Improper Integrals

Integrating to Infinity

An ordinary definite integral has finite limits and a bounded integrand. An improper integral relaxes one of these: either the interval is infinite (a limit is $\pm\infty$) or the integrand becomes unbounded (a vertical asymptote in the interval). Remarkably, such integrals can still yield a finite value — an infinite region with a finite area.

Improper integrals are defined as limits of proper ones. For an infinite upper limit: $$\int_a^{\infty} f(x)\, dx = \lim_{b \to \infty} \int_a^b f(x)\, dx.$$ If the limit exists and is finite, the integral converges to that value; otherwise it diverges. The same limit device handles an integrand that blows up at an endpoint (approach the bad point with a limit).

The behavior can be surprising. Consider the family $\int_1^{\infty} \frac{1}{x^p}\, dx$:

• It converges when $p > 1$ (the tail shrinks fast enough).
• It diverges when $p \le 1$.

So $\int_1^{\infty} \frac{1}{x^2}\, dx = 1$ (finite!) but $\int_1^{\infty} \frac{1}{x}\, dx = \infty$ — even though both integrands go to zero, only the first decays quickly enough for a finite total. Convergence depends delicately on how fast the function decays.

When an integral is hard to evaluate exactly, comparison settles convergence: if $0 \le f \le g$ and $\int g$ converges, then $\int f$ converges too (and if $\int f$ diverges, so does $\int g$). Improper integrals matter throughout probability (total probability over an infinite range), physics (fields extending to infinity), and the theory of series.

Common pitfall: assuming that because the integrand goes to zero, the improper integral must converge. Decay to zero is necessary but not sufficient — $\frac{1}{x} \to 0$ yet $\int_1^{\infty}\frac{1}{x}\, dx$ diverges. Convergence depends on how fast the function decays (for $\frac{1}{x^p}$, you need $p > 1$). Never conclude convergence from "the terms get small" alone; evaluate the limit or compare.