Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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1/x1/x on [−1,1][-1,1] has principal value 00 but no improper integral

Example

The function f(x)=1/xf(x)=1/x on [−1,1]∖{0}[-1,1]\setminus\{0\} has PV⁡ ⁣∫−11dxx=0,\operatorname{PV}\!\int_{-1}^1\frac{dx}{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 pp-test at p=1p=1 says ∫01x−1dx\int_0^1x^{-1}dx diverges (The improper pp-test for rational exponents).

Verification

technique · computation
1.1

Substitution x=−tx=-t gives [L1] ∫−1−εdxx=−∫ε1dtt.\int_{-1}^{-\varepsilon}\frac{dx}{x}=-\int_\varepsilon^1\frac{dt}{t}. Thus the symmetric sum is exactly zero for every ε>0\varepsilon>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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 79 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources