Improper integrals

Even when the interval of integration runs off to infinity, the area can still be finite1. We compare y=1x2y = \dfrac{1}{x^2} and y=1xy = \dfrac{1}{x} integrated from 11 onward.

When it converges

An infinite interval is defined by a limit. Integrate to a finite bb first, then let b→∞b \to \infty.

∫1bdxx2=[−1x]1b=1−1b⟶1\int_1^b \frac{dx}{x^2} = \left[ -\frac{1}{x} \right]_1^b = 1 - \frac{1}{b} \longrightarrow 1

The area settles at 11. The interval stretches without bound, but the curve falls so quickly that each new contribution is smaller than the last.

When it diverges

∫1bdxx=[log⁡x]1b=log⁡b⟶∞\int_1^b \frac{dx}{x} = \left[ \log x \right]_1^b = \log b \longrightarrow \infty

The area grows without bound. The growth of log⁡b\log b is slow, but it never stops.

bb∫1bdxx2\int_1^b \dfrac{dx}{x^2}∫1bdxx\int_1^b \dfrac{dx}{x}
10100.90.92.302.30
10310^30.9990.9996.916.91
10610^60.9999990.99999913.8213.82

The boundary is p=1p = 1

For y=1xpy = \dfrac{1}{x^p} integrated from 11 to ∞\infty, the fate is decided by pp.

Range of ppResult
p>1p > 1converges
p≤1p \leq 1diverges

Both curves approach the xx-axis on the graph, and yet only one converges. What matters is how fast they approach.

The type that blows up at an end

Even over a finite range, an end where the function diverges is treated the same way.

IntegralResult
∫01dxx\int_0^1 \dfrac{dx}{\sqrt{x}}22, converges
∫01dxx\int_0^1 \dfrac{dx}{x}diverges

For either type one integrates while avoiding the end, then takes the limit as the end is approached. A finite limit means convergence; anything else means divergence.

The steeply falling curve on the graph is y=1x2y = \dfrac{1}{x^2}, the gentler one is y=1xy = \dfrac{1}{x}, and the large dot is (1,1)(1, 1), where the two curves meet.

  1. Improper integral, Wikipedia