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pp-test

Proposition

Let aa be a positive real number. The improper integral ∫a+∞t−p dt\displaystyle\int_{a}^{+\infty}t^{-p}\,\mathrm{d}t converges if and only if p>1;p>1\text{;} the improper integral ∫0at−p dt\displaystyle\int_{0}^{a}t^{-p}\,\mathrm{d}t converges if and only if p<1.p< 1\text{.}

Interpretation

The pp-test settles the improper integrals of powers t−pt^{-p}, which serve as the reference integrals in the comparison test. Towards infinity, t−pt^{-p} must decay fast enough, which needs p>1p>1; towards 00, it must not blow up too fast, which needs p<1p<1.

The curves 1 over x squared, 1/x and 1 over the square root of x from x = 1 onwards; only the area under 1 over x squared, which is shaded, is finite.

All three curves decrease to 00, but only the area under 1x2\frac{1}{x^2} is finite, and it equals 11. Under 1x\frac{1}{x} the area from 11 to bb is ln⁡b\ln b, and under 1x\frac{1}{\sqrt{x}} it is 2b−22\sqrt{b}-2; both grow without bound as b→∞b\to\infty.

The test is a direct computation. For p≠1p\ne1,

∫1bt−p dt=b1−p−11−p,\int_1^b t^{-p}\,\mathrm{d}t=\frac{b^{1-p}-1}{1-p},

which has a finite limit as b→∞b\to\infty exactly when 1−p<01-p<0. In the same way, ∫ε1t−p dt=1−ε1−p1−p\int_\varepsilon^1 t^{-p}\,\mathrm{d}t=\frac{1-\varepsilon^{1-p}}{1-p} has a finite limit as ε↓0\varepsilon\downarrow0 exactly when 1−p>01-p>0, and for p=1p=1 both limits are infinite because of ln⁡\ln. The value of a>0a>0 plays no role: moving it changes the integral only by an ordinary integral over a bounded interval.

The two conditions point in opposite directions, so no power is integrable over all of (0,∞)(0,\infty): ∫0∞t−p dt\int_0^\infty t^{-p}\,\mathrm{d}t diverges for every pp. Through the integral test, the first statement settles the pp-series, and the second shows where the integral defining the gamma function converges near 00.

Key Results

  • The convergence depends solely on the interval and the value of p.p\text{.}
  • Behaviour near infinity and near zero are treated separately and differently.

Examples

  • ∫1∞1x2 dx\displaystyle \int_1^\infty \frac{1}{x^2}\,\mathrm{d}x converges.
  • ∫011x dx\displaystyle \int_0^1 \frac{1}{x}\,\mathrm{d}x diverges.

Prerequisites

Sources

Aliases: p test Symbols: epsilon, infinity, integral, square root

Last updated: 20 October 2026 03:02