Improper integral

Definition

Let $f \colon [a, \infty) \to \mathbb{R}$ be Riemann integrable on $[a, b]$ for each $b > a$. The improper integral of $f$ over $[a, \infty)$ converges if the following limit exists in $\mathbb{R}$: $$\int_a^\infty f(x) \, dx = \lim_{b \to \infty} \int_a^b f(x) \, dx.$$ Similarly, let $f \colon (a, b] \to \mathbb{R}$ be Riemann integrable on $[c, b]$ for each $c \in (a, b)$. Then $\int_a^b f(x) \, dx = \lim_{c \to a^+} \int_c^b f(x) \, dx$, if this one-sided limit exists.

Remarks

If the limit does not exist, the improper integral diverges. For an integral over $(-\infty, \infty)$, or with several bad points, we split the interval. Each piece must converge separately. For example, $\int_1^\infty \frac{dx}{x^2} = 1$, and $\int_0^1 \frac{dx}{\sqrt{x}} = 2$. But $\int_1^\infty \frac{dx}{x}$ diverges.

Explanations

L1 human

The Riemann integral needs a bounded interval and a bounded function. An improper integral removes these limits by a limit of ordinary integrals.