AP Calculus BC

Infinite Limit vs Infinite Integrand

Q: How do I know if an integral is improper?

Check the bounds for \infty, then find the points where the integrand is undefined on the closed interval and check whether it is unbounded near them. A point where the function is undefined but stays bounded, such as \frac{x^2 - 1}{x - 1} at x = 1, leaves an ordinary integral worth 4 on [0, 2]. If neither test turns anything up, the fundamental theorem applies in the usual way.

Q: Can an integral be improper for both reasons at once?

Yes. \int_0^{\infty} \frac{1}{\sqrt{x}}\,dx has an infinite upper limit and an integrand that blows up at 0. Split it at a convenient point such as x = 1 and test each piece on its own; here the piece from 1 to \infty diverges, so the whole integral does.

Type one has an infinite limit of integration; type two has a finite interval but an integrand that blows up at an endpoint or somewhere inside it.

Infinite limit of integration

Use when: At least one bound is infinite, so replace it by a variable and take the limit as that variable runs off.

Infinite integrand

Use when: The interval is finite but the integrand is unbounded at an endpoint or an interior point, so replace a bad endpoint with a variable and take one one sided limit; if the bad point is interior, split there first and take a one sided limit on each piece.

Side by side

	Infinite limit of integration	Infinite integrand
What is infinite	A limit of integration	The integrand, which is unbounded at a point of the interval
How you spot it	An $\infty$ written in a bound	A denominator that vanishes, or $\ln$ or $\tan$ blowing up, at or between the bounds
Limit you take	$\lim_{b \to \infty}$ on the bound you replaced	$\lim_{c \to a^{+}}$ or $\lim_{c \to b^{-}}$ at the bad point
Splitting required	When both bounds are infinite, and also whenever the integrand blows up somewhere on the interval, as in $\int_0^{\infty} \frac{dx}{\sqrt{x}}$ , which is split at $1$	Always when the bad point lies inside the interval
Cost of missing it	Almost impossible to miss	When the bad point is interior and the antiderivative jumps across it, the fundamental theorem can return a finite number that is wrong, such as $-2$ for $\int_{-1}^{1} \frac{dx}{x^2}$

An improper integral is one with an infinite bound or an unbounded integrand, and it has to be defined by a limit rather than evaluated straight from the fundamental theorem. The first reason is written on the page in plain sight. The second hides in the algebra, and catching it means finding where the integrand is undefined before any antiderivative gets written down, then checking whether the function actually blows up there. A hole where the function stays bounded, as in $\frac{x^2 - 1}{x - 1}$ at $x = 1$ , leaves an ordinary integral.

\underbrace{\int_a^{\infty} f\,dx = \lim_{b \to \infty} \int_a^{b} f\,dx}_{\text{infinite limit}} \qquad\qquad \underbrace{\int_a^{b} f\,dx = \lim_{c \to a^{+}} \int_c^{b} f\,dx}_{f \text{ unbounded at } a}

Watch the second type do its damage, which it does when the bad point is interior. The integrand $\frac{1}{x^2}$ is positive wherever it is defined, so $\int_{-1}^{1} \frac{1}{x^2}\,dx$ cannot be negative. Applying the fundamental theorem anyway gives $\left[-\frac{1}{x}\right]_{-1}^{1} = -1 - 1 = -2$ , a negative number for a positive integrand. The antiderivative $-\frac{1}{x}$ is not continuous on $[-1, 1]$ , so the theorem never applied. Splitting at $0$ shows $\int_0^{1} \frac{1}{x^2}\,dx$ diverges, and the whole integral diverges with it.

The check that prevents the wrong answer

Before writing any antiderivative, find where the integrand is undefined, ask whether any of those points lie in $[a, b]$ , endpoints included, and check whether the function blows up there rather than merely having a hole. Zeros of a denominator, $\tan x$ at $\frac{\pi}{2}$ , and $\ln x$ at $0$ are the usual culprits. Skip that check and, when the blow up sits inside the interval, the fundamental theorem can hand back a tidy finite number with nothing to warn you that it is meaningless.

Frequently asked questions

How do I know if an integral is improper?

Check the bounds for $\infty$ , then find the points where the integrand is undefined on the closed interval and check whether it is unbounded near them. A point where the function is undefined but stays bounded, such as $\frac{x^2 - 1}{x - 1}$ at $x = 1$ , leaves an ordinary integral worth $4$ on $[0, 2]$ . If neither test turns anything up, the fundamental theorem applies in the usual way.

Do I split the integral at the vertical asymptote?

Yes, whenever the asymptote falls strictly inside the interval. Write the integral as two pieces meeting at that point and take a one sided limit for each. If the asymptote sits at an endpoint instead, there is nothing to split: replace that bound by a variable and take a single one sided limit. If either piece diverges the whole integral diverges, and you can stop there.

Can an integral be improper for both reasons at once?

Yes. $\int_0^{\infty} \frac{1}{\sqrt{x}}\,dx$ has an infinite upper limit and an integrand that blows up at $0$ . Split it at a convenient point such as $x = 1$ and test each piece on its own; here the piece from $1$ to $\infty$ diverges, so the whole integral does.

In the CED: Unit 6: Integration and Accumulation