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A Dirichlet-type transfer criterion for divergence

Statement

Let $u$ be continuous on $[a,\infty)$ and suppose $\int_a^\infty u$ diverges. Let $v>0$ be differentiable, satisfy $v(x)\to\infty$ , and suppose $\int_a^\infty\frac{|v'(x)|}{v(x)^2}\,dx$ converges. Then $\int_a^\infty u(x)v(x)\,dx$ diverges.

Facts & Assumptions

Given: Functions $u,v$ satisfying the statement.

[L1]

The reciprocal rule gives $(1/v)'=-v'/v^2$ (Sums, scalar multiples, products and quotients: $(f+g)'(c) = f'(c) + g'(c)$ , $(\alpha f)'(c) = \alpha f'(c)$ , $(fg)'(c) = f'(c)g(c) + f(c)g'(c)$ , and $(f/g)'(c) = \bigl(f'(c)g(c) - f(c)g'(c)\bigr)/g(c)^{2}$ when $g(c) \ne 0$ ).

[L2]

The differentiable-multiplier clause of Dirichlet's test applies to a continuous function with bounded truncation primitive (Dirichlet's test for improper integrals).

[L3]

Convergence of an improper integral bounds its truncation primitive near infinity, while on the remaining compact interval the integral function is Lipschitz and hence bounded (Improper integrals over unbounded intervals, The integral function of a bounded integrable $f$ is Lipschitz, hence uniformly continuous).

Proof

technique · contradiction

1.1

Suppose for contradiction that $\int_a^\infty uv$ converges. Then its truncation primitive is bounded by [L3], and $uv$ is continuous. Put $g=1/v$ . Positivity and $v\to\infty$ give $g\to0$ : for $\varepsilon>0$ , eventually $v>1/\varepsilon$ , hence $0<g<\varepsilon$ . Also [L1] and the hypothesis give absolute convergence of $\int g'$ .

L3L1assume-contra

2.1

Apply [L2] with the continuous function $uv$ and multiplier $g$ . It yields convergence of $\int(uv)g=\int u$ , contradicting the hypothesis. Hence $\int uv$ diverges.

step 1.1L2discharge-contradiction∎

Depends on

Used by

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Dependency tree · next 3 levels

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Sources

William F. Trench, Introduction to Real Analysis, Theorem 3.4.11 (standard reference, not scraped)