The great British mathematician G. H. Hardy (1877–1947) observed that in calculus, we learn to ask, not “ What is it?” but rather “How shall we define it?” We saw that tangent lines and areas under curves have no clear meaning until we define them precisely using limits. Here again, the key question is “How shall we define the area of an unbounded region?”

DEFINITION Improper Integral

Fix a number \(a\) and assume that \(f(x)\) is integrable over \([a,b]\) for all \(b> a\). The improper integral of \(f(x)\) over \([a,\infty)\) is defined as the following limit (if it exists): \[ \int_a^\infty f(x)\,dx = \lim_{R\to\infty} \int_a^R f(x)\,dx \]

We say that the improper integral converges if the limit exists (and is finite) and that it diverges if the limit does not exist.

EXAMPLE 1

Show that \(\int_2^{\infty} \dfrac{dx}{x^3}\) converges and compute its value.

Solution

Figure 7.10: The area over \([2,\infty)\) is equal to \(\frac18\).

Step 1. Integrate over a finite interval \([2,R]\). \[ \int_2^R \frac{dx}{x^3} = -\frac12x^{-2}\bigg|_2^R= -\frac12\big(R^{-2}\big) + \frac12\big(2^{-2}\big) = \frac18-\frac1{2R^2} \]

Step 2. Compute the limit as \(R\to\infty\). \[ \int_2^{\infty} \frac{dx}{x^3} = \lim_{R\to\infty} \int_2^R \frac{dx}{x^3} = \lim_{R\to\infty}\left( \frac18-\frac1{2R^2}\right) = \frac18 \]

We conclude that the infinite shaded region in Figure 7.10 has area \(\frac18\).

EXAMPLE 2

Determine whether \(\displaystyle \int_{-\infty}^{-1} \dfrac{dx}x\) converges.

Solution First, we evaluate the definite integral over a finite interval \([R,-1]\) Since the lower limit of the integral is \(-\infty\), we take \(R<-1\): \[ \int_{R}^{-1} \frac{dx}x = \ln |x|\bigg|_{R}^{-1} = \ln \left|-1\right| - \ln |R| = -\ln |R| \]

Then we compute the limit as \(R\to -\infty\): \[ \lim_{R\to-\infty} \int_{R}^{-1} \frac{dx}x = \lim_{R\to-\infty} \left( -\ln |R| \right) = - \lim_{R\to-\infty} \ln |R| = -\infty \]

Figure 7.11: The integral of \(f(x)=x^{-1}\) over \((-\infty,-1]\) is infinite.

The limit is infinite, so the improper integral diverges. We conclude that the area of the unbounded region in Figure 7.11 is infinite.

CONCEPTUAL INSIGHT

If you compare the unbounded shaded regions in Figures 3 and 4, you may wonder why one has finite area and the other has infinite area. Convergence of an improper integral depends on how rapidly the function \(f(x)\) tends to zero as \(x\to\infty\) (or \(x\to -\infty\)). Our calculations show that \(x^{-2}\) decreases rapidly enough for convergence, whereas \(x^{-1}\) does not.

An improper integral of a power function \(f(x) = x^{-p}\) is called a p-integral. Note that \(f(x)=x^{-p}\) decreases more rapidly as \(p\) gets larger. Interestingly, our next theorem shows that the exponent \(p=-1\) is the dividing line between convergence and divergence.

THEOREM 1

The \(p\)-Integral over \([a,\infty)\)] For \(a>0\), \[ \begin{align*} \int_a^{\infty} \frac{dx}{x^p} = \begin{cases} \dfrac{a^{1-p}}{p-1}& \text{if \(p > 1\)} \\ \textrm{diverges}& \text{if \(p \le 1\)} \end{cases} \end{align*} \]

\(p\)-integrals are particularly important because they are often used to determine the convergence or divergence of more complicated improper integrals by means of the Comparison Test (see Example 8).

Proof

Denote the \(p\)-integral by \(J\). Then \[ J= \lim_{R\to\infty}\int_a^Rx^{-p}\,dx =\lim_{R\to\infty}\frac{x^{1-p}}{1-p}\bigg|_a^R =\lim_{R\to\infty}\left(\frac{R^{1-p}}{1-p} -\frac{a^{1-p}}{1-p}\right) \]

If \(p>1\), then \(1-p<0\) and \(R^{1-p}\) tends to zero as \(R\to\infty\). In this case, \(J= \frac{a^{1-p}}{p-1}\). If \(p<1\), then \(1-p>0\) and \(R^{1-p}\) tends to \(\infty\). In this case, \(J\) diverges. If \(p=1\), then \(J\) diverges because \(\lim\limits_{R\to\infty}\int_a^R x^{-1}\,dx = \lim\limits_{R\to\infty}(\ln R - \ln a) = \infty\).

