value is essential on the left piece, where $x - 1 < 0$ :

\begin{aligned} \int_{0}^{1-\varepsilon}\frac{dx}{x-1} &= \Bigl[\ln|x-1|\Bigr]_{0}^{1-\varepsilon}\\ &= \ln\varepsilon-\ln 1\\ &= \ln\varepsilon,\\ \int_{1+\varepsilon}^{2}\frac{dx}{x-1} &= \Bigl[\ln|x-1|\Bigr]_{1+\varepsilon}^{2}\\ &= \ln 1-\ln\varepsilon\\ &= -\ln\varepsilon. \end{aligned}

Their sum is $\ln ε - \ln ε = 0$ for every admissible $ε$ , so

P . V . \int_{0}^{2} \frac{d x}{x - 1} = 0.

This is the first worked example translated one unit to the right: the substitution $u = x - 1$ turns the integral into $P . V . \int_{- 1}^{1} d u / u$ . The ordinary integral diverges, since $\int_{1}^{2} d x / (x - 1) = lim_{ε \to 0^{+}} (- \ln ε) = + \infty$ .

Compute $P . V . \int_{- \infty}^{\infty} \frac{d x}{1 + x^{2}}$ , and show that the ordinary improper integral converges to the same number.

Answer

Both equal $π$ .

Solution

Cauchy principal value. An antiderivative of $1 / (1 + x^{2})$ is $\arctan x$ , so

\begin{aligned} \int_{-R}^{R}\frac{dx}{1+x^{2}} &= \Bigl[\arctan x\Bigr]_{-R}^{R}\\ &= \arctan R-\arctan(-R)\\ &= 2\arctan R. \end{aligned}

Since $\arctan R \to π / 2$ as $R \to \infty$ ,

P . V . \int_{- \infty}^{\infty} \frac{d x}{1 + x^{2}} = lim_{R \to \infty} 2 \arctan R = 2 \cdot \frac{π}{2} = π .

Ordinary integral. Split at $0$ and test each half on its own:

\begin{aligned} \int_{0}^{\infty}\frac{dx}{1+x^{2}} &= \lim_{b\to\infty}\Bigl[\arctan x\Bigr]_{0}^{b}\\ &= \lim_{b\to\infty}\arctan b\\ &= \frac{\pi}{2},\\ \int_{-\infty}^{0}\frac{dx}{1+x^{2}} &= \lim_{a\to-\infty}\Bigl[\arctan x\Bigr]_{a}^{0}\\ &= \lim_{a\to-\infty}(-\arctan a)\\ &= \frac{\pi}{2}. \end{aligned}

Both halves converge, so the ordinary improper integral converges and equals $\frac{π}{2} + \frac{π}{2} = π$ .

This is the expected outcome: whenever the ordinary integral converges, the principal value exists and agrees with it, so writing $P . V .$ in front of a convergent integral changes nothing. The symbol earns its keep only when the ordinary integral fails.

Compute $P . V . \int_{- 1}^{3} \frac{d x}{x - 1}$ .

Answer

$0$ . The ordinary improper integral diverges.

Solution

The singularity is at $c = 1$ . It is not at the origin, so the first thing to check is where it sits inside $[- 1, 3]$ : the distance to the left endpoint is $1 - (- 1) = 2$ and the distance to the right endpoint is $3 - 1 = 2$ . The interval is symmetric about the singularity, which is what matters.

Delete $(1 - ε, 1 + ε)$ with $0 < ε < 2$ :

\begin{aligned} \int_{-1}^{1-\varepsilon}\frac{dx}{x-1} &= \Bigl[\ln|x-1|\Bigr]_{-1}^{1-\varepsilon}\\ &= \ln\varepsilon-\ln 2,\\ \int_{1+\varepsilon}^{3}\frac{dx}{x-1} &= \Bigl[\ln|x-1|\Bigr]_{1+\varepsilon}^{3}\\ &= \ln 2-\ln\varepsilon. \end{aligned}

Adding, both the $\ln ε$ terms and the $\ln 2$ terms cancel:

(\ln ε - \ln 2) + (\ln 2 - \ln ε) = 0,

P . V . \int_{- 1}^{3} \frac{d x}{x - 1} = 0.

Equivalently, substitute $u = x - 1$ : the integral becomes $P . V . \int_{- 2}^{2} d u / u$ , which is $0$ by the general odd-function result with $a = 2$ . The ordinary integral diverges, because $\int_{1}^{3} d x / (x - 1) = + \infty$ .

