Concept preview

What you will be able to do

1. 01Recognise the two ways a definite integral can be improper — an infinite limit of integration, or an integrand that blows up to $\pm\infty$ somewhere on the interval — and rewrite it honestly as a LIMIT such as $\displaystyle\lim_{b\to\infty}\int_a^b f(x)\,dx$.
2. 02Evaluate an improper integral by computing the ordinary integral to a movable boundary and then taking the limit, deciding whether the integral CONVERGES (the limit is a finite number) or DIVERGES (the limit is infinite or does not exist).
3. 03Apply the $p$-integral benchmarks — $\int_1^\infty x^{-p}\,dx$ converges iff $p>1$, and $\int_0^1 x^{-p}\,dx$ converges iff $p<1$ — and the exponential-decay result $\int_0^\infty e^{-x}\,dx=1$ to classify and evaluate standard cases.

A definite integral $\int_a^b f(x)\,dx$ is called improper when it violates one of the assumptions that made the Fundamental Theorem apply cleanly. There are two flavours. The first is an infinite limit of integration: at least one endpoint is $\pm\infty$, as in $\int_1^\infty \frac{1}{x^2}\,dx$ or $\int_{-\infty}^{0} e^{x}\,dx$. The region stretches forever in the horizontal direction. The second is an unbounded integrand: the function itself has a vertical asymptote somewhere on the interval, as in $\int_0^1 \frac{1}{\sqrt{x}}\,dx$, where $1/\sqrt{x}\to+\infty$ as $x\to 0^+$. The region stretches forever in the vertical direction. Either way, the area is not obviously finite, and we are not allowed to simply plug an infinite number into an antiderivative.