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A rational-kernel Frullani integral

Example

For a,b>0a,b>0, ∫0∞(1+ax)−1−(1+bx)−1x dx=∫abdtt.\int_0^\infty\frac{(1+ax)^{-1}-(1+bx)^{-1}}x\,dx=\int_a^b\frac{dt}{t}.

Facts & Assumptions

Given: Positive a,ba,b and f(t)=1/(1+t)f(t)=1/(1+t).

[L1]

Frullani's formula gives (f(0)−L)∫abdt/t(f(0)-L)\int_a^b dt/t when f(t)→Lf(t)\to L (Frullani's formula with its proper integral factor).

[L2]

The pp-test gives convergence of ∫1∞x−2dx\int_1^\infty x^{-2}dx (The improper pp-test for rational exponents).

Verification

technique · computation
1.1

Here ff is continuous and f(0)=1f(0)=1. Also 0<f(t)=1/(1+t)<ε0<f(t)=1/(1+t)<\varepsilon whenever t>1/εt>1/\varepsilon, so f(t)→0f(t)\to0 directly from the definition of the limit at infinity. Thus [L1] gives exactly the displayed identity.

L1
1.2

For x>0x>0, the integrand simplifies to [L2] b−a(1+ax)(1+bx).\frac{b-a}{(1+ax)(1+bx)}. It has finite limit b−ab-a at zero and is bounded in absolute value by a constant multiple of x−2x^{-2} for x≥1x\ge1. This independently confirms local convergence at zero and tail convergence by [L2], without replacing the proper factor by a logarithm. ∎

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