Contrast this with the earlier asymmetric example $P . V . \int_{- 1}^{2} d x / x = \ln 2$ : there the singularity sat off-centre and a constant survived the cancellation. Here it sits dead centre and nothing survives.

Answer both parts.

Suppose $P . V . \int_{a}^{b} f$ and $P . V . \int_{a}^{b} g$ both exist, with the same singular point $c$ , and let $α, β$ be constants. Show that $P . V . \int_{a}^{b} (α f + β g)$ exists and equals $α P . V . \int_{a}^{b} f + β P . V . \int_{a}^{b} g$ .
Show by example that $P . V . \int_{a}^{b} (f + g)$ can exist even when neither $P . V . \int_{a}^{b} f$ nor $P . V . \int_{a}^{b} g$ exists.

Answer

(a) Follows from linearity of the ordinary integral and of limits. (b) On $[- 1, 1]$ take $f (x) = 1 / x^{2}$ and $g (x) = 1 / x - 1 / x^{2}$ : neither principal value exists, but $P . V . \int_{- 1}^{1} (f + g) = P . V . \int_{- 1}^{1} d x / x = 0$ .

Solution

(a) For each $ε > 0$ small enough, write

I_{ε} (h) = \int_{a}^{c - ε} h (x) d x + \int_{c + ε}^{b} h (x) d x,

so that $P . V . \int_{a}^{b} h = lim_{ε \to 0^{+}} I_{ε} (h)$ whenever the limit exists. Ordinary definite integrals are linear on each of the two pieces, so

\begin{aligned} I_{\varepsilon}(\alpha f+\beta g) &= \int_{a}^{c-\varepsilon}(\alpha f+\beta g)+\int_{c+\varepsilon}^{b}(\alpha f+\beta g)\\ &= \alpha\left(\int_{a}^{c-\varepsilon}f+\int_{c+\varepsilon}^{b}f\right)+\beta\left(\int_{a}^{c-\varepsilon}g+\int_{c+\varepsilon}^{b}g\right)\\ &= \alpha I_{\varepsilon}(f)+\beta I_{\varepsilon}(g). \end{aligned}

By hypothesis $I_{ε} (f)$ and $I_{ε} (g)$ have limits as $ε \to 0^{+}$ . The limit of a linear combination is the linear combination of the limits, so $I_{ε} (α f + β g)$ converges and

P . V . \int_{a}^{b} (α f + β g) = α P . V . \int_{a}^{b} f + β P . V . \int_{a}^{b} g .

(b) Work on $[- 1, 1]$ with the singularity at $c = 0$ , and take

f (x) = \frac{1}{x^{2}}, g (x) = \frac{1}{x} - \frac{1}{x^{2}} .

For $f$ , using the antiderivative $- 1 / x$ ,

\begin{aligned} I_{\varepsilon}(f) &= \left[-\frac{1}{x}\right]_{-1}^{-\varepsilon}+\left[-\frac{1}{x}\right]_{\varepsilon}^{1}\\ &= \left(\frac{1}{\varepsilon}-1\right)+\left(-1+\frac{1}{\varepsilon}\right)\\ &= \frac{2}{\varepsilon}-2, \end{aligned}

which tends to $+ \infty$ . So $P . V . \int_{- 1}^{1} f$ does not exist. For $g$ , the $1 / x$ part contributes $0$ by the first worked example, and the $- 1 / x^{2}$ part contributes the negative of the quantity just computed:

I_{ε} (g) = 0 - (\frac{2}{ε} - 2) = 2 - \frac{2}{ε} ⟶ - \infty,

so $P . V . \int_{- 1}^{1} g$ does not exist either. Yet

f (x) + g (x) = \frac{1}{x^{2}} + \frac{1}{x} - \frac{1}{x^{2}} = \frac{1}{x},

and $P . V . \int_{- 1}^{1} d x / x = 0$ exists.

The point of part (a) is that linearity holds only when the pieces are already known to exist. Part (b) shows that the hypothesis cannot be dropped: splitting a principal value into two principal values is legitimate only after each half has been checked.

Let $f$ be continuous on $[a, b] ∖ {c}$ with an infinite discontinuity at the interior point $c$ . Prove that if the ordinary improper integral $\int_{a}^{b} f (x) d x$ converges, then $P . V . \int_{a}^{b} f (x) d x$ exists and equals it. Then explain why the converse is false.

Answer

Take $δ = η = ε$ in the two independent limits; the sum of the limits is the limit of the sum. The converse fails: $P . V . \int_{- 1}^{1} d x / x = 0$ while $\int_{- 1}^{1} d x / x$ diverges.

