Recognize standard convergence thresholds

Where the exponent tips the balance

The curve is \(f(x) = 1/x^p\) for \(x \ge 1\), and the first slider sets the exponent \(p\). The teal region lies under the curve from \(x = 1\) to the orange edge \(b\).

The first readout shows the area from \(x = 1\) to \(x = 100\): 0.99 for \(p = 2\), about 4.6 for \(p = 1\), and 18 for \(p = 0.5\).

Drag \(p\) slowly from 3 down to 0.5. The lower graph plots the area against \(b\) on a squeezed axis whose right end stands for \(\infty\).

If \(p\) is greater than 1, the graph levels off at a dashed line, so the area has a limit. If \(p\) is 1 or less, it rises without bound, so the area has none.

One threshold, two directions

The curve \(1/x^p\) is split at \(x = 1\). The right piece runs to infinity, and the left piece runs toward the vertical axis, where the curve has an asymptote. Both places are trouble spots, which make the integral improper.

Increase p. The right piece falls faster, so its region becomes thinner and can have a finite area. The left piece rises faster along the vertical axis, so its region becomes taller and can have an infinite area.

Press the button to move both edges toward their trouble spots.

Both pieces have the same threshold, \(p = 1\), the exponent that separates a finite area from an infinite one, but the direction differs. Toward infinity the area is finite when \(p\) is greater than 1, and toward 0 when \(p\) is less than 1.

Notation

1. Start with the region from 1 to \(\cy{t}\). For \(p \ne 1\), an antiderivative of \(x^{-p}\) is \(x^{1-p}/(1-p)\).

   \[\int_1^{\cy{t}} x^{-p}\,dx = \frac{t^{1-p} - 1}{1 - p}\]

2. If \(p > 1\), the exponent \(1 - p\) is negative, so \(t^{1-p}\) approaches 0 as \(t\) grows. The area approaches \((0 - 1)/(1 - p) = 1/(p - 1)\), so the integral converges.

   \[\int_1^{\infty} \frac{dx}{x^p} = \cb{\frac{1}{p-1}} \qquad (p > 1)\]

3. If \(p < 1\), the exponent \(1 - p\) is positive, so \(t^{1-p}\) grows without bound and the integral diverges.

4. Near 0, the same antiderivative gives the region from \(\cy{s}\) to 1 the area \((1 - s^{1-p})/(1-p)\). If \(p < 1\), the term \(s^{1-p}\) approaches 0 as \(s\) does, so the integral converges. If \(p > 1\), the term grows without bound, so it diverges.

   \[\int_0^{1} \frac{dx}{x^p} = \cb{\frac{1}{1-p}} \qquad (p < 1)\]

5. The same thresholds hold at any trouble spot \(c\), because the substitution \(u = x - c\) turns each integral into one of \(1/u^p\).

   \[\begin{aligned} \int_c^{c+1} \frac{dx}{(x-c)^p} &\text{ converges if } p < 1 \\ \int_{c+1}^{\infty} \frac{dx}{(x-c)^p} &\text{ converges if } p > 1 \end{aligned}\]

When it fails

At \(p = 1\), the formula divides by \(1 - p = 0\), so it does not apply.

The antiderivative is \(\ln x\), so the area from 1 to \(t\) is \(\ln t\) and from \(s\) to 1 it is \(-\ln s\). Both grow without bound, so the integral diverges at both trouble spots.

Example 1: toward infinity

Decide whether each integral converges, and find the value if it does. (a) \(\displaystyle\int_1^{\infty} \frac{dx}{x^{4/3}}\) (b) \(\displaystyle\int_1^{\infty} \frac{dx}{x^{3/4}}\)

1. Both integrals have their trouble spot at infinity, so each converges only if p > 1. The lower graphs show the area under the brighter curve (a) on the left and under the fainter curve (b) on the right.

2. In (a), \(p = \tfrac43\) is greater than 1, so the integral converges to \(1/(p-1) = 3\).

   \[\begin{aligned} \int_1^{t} x^{-4/3}\,dx &= 3 - 3t^{-1/3} \\ &\to \cb{3} \end{aligned}\]

3. In (b), \(p = \tfrac34\) is less than 1, so the integral diverges.

   \[\begin{aligned} \int_1^{t} x^{-3/4}\,dx &= 4t^{1/4} - 4 \\ &\to \infty \end{aligned}\]

Example 2: toward 0

Decide whether \(\displaystyle\int_0^{1} \frac{dx}{x^{5/3}}\) converges.

1. The curve is unbounded at the left end, so the trouble spot is 0.

2. A tempting wrong move: The exponent is \(\tfrac53\), which is greater than 1, so the integral converges.

   \[p = \tfrac53 > 1 \;\Rightarrow\; \text{converges}\]

   This is the threshold for infinity, and it would give \(\tfrac{1}{p-1} = \tfrac32\), the red dashed line. But the region lies next to the vertical axis, and its area rises past the red line without bound as \(s\) approaches 0.

3. At the trouble spot 0, the integral converges only if \(p < 1\). Here \(p = \tfrac53\) is not less than 1, so the integral diverges, and the area from \(s\) to 1 grows without bound.

   \[\begin{aligned} \int_s^{1} x^{-5/3}\,dx &= \tfrac32\left(s^{-2/3} - 1\right) \\ &\to \infty \end{aligned}\]

Practice

Decide whether \(\displaystyle\int_3^{7} \frac{dx}{(x-3)^{5/4}}\) converges.

Find the point where the curve is unbounded. Then compare the exponent with the threshold for that trouble spot.

Questions

The graph shows \(f(x) = 1/x^{3/2}\). The lower graphs show the teal area from 1 to \(\cy{t}\) on the right and from \(\cy{s}\) to 1 on the left.

1. What is \(\displaystyle\int_1^{\infty} f(x)\,dx\)?

   • 2
   • 1.5
   • The integral diverges
   Answer

   2

2. What can you say about \(\displaystyle\int_0^{1} f(x)\,dx\)?

   • It diverges
   • It converges to 2, as the integral from 1 to infinity does
   • It converges, because the curve is finite at x = 1
   Answer

   It diverges

3. What is \(\displaystyle\int_0^{1} x^{-4/5}\,dx\)?

   • 5
   • \(-5\)
   • The integral diverges
   Answer

   5

Summary

At a trouble spot, you compare the exponent \(p\) of \(1/x^p\) with 1: toward infinity the integral converges for \(p > 1\), and toward 0 for \(p < 1\).

This works because the antiderivative contains \(t^{1-p}\), and the sign of \(1 - p\) decides whether this term approaches 0 or grows without bound.

With these thresholds you decide whether an integral of a power converges before you compute anything.

This is one of five skills in topic 6.13 of Which & Why: AP Calculus.