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$1 / x$ on $[- 1, 1]$ has principal value $0$ but no improper integral

Example

The function $f (x) = 1 / x$ on $[- 1, 1] ∖ {0}$ has $PV ⁣ \int_{- 1}^{1} \frac{d x}{x} = 0,$ but its two one-sided improper integrals do not converge.

Facts & Assumptions

Given: The reciprocal function away from zero.

[L1]

Principal value uses equal truncations on the two sides (Cauchy principal values at a finite singularity and on the real line).

[L2]

The rational $p$ -test at $p = 1$ says $\int_{0}^{1} x^{- 1} d x$ diverges (The improper $p$ -test for rational exponents).

Verification

technique · computation

1.1

Substitution $x = - t$ gives [L1] $\int_{- 1}^{- ε} \frac{d x}{x} = - \int_{ε}^{1} \frac{d t}{t} .$ Thus the symmetric sum is exactly zero for every $ε > 0$ , and [L1] gives principal value zero.

2.1

By [L2], the right-hand integral diverges to $+ \infty$ ; the identity in step 1.1 makes the left-hand one diverge to $- \infty$ . Hence separate improper convergence fails, showing that the converse of the principal-value theorem is false.

L2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

William F. Trench, Introduction to Real Analysis, Section 3.4 (standard reference, not scraped)

1/x1/x on [−1,1][−1,1] has principal value 00 but no improper integral