Separate limits at every infinite endpoint and singularity

Goal and scope

Compute three explicit power-law improper integrals and distinguish principal value.

Only complete piecewise power laws, not general integrability from finite cutoffs.

Givens and symbols

• algebra
• epsilon
• FTC
• definition_here
• log_integral_boundary

• G1 — One side: f(x)=3/x^2 for x≥2. Two sides: g(x)=2/|x|^2 for x≤-1, zero for -1<x<1, and 3/x^2 for x≥1.
• G2 — Singularity example: h(x)=(-x)^(-1/2) for -1≤x<0 and 2*x^(-1/2) for 0<x≤1, split at zero.

• A,B,b — independent infinite-end cutoffs
• \delta — one-sided singular cutoff
• \operatorname{PV} — symmetric principal value

Rules and sources

• mko_definition · cited — A one-sided integral int_a^infinity f, from a to infinity, is the limit of finite-cutoff integrals as b→infinity; this slice declares f(x)=scale/x^p on [lower,infinity). The two-sided exact family declares f(x)=left_scale/|x|^p on x≤−1, f(x)=0 on −1<x<1 and f(x)=right_scale/x^p on x≥1; its tails must converge separately. An interior-singularity family explicitly gives both power-law pieces on [c−1,c+1] and splits at c. A symmetric principal value is insufficient. Absolute convergence is sufficient but not necessary, and apparently stable finite cutoffs do not prove convergence. · mko-improper-integral
• algebra · prerequisite — Use real-field identities; division requires a nonzero denominator and multiplication by a positive number preserves inequalities.
• epsilon · prerequisite — Convergence requires for every positive epsilon an integer N controlling every n≥N by a strict error bound; use the Archimedean property of the reals.
• FTC · prerequisite — For continuous power laws on finite closed intervals use the fundamental theorem of calculus; the derivative of -c/x is c/x^2, and that of 2sqrt(x) is x^(-1/2) for x>0.
• definition_here · prerequisite — An ordinary improper integral requires a finite limit of finite-cutoff integrals separately at each infinite endpoint and each side of an interior singularity, then adds them; principal value separately binds symmetric cutoffs.
• log_integral_boundary · prerequisite — For nonzero reals the derivative of log|x| is 1/x, and log B tends to infinity as B tends to infinity.

Main derivation

b>2

\int_2^b3x^{-2}\,dx=[-3/x]_2^b=3/2-3/b

3/b\to0\quad(b\to\infty)

\int_2^\infty3x^{-2}\,dx=3/2

A>1,\ B>1

\int_{-A}^{-1}g=2-2/A,\quad\int_1^B g=3-3/B

A\to\infty\text{ independently of }B\to\infty

\int_{-\infty}^{\infty}g=2+0+3=5

0<\delta<1

\int_{-1}^{-\delta}h=2-2\sqrt\delta,\quad\int_{\delta}^{1}h=4-4\sqrt\delta

\sqrt\delta\to0\quad(\delta\downarrow0)

\int_{-1}^{1}h=2+4=6

j(x)=1/x\quad(|x|\ge1),\qquad j(x)=0\quad(|x|<1)

\int_{-A}^{-1}j=-\log A,\quad\int_1^B j=\log B

A=B\Longrightarrow-\log A+\log A=0

\operatorname{PV}\int j=0\text{ but the ordinary improper integral does not exist}

Substeps and lemmas

No substeps in this guide.

Reverse audits

• S08 · premise_audit · S07 — The ordinary integral is rejected without denying the symmetric limit is zero.
• S07 · premise_audit · G1, log_integral_boundary — Only the stated logarithmic-antiderivative premise for this counterexample is used, not a new exp/log theory.
• S06 · premise_audit · S05 — A missing value at zero is not the obstruction; the one-sided integrability limits are the obligations.
• S05 · premise_audit · G2 — Using one symbol delta is convenient notation; the two one-sided limits must still exist separately.
• S04 · premise_audit · S03 — There is no symmetric cancellation; each side is independently qualified.
• S03 · premise_audit · G1 — Qualification must not rely on tying A and B together.
• S02 · premise_audit · S01 — This is not a conclusion from several close cutoff values.
• S01 · premise_audit · G1 — b>2 keeps the integrand continuous and denominators positive.

Error controls and negative controls

Separate endpoint accounts

Record each infinite endpoint and singularity side separately.

Principal-value separation

Symmetric cutoffs certify principal value, not independent tails.

• Divergent sides cancel into an ordinary integral. → reject — Neither one-sided limit of j is finite.
• Only one side of an interior singularity must converge. → reject — An ordinary improper integral requires both one-sided limits at each singularity.
• Several close large-cutoff values prove convergence. → reject — No bound for every later cutoff is supplied.

Result and limitations

The three qualified improper integrals are 3/2, 5 and 6; contrast j has only symmetric principal value zero.

Verification and source binding

1 declared checks for this guide; only the listed expressions are covered.

• source commit: b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f
• adopted source SHA-256: fb00245eacaaeb6218a92f36d27aed6067e63e73e73270089daeadae5933b7f9
• original source SHA-256: fb00245eacaaeb6218a92f36d27aed6067e63e73e73270089daeadae5933b7f9
• Lean: NOT_RUN · independent AI: SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW

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