Solution

Suppose $\int_{a}^{b} f$ converges in the ordinary sense. By definition this means that the two one-sided limits

L_{1} = lim_{δ \to 0^{+}} \int_{a}^{c - δ} f (x) d x, L_{2} = lim_{η \to 0^{+}} \int_{c + η}^{b} f (x) d x

both exist as finite numbers, and that $\int_{a}^{b} f (x) d x = L_{1} + L_{2}$ .

Now form the principal-value expression with a single parameter $ε$ :

I_{ε} = \int_{a}^{c - ε} f (x) d x + \int_{c + ε}^{b} f (x) d x .

The first term is the function of $δ$ above evaluated at $δ = ε$ , and the second is the function of $η$ evaluated at $η = ε$ . A limit that exists as $δ \to 0^{+}$ certainly exists along the particular choice $δ = ε \to 0^{+}$ , so each term separately has a limit, namely $L_{1}$ and $L_{2}$ . Since both limits exist, the limit of the sum is the sum of the limits:

\begin{aligned} \operatorname{P.V.}\!\int_{a}^{b}f(x)\,dx &= \lim_{\varepsilon\to0^{+}}I_{\varepsilon}\\ &= \lim_{\varepsilon\to0^{+}}\int_{a}^{c-\varepsilon}f(x)\,dx+\lim_{\varepsilon\to0^{+}}\int_{c+\varepsilon}^{b}f(x)\,dx\\ &= L_{1}+L_{2}\\ &= \int_{a}^{b}f(x)\,dx. \end{aligned}

So the principal value exists and equals the integral. The identical argument with $δ = η = ε$ replaced by $a = - R$ , $b = R$ proves the corresponding statement for doubly infinite integrals.

Why the converse fails. The step that cannot be reversed is "both limits exist separately, so the limit of the sum splits". If the pieces do not have individual limits, the sum may still settle down because the divergences cancel, and then no conclusion about the pieces can be drawn. The standard counterexample is $f (x) = 1 / x$ on $[- 1, 1]$ : the pieces are $\ln ε \to - \infty$ and $- \ln ε \to + \infty$ , neither of which converges, yet their sum is identically $0$ . Hence $P . V . \int_{- 1}^{1} d x / x = 0$ exists while $\int_{- 1}^{1} d x / x$ diverges.

Show that $P . V . \int_{- 1}^{1} \frac{d x}{x^{2}}$ does not exist. Why does the cancellation that saved $\int_{- 1}^{1} d x / x$ not help here?

Answer

The principal value does not exist: the symmetric sum equals $\frac{2}{ε} - 2 \to + \infty$ . The integrand is even and positive, so the two pieces add instead of cancelling.

Solution

The singularity is at $x = 0$ , interior to $[- 1, 1]$ . Delete the symmetric gap and use the antiderivative $\int x^{- 2} d x = - \frac{1}{x} + C$ :

\begin{aligned} \int_{-1}^{-\varepsilon}\frac{dx}{x^{2}} &= \left[-\frac{1}{x}\right]_{-1}^{-\varepsilon}\\ &= \frac{1}{\varepsilon}-1,\\ \int_{\varepsilon}^{1}\frac{dx}{x^{2}} &= \left[-\frac{1}{x}\right]_{\varepsilon}^{1}\\ &= -1+\frac{1}{\varepsilon}\\ &= \frac{1}{\varepsilon}-1. \end{aligned}

The two pieces are equal, not opposite, so adding them doubles rather than destroys:

I_{ε} = (\frac{1}{ε} - 1) + (\frac{1}{ε} - 1) = \frac{2}{ε} - 2.

As $ε \to 0^{+}$ this increases without bound, so the limit does not exist as a finite number and

P . V . \int_{- 1}^{1} \frac{d x}{x^{2}} does not exist.

Why the cancellation fails. Cancellation needs opposite signs. The function $1 / x$ is odd, so the area to the left of the gap is negative and the area to the right is positive, and the two destroy each other. The function $1 / x^{2}$ is even and strictly positive, so both areas are positive and they reinforce each other. A symmetric limit can only cancel what has opposite signs; it can never tame a positive integrand.

This also settles a question raised earlier. Writing $[- 1 / x]_{- 1}^{1} = - 1 - 1 = - 2$ produces a negative number for the integral of a positive function, which is absurd. The Fundamental Theorem simply does not apply across the singularity, and here not even the principal value rescues the expression: both the ordinary integral and the principal value fail to exist.

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