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The substitution $x = 1 / t$ exchanges the two rational $p$ -tests

Example

For rational $p$ , the decreasing substitution $x = 1 / t$ gives $\int_{1}^{\infty} x^{- p} d x = \int_{0}^{1} t^{p - 2} d t$ whenever either improper integral converges, and convergence occurs exactly when $p > 1$ .

Facts & Assumptions

Given: The map $ϕ (t) = 1 / t$ from $(0, 1]$ onto $[1, \infty)$ .

[L1]

The reciprocal derivative is $ϕ^{'} (t) = - t^{- 2}$ (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[L2]

Rational exponent laws give $(1 / t)^{- p} t^{- 2} = t^{p - 2}$ (Laws of rational exponents).

[L3]

Improper substitution preserves convergence and value (Change of variable in an improper integral).

Verification

technique · computation

1.1

The map $ϕ$ is decreasing, so [L3] uses $∣ ϕ^{'} ∣ = t^{- 2}$ . By [L1] and [L2], the transformed integrand is $t^{p - 2}$ , proving the identity.

L3L1L2

2.1

Write $t^{p - 2} = t^{- (2 - p)}$ . The finite-endpoint $p$ -test says this converges exactly when $2 - p < 1$ , namely $p > 1$ , which is also precisely the infinite-endpoint threshold for the original integral.

given∎

Depends on

Used by

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Sources

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

The substitution x=1/tx=1/t exchanges the two rational pp-tests