EXAMPLE 3 Using L'Hôpital's Rule

Calculate \(\displaystyle\int_{0}^{\infty} xe^{-x}\, dx\).

Solution First, use Integration by Parts with \(u=x\) and \(v'=e^{-x}\): \[ \begin{eqnarray*} \int xe^{-x}\, dx &=& -xe^{-x} + \int e^{-x}\, dx = -xe^{-x}-e^{-x} = -(x+1)e^{-x} + C\\ \int_0^R xe^{-x}\, dx &=&-(x+1)e^{-x}\bigg|_0^R = -(R+1)e^{-R} + 1 = 1-\frac{R+1}{e^R} \end{eqnarray*} \]

Then compute the improper integral as a limit using L'Hôpital's Rule: \[ \int_0^{\infty} xe^{-x}\, dx = 1 - \lim_{R\to \infty} \frac{R+1}{e^R} = 1 - \underbrace{\lim_{R\to \infty} \frac{1}{e^R}}_{\textrm{L'Hôpital's Rule}} =1-0=1 \]

In physics, we speak of moving an object “infinitely far away.” In practice this means “very far away,” but it is more convenient to work with an improper integral.

EXAMPLE 4 Escape Velocity

The earth exerts a gravitational force of magnitude \(F(r)= {GM_em}/{r^2}\) on an object of mass \(m\) at distance \(r\) from the center of the earth.

1. Find the work required to move the object infinitely far from the earth.
2. Calculate the escape velocity \(v_{\text{esc}}\) on the earth's surface.

Solution This amounts to computing a \(p\)-integral with \(p=2\). Recall that work is the integral of force as a function of distance (Section 6.5).

(a) The work required to move an object from the earth's surface (\(r=r_e\)) to a distance \(R\) from the center is \[ \int_{r_e}^{R} \frac{GM_em}{r^2} \, dr = -\frac{GM_em}{r}\bigg|_{r_e}^{R} = GM_em\left(\frac{1}{r_e} - \frac{1}{R}\right)~\mathrm{joules} \]

The work moving the object “infinitely far away” is the improper integral \[ GM_em\int_{r_e}^{\infty} \frac{dr}{r^2} = \lim_{R\to\infty} GM_em \left(\frac{1}{r_e} - \frac{1}{R}\right) = \frac{GM_em}{r_e}~\mathrm{joules} \]

(b) By the principle of Conservation of Energy, an object launched with velocity \(v_0\) will escape the earth's gravitational field if its kinetic energy \(\frac12m v_0^2\) is at least as large as the work required to move the object to infinity—that is, if \[ \frac{1}{2}mv_0^2 \ge \frac{GM_em}{r_e}\quad\Rightarrow\quad v_0 \ge \left(\frac{2GM_e}{r_e}\right)^{1/2} \]

Using the values recalled in the marginal note, we find that \(v_0\ge 11{,}200~\mathrm{m\text{/}s}\). The minimal velocity is the escape velocity \(v_{\text{esc}} = 11{,}200~\text{m/s}\).

Escape velocity in miles per hour is approximately 25,000 mph.

In practice, the word “forever” means “a long but unspecified length of time.” For example, if the investment pays out dividends for 100 years, then its present value is \[ \int_0^{100}6000e^{-0.04t}\,dt \approx \$147{,}253 \]

The improper integral ($150,000) gives a useful and convenient approximation to this value.

EXAMPLE 5 Perpetual Annuity

An investment pays a dividend continuously at a rate of \(\$6000/\mathrm{year}\). Compute the present value of the income stream if the interest rate is 4% and the dividends continue forever.

Solution Recall from Section 5.8 that the present value (PV) after \(T\) years at interest rate \(r=0.04\) is \(\displaystyle\int_0^{T} 6000e^{-0.04t}\, dt\). Over an infinite time interval, \[ PV = \int_0^{\infty} 6000e^{-0.04t}\, dt = \lim_{T\to\infty} \frac{6000e^{-0.04t}}{- 0.04}\bigg|_0^{T} = \frac{6000}{0.04} = \$ 150{,}000 \]

Although an infinite number of dollars are paid out during the infinite time interval, their total present value is finite.

DEFINITION Integrands with Infinite Discontinuities

If \(f(x)\) is continuous on \([a,b)\) but discontinuous at \(x=b\), we define \[ \int_a^b f(x)\, dx = \lim_{R\to b-} \int_a^R f(x)\, dx \]

Similarly, if \(f(x)\) is continuous on \((a,b]\) but discontinuous at \(x=a\), \[ \int_a^b f(x)\, dx = \lim_{R\to a+} \int_R^b f(x)\, dx \]

In both cases, we say that the improper integral converges if the limit exists and that it diverges otherwise.

