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Convergence range of $x^{-p}(1+x)^{-q}$ on $(0,\infty)$ for rational exponents

Example

For rational $p,q$ , the positive integral $\int_0^\infty x^{-p}(1+x)^{-q}dx$ converges exactly when $p<1\quad\text{and}\quad p+q>1.$

Facts & Assumptions

Given: Rational exponents $p,q$ and the positive-domain kernel.

[L1]

Positive rational powers obey the product, quotient, and monotonicity laws (Rational powers $a^r$ of a positive base, Laws of rational exponents, Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ ).

[L2]

Two-sided eventual comparison by positive constant multiples gives equivalent improper convergence, using comparison in each direction and linearity for the constant multiples (Comparison tests for improper integrals, Linearity of convergent improper integrals).

[L3]

The rational $p$ -test gives the exact thresholds at zero and infinity (The improper $p$ -test for rational exponents).

Verification

technique · direct

1.1

For $0<x\le1$ , one has $1\le1+x\le2$ . If $q\ge0$ , [L1] gives $2^{-q}\le(1+x)^{-q}\le1$ ; if $q<0$ , it gives $1\le(1+x)^{-q}\le2^{-q}$ . Thus the kernel is bounded above and below by positive constant multiples of $x^{-p}$ , so [L2] and [L3] give convergence at zero exactly when $p<1$ .

L1L2L3

1.2

For $x\ge1$ , $1/2\le x/(1+x)<1$ . By [L1], the quotient of the kernel by $x^{-(p+q)}$ is $(x/(1+x))^q$ and lies between $2^{-q}$ and $1$ when $q\ge0$ , and between $1$ and $2^{-q}$ when $q<0$ . Hence [L2] and [L3] give convergence at infinity exactly when $p+q>1$ .

L1L2L3

2.1

The mixed definition requires both ends separately, so the full integral converges exactly under the two simultaneous inequalities. Positive bases ensure every rational power used above is defined, regardless of the signs of $p$ and $q$ .

given∎

Depends on

Used by

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Sources

William F. Trench, Introduction to Real Analysis, comparison-test examples (standard reference, not scraped)

Convergence range of x−p(1+x)−qx^{-p}(1+x)^{-q}x−p(1+x)−q on (0,∞)(0,\infty)(0,∞) for rational exponents