Decide convergence or divergence
Two curves that fall, two different areas
Two curves fall toward the x-axis as \(x\) grows. The upper curve is \(f(x) = 2/(x+3)\), and the lower curve is \(g(x) = 2/(x+3)^2\).
Drag the orange edge \(\cy{b}\) to the right. The lower graphs plot the area under each curve from 0 to \(b\), on axes squeezed so that the right end stands for \(\infty\).
The area under \(g\) rises toward the dashed line at height \(\tfrac23\) and does not pass it.
The area under \(f\) rises without bound. It is \(2\ln\dfrac{b+3}{3}\), which is about 4.6 at \(b = 27\) and about 9.2 at \(b = 297\).
Both curves approach 0, but only one area is finite, so the height of the curve does not decide the question. The limit of the areas decides it.
Three ways a limit can end
Press each button and drag the edge \(\cy{t}\) toward the trouble spot, which is the place that makes the integral improper. The lower graph shows the area \(A(t)\) from \(a\) to \(t\).
In the first setting, \(A(t)\) approaches a number \(L\), so the integral converges and its value is \(L\).
In the second setting, \(A(t)\) passes every height. In the third setting, it rises and falls again and again. In both, the limit is not a number, so the integral diverges.
A divergent integral has no value.
Notation
An improper integral converges if the limit of its finite integrals is a number. The integral then equals that number.
\[\lim_{t \to \infty} \int_a^{t} f(x)\,dx = \cb{L}\]It diverges if the limit is \(\infty\) or \(-\infty\), or if the limit does not exist. In the figure, the curve changes to one whose area on the left falls without bound and whose area on the right rises without bound.
To decide, find an antiderivative, evaluate it between \(a\) and \(t\), and take the limit of the result.
If an integral is split at its trouble spots, it converges only if every piece converges. One divergent piece makes the whole integral diverge, even when the other piece is finite.
Each trouble spot has its own limit. The limits are taken separately, so one piece cannot cancel another.
\[\begin{aligned} \int_{-\infty}^{\infty} f(x)\,dx &= \lim_{\cy{s} \to -\infty} \int_{\cy{s}}^{0} f(x)\,dx \\ &\quad + \lim_{\cy{t} \to \infty} \int_{0}^{\cy{t}} f(x)\,dx \end{aligned}\]
When it fails
The teal region lies above the x-axis, and the red region lies below it, so it counts as negative. If both edges move out by the same amount, the teal and red areas cancel, and the net area is 0.
This is not the integral. The teal area alone rises without bound, and the red area falls without bound. Infinity minus infinity is not a number, so the integral diverges.
Example 1: a limit that is a number
Decide whether \(\displaystyle\int_0^{\infty} \frac{8x}{(x^2+4)^2}\,dx\) converges.
Replace the infinite limit by \(\cy{t}\).
\[\lim_{t \to \infty} \int_0^{\cy{t}} \frac{8x}{(x^2+4)^2}\,dx\]Substitute \(u = x^2 + 4\), so that \(du = 2x\,dx\) and \(8x\,dx = 4\,du\). The integral becomes \(\int 4/u^2\,du\), whose antiderivative is \(-4/u\). Return to \(x\).
\[\int \frac{8x}{(x^2+4)^2}\,dx = -\frac{4}{x^2+4}\]Evaluate between 0 and \(\cy{t}\). At \(x = 0\) the antiderivative is \(-4/4 = -1\), and subtracting \(-1\) gives the 1.
\[\left[-\frac{4}{x^2+4}\right]_0^{t} = \cb{1 - \frac{4}{t^2+4}}\]As \(t\) grows, the fraction \(\dfrac{4}{t^2+4}\) approaches 0. The limit is the number 1, so the integral converges.
\[\lim_{t \to \infty} \left(1 - \frac{4}{t^2+4}\right) = \cb{1}\]
Example 2: a curve that falls, an area that does not stop
Decide whether \(\displaystyle\int_2^{\infty} \frac{dx}{x \ln x}\) converges.
Replace the infinite limit by \(\cy{t}\). Substitute \(u = \ln x\), so that \(du = dx/x\) and the integral becomes \(\int du/u\). An antiderivative is \(\ln(\ln x)\), with no absolute value because \(\ln x > 0\) for \(x \ge 2\).
\[\int_2^{\cy{t}} \frac{dx}{x \ln x} = \cb{\ln(\ln t) - \ln(\ln 2)}\]A tempting wrong move: The curve approaches the x-axis as \(x\) grows, so the area on the right is small and the integral converges.
\[\frac{1}{x \ln x} \to 0 \quad \Rightarrow \quad \text{converges}\]In the picture, the area continues to rise. It is about 1.2 up to \(t = 10\), and about 3.0 up to \(t = 10^6\). It passes every number, but very slowly, so the height of the curve does not decide the question.
Take the limit of the area. As \(t\) grows without bound, so does \(\ln t\), and so does \(\ln(\ln t)\). The limit is infinite, so the integral diverges.
\[\lim_{t \to \infty} \left[\ln(\ln t) - \ln(\ln 2)\right] = \infty\]
Practice
Decide whether \(\displaystyle\int_1^{5} \frac{dx}{x-3}\) converges.
Find the point where the curve is unbounded, and split there. Then study each piece alone.
Solution
The curve is unbounded at \(x = 3\), between 1 and 5. Split the integral there.
\[\int_1^5 = \int_1^3 + \int_3^5\]An antiderivative is \(\ln|x-3|\). As \(s\) approaches 3 from the right, \(\ln(s-3)\) falls to \(-\infty\), so the area from \(s\) to 5 rises without bound.
\[\begin{aligned} &\lim_{s \to 3^+} \int_{s}^{5} \frac{dx}{x-3} \\ &= \lim_{s \to 3^+} \left[\ln 2 - \ln(s-3)\right] \\ &= \infty \end{aligned}\]The left piece is the red region, and it diverges to \(-\infty\). Neither piece is a number, so they cannot cancel, and the whole integral diverges.
\[\begin{aligned} &\lim_{t \to 3^-} \int_1^{t} \frac{dx}{x-3} \\ &= \lim_{t \to 3^-} \left[\ln|t-3| - \ln 2\right] \\ &= -\infty \end{aligned}\]
Questions
The graph shows two curves, \(f(x) = 3e^{-x/2}\) (brighter) and \(g(x) = 3/(x+1)\) (fainter). The lower graphs show the area from 0 to \(\cy{b}\) under each curve: the left graph for \(f\), and the right graph for \(g\).
What is the value of \(\displaystyle\int_0^{\infty} f(x)\,dx\)?
- 6
- 3
- The integral diverges
Answer
6
The area under \(g\) is 6.9 at \(b = 9\), 13.8 at \(b = 99\), and 20.7 at \(b = 999\). What does this show about \(\displaystyle\int_0^{\infty} g(x)\,dx\)?
- It diverges, because the area rises by about 6.9 each time b is multiplied by 10
- It converges to about 20.7
- It converges, because \(g(x)\) approaches 0
Answer
It diverges, because the area rises by about 6.9 each time b is multiplied by 10
What can you say about \(\displaystyle\int_{-3}^{3} \frac{dx}{x^3}\)?
- It diverges
- It equals 0, because the pieces cancel
- It equals \(\tfrac19\)
Answer
It diverges
Summary
You find the limit of the finite integrals, and you call the integral convergent only if that limit is a number.
This works because the integral is defined as that limit, so the height of the curve and the canceling of pieces do not matter.
You use it for every improper integral, and you treat each trouble spot of a split integral separately.
This is one of five skills in topic 6.13 of Which & Why: AP Calculus.