Improper Integrals

Integrating to Infinity

An ordinary definite integral has finite limits. An improper integral relaxes this: either the interval is infinite ($\pm\infty$) or the integrand is unbounded (a vertical asymptote).

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; otherwise it diverges.

Common pitfall: assuming that because the integrand goes to zero, the integral must converge. Decay to zero is necessary but not sufficient.

The $p$-Test & Comparison

Behavior depends delicately on decay speed. Consider the family $\int_1^{\infty} \frac{1}{x^p}\, dx$:

For example, $\int_1^{\infty} \frac{1}{x^2}\, dx = 1$ (finite), but $\int_1^{\infty} \frac{1}{x}\, dx = \infty$, even though both integrands approach zero.

When exact evaluation fails, use comparison: if $0 \le f \le g$ and $\int g$ converges, $\int f$ converges too. If $\int f$ diverges, $\int g$ diverges.