Identify why an integral is improper
A region that stops being finite
The curve is \(f(x) = 1/(x-3)^2\). The teal region lies under it from \(x = 0\) to the orange line at \(x = b\).
Drag \(b\) toward 3. The curve is higher near \(x = 3\), so the region becomes taller at its right edge and the height \(f(b)\) grows.
The dashed line at \(x = 3\) is a vertical asymptote. The curve has no value there, because the denominator is 0, and its height grows without bound as \(x\) approaches 3.
Drag \(b\) past 3. The interval from 0 to \(b\) now contains the asymptote.
An ordinary definite integral needs a finite interval and a finite height at every point of it. Here the curve has no finite height at \(x = 3\).
Three places to look
Press each button in turn. In the first setting, the region continues to the right with no end, so the interval is infinite.
In the second setting, the interval is finite, but the curve rises without bound at its right end.
In the third setting, both ends are finite and the curve has a finite height at both, but the asymptote lies between them.
Each setting breaks one condition for an ordinary integral, and one bad point is enough.
Notation
An integral is ordinary when two conditions hold. The interval from \(a\) to \(b\) is finite, and the curve has a finite height at every point of the interval, the ends included.
\[\int_a^b f(x)\,dx\]The first reason for an improper integral is that a limit of integration is infinite. Each of these forms is improper.
\[\begin{gathered} \int_a^{\cy{\infty}} f(x)\,dx \\ \int_{\cy{-\infty}}^{b} f(x)\,dx \\ \int_{\cy{-\infty}}^{\cy{\infty}} f(x)\,dx \end{gathered}\]The second reason is that the integrand is unbounded at a point \(c\) of the interval. This means that \(|f(x)|\) grows without bound as \(x\) approaches \(c\), and \(c\) can be an end or lie between the ends.
\[\lim_{x \to \cy{c}} |f(x)| = \infty\]To find such a point, solve for the values of \(x\) where the integrand is undefined. Then ask whether each one lies between \(a\) and \(b\).
The integral is improper when the interval is infinite or when the curve is unbounded at some point of it.
When it fails
Drag the slider to move the asymptote. If it lies to the right of 4, the curve has a finite height from 0 to 4, so the integral is ordinary.
When the asymptote moves into the interval, the integral becomes improper. Both ends are still finite, but the curve is unbounded at one point between them.
Example 1: an asymptote at an end
Is \(\displaystyle\int_0^2 \frac{dx}{4 - x^2}\) improper? Give every reason.
The interval from 0 to 2 is finite, so no limit of integration is infinite.
The integrand is undefined where the denominator is 0. Solve \(4 - x^2 = 0\).
\[x = \cy{-2} \quad\text{or}\quad x = \cy{2}\]The point 2 is the right end of the interval, and the curve rises without bound as \(x\) approaches 2 from the left.
\[\lim_{x \to 2^-} \frac{1}{4 - x^2} = \infty\]The point \(-2\) lies outside the interval, so it does not matter. The integral is improper, and the only reason is the asymptote at the end \(x = 2\).
Example 2: an asymptote inside
Is \(\displaystyle\int_0^{\pi} \tan x\,dx\) improper?
The interval from 0 to \(\pi\) is finite. The red part of the region lies below the x-axis, so it counts as negative. Find the points where \(\tan x\) is undefined.
\[\tan x = \frac{\sin x}{\cos x}\]The denominator \(\cos x\) is 0 at \(x = \pi/2\), which lies between 0 and \(\pi\).
\[\cos\frac{\pi}{2} = 0\]A tempting wrong move: The curve has a finite height at both ends, because \(\tan 0 = 0\) and \(\tan \pi = 0\). Evaluate the antiderivative \(-\ln|\cos x|\) at the two ends.
\[\begin{aligned} -\ln|\cos \pi| + \ln|\cos 0| \\ = 0 - 0 = 0 \end{aligned}\]The curve is unbounded near \(\pi/2\), and \(-\ln|\cos x|\) is not defined there. An antiderivative cannot be used across a point where it is undefined, so the value 0 is not valid.
Test every point of the interval, not only the ends. The curve is unbounded at \(\pi/2\), between the ends, so the integral is improper.
\[\lim_{x \to \pi/2^-} \tan x = \infty\]
Practice
Is \(\displaystyle\int_0^4 \frac{dx}{(x-1)(x-5)}\) improper? Give every reason.
Mark each point where the denominator is 0. Then drag the orange edge to 4 and see which of these points lie between the edges.
Solution
The interval from 0 to 4 is finite.
The denominator is 0 where one of its factors is 0.
\[\begin{aligned} (1-1)(1-5) &= 0 \\ (5-1)(5-5) &= 0 \end{aligned}\]The point 1 lies between 0 and 4, so the integral is improper. The point 5 lies outside, so it does not matter, and the only reason is the asymptote at \(x = 1\).
Questions
The graph shows \(f(x) = 1/(x-1)^2\). The dashed line at \(x = 1\) is its vertical asymptote.
Is \(\displaystyle\int_2^6 f(x)\,dx\) improper?
- No, because the asymptote lies outside the interval
- Yes, because the curve is unbounded at \(x = 1\)
- Yes, because the curve comes close to the x-axis on the right
Answer
No, because the asymptote lies outside the interval
Which of these integrals is improper?
- \(\displaystyle\int_{1.5}^{5} f(x)\,dx\)
- \(\displaystyle\int_{3}^{8} f(x)\,dx\)
- \(\displaystyle\int_{0}^{3} f(x)\,dx\)
Answer
\(\displaystyle\int_{0}^{3} f(x)\,dx\)
Is \(\displaystyle\int_0^3 \frac{dx}{x^2 - 5x + 4}\) improper?
- Yes, because the denominator is 0 at \(x = 1\), which lies between 0 and 3
- No, because the denominator is not 0 at \(x = 0\) or at \(x = 3\)
- Yes, because the denominator is 0 at \(x = 4\)
- No, because the interval is finite
Answer
Yes, because the denominator is 0 at \(x = 1\), which lies between 0 and 3
Summary
You test each limit of integration for infinity, and you ask whether the integrand is undefined at any point of the interval.
This works because an integral is ordinary only when the interval is finite and the curve has a finite height at every point of it.
You use this test before you integrate, in particular when the integrand has a denominator.
This is one of five skills in topic 6.13 of Which & Why: AP Calculus.