Improper Integrals

An improper integral is simply an integral over a region that goes on forever, either because an endpoint is $\infty$ or because the function shoots up to infinity somewhere. "Improper" doesn't mean wrong; it just means the integral breaks the usual rules (a finite interval and a function that stays finite), so it needs an extra step. It's good for questions like the total work to escape Earth's gravity or a drug's total exposure over all time. All we're really asking with $\int_a^\infty f(x)\,dx = \lim_{t\to\infty}\int_a^t f(x)\,dx$ is: take $\int_a^t f(x)\,dx$ (an ordinary area, stopping at some finite point $t$), push $t$ farther and farther out ($\lim_{t\to\infty}$), and see whether the area levels off at a number (that's the answer) or keeps growing (no answer). It's an idea you already know from limits, just applied to area; bear with it.

The Fundamental Theorem of Calculus needs a finite interval and a function that doesn't blow up, so neither $\int_1^\infty$ nor an integrand with a vertical asymptote fits. But you already know how calculus handles "you can't plug that in": take a limit. Integrate up to a point where everything is fine, then let that point slide toward the trouble.

Type I: an infinite limit. Integrate to a finite $t$, then take the limit:

Think about how strange that is. The region under $1/x^2$ is infinitely long, yet its area is exactly 1. Now compare $1/x$:

so $\int_1^\infty \frac{dx}{x}$ diverges even though $1/x \to 0$. Getting smaller isn't enough; the function has to shrink fast enough that the extra slivers of area stop adding up. There's an animation further down that shows the two areas side by side as $t$ grows.

If both limits are infinite, split at any convenient point: $\int_{-\infty}^\infty f = \int_{-\infty}^0 f + \int_0^\infty f$. The whole thing converges only if both pieces do.

Type II: a vertical asymptote. If $f$ blows up at the endpoint $b$, approach it from the inside:

If $f$ blows up somewhere inside $[a, b]$, split there.

A common mistake. For $\int_{-1}^{1} \frac{dx}{x^2}$, it's tempting to just plug in the limits and get $\left[-\frac1x\right]_{-1}^{1} = -2$, because it looks like an ordinary definite integral. It isn't. There's an asymptote at $x = 0$, right in the middle, and the Fundamental Theorem doesn't apply across it. You can tell something's off anyway: the integrand is positive, so a negative answer is impossible. Split at $0$ and each half diverges, so the integral diverges.

Escape velocity comes from an improper integral: the work to lift a rocket from Earth's surface to infinity, $\int_R^\infty \frac{GMm}{r^2}\,dr = \frac{GMm}{R}$, is finite, so a fast enough rocket never falls back. In medicine, a drug's total exposure is the area under its concentration curve from 0 to ∞.