EXAMPLE 6

Calculate: (a) \(\displaystyle\int_0^9\dfrac{dx}{\sqrt x}\quad\) and\(\quad\)(b) \(\displaystyle\int_0^{1/2}\dfrac{dx}{x}\).

Solution Both integrals are improper because the integrands have infinite discontinuities at \(x=0\). The first integral converges: \[ \begin{align*} \int_0^9 \frac{dx}{\sqrt{x}} &= \lim_{R\to 0+}\int_R^9 \,x^{-1/2} dx = \lim_{R\to 0+}2x^{1/2}\bigg|_R^9\\ &= \lim_{R\to 0+}(6-2R^{1/2})=6 \end{align*} \]

The second integral diverges: \[ \begin{align*} \int_0^{1/2}\dfrac{dx}{x} &= \lim_{R\to 0+} \int_R^{1/2}\frac{dx}{x} = \lim_{R\to 0+} \left(\ln \frac12 - \ln R \right) \\ &=\ln \frac12 - \lim_{R\to 0+} \ln R = \infty \end{align*} \]

Theorem 2 is valid for all exponents \(p\). However, the integral is not improper if \(p<0\).

THEOREM 2 The \(p\)-Integral over \(\lbrack 0,a\rbrack\)]

For \(a >0\), \[ \int_0^a \frac{dx}{x^p} = \begin{cases} \dfrac{a^{1-p}}{1-p}& \text{if \(p < 1\)} \\ \textrm{diverges} &\text{if \(p \ge 1\)} \end{cases} \]

GRAPHICAL INSIGHT

The \(p\)-integrals \(\int_a^\infty\,x^{-p}\, dx\) and \(\int_0^a\,x^{-p}\, dx\) have opposite behavior for \(p\ne 1\). The first converges only for \(p>1\), and the second converges only for \(p<1\) (both diverge for \(p=1\)). This is reflected in the graphs of \(y=x^{-p}\) and \(y = x^{-q}\), which switch places at \(x=1\) (Figure 7.13). We see that a large value of \(p\) helps \(\int_a^\infty\,x^{-p}\, dx\) to converge but causes \(\int_0^a\,x^{-p}\, dx\) to diverge.

EXAMPLE 7

Evaluate \(\displaystyle\int_0^1\dfrac{dx}{\sqrt{1 - x^2}}\).

Solutionm This integral is improper with an infinite discontinuity at \(x=1\) (Figure 7.14). Using the formula \({\int {dx}/{\sqrt{1-x^2}}=\sin^{-1}x + C}\), we find \[ \begin{align*} \int_0^1\frac{dx}{\sqrt{1-x^2}} &= \lim_{R\to 1-}\int_0^R\frac{dx}{\sqrt{1-x^2}}\\ &=\lim_{R\to 1-}(\sin^{-1}R-\sin^{-1}0)\\ &= \sin^{-1}1-\sin^{-1}0 =\frac{\pi}2-0=\frac{\pi}2 \end{align*} \]

Figure 7.14: The infinite shaded region has area \(\frac{\pi}{2}\).

THEOREM 3 Comparison Test for Improper Integrals

Assume that \(f(x) \ge g(x) \geq 0\) for \(x \ge a\).

• If \(\displaystyle\int_a^{\infty} f(x)\, dx\) converges, then \(\displaystyle\int_a^{\infty} g(x)\, dx\) also converges.
• If \(\displaystyle\int_a^{\infty} g(x)\, dx\) diverges, then \(\displaystyle\int_a^{\infty} f(x)\, dx\) also diverges.

The Comparison Test is also valid for improper integrals with infinite discontinuities at the endpoints.

What the Comparison Test says (for nonnegative functions):

• If the integral of the bigger function converges, then the integral of the smaller function also converges.
• If the integral of the smaller function diverges, then the integral of the larger function also diverges.

EXAMPLE 8

Show that \(\int_1^{\infty} \dfrac{dx}{\sqrt{x^3 + 1}}\) converges.

Solution We cannot evaluate this integral, but we can use the Comparison Test. To show convergence, we must compare the integrand \((x^3+1)^{-1/2}\) with a larger. function whose integral we can compute.

It makes sense to compare with \(x^{-3/2}\) because \(\sqrt{x^3} \le \sqrt{x^3 + 1}\), and therefore \[ \frac{1}{\sqrt{x^3 + 1}}\le \frac{1}{\sqrt{x^3}}={x^{-3/2}} \]

The integral of the larger function converges, so the integral of the smaller function also converges: \[ \underbrace{\int_1^{\infty} \dfrac{dx}{x^{3/2}}}_{\textrm{\(p\)-integral with \(p>1\)}}~ \textrm{converges}\quad \Rightarrow\quad \underbrace{\int_1^{\infty} \dfrac{dx}{\sqrt{x^3 + 1}}}_{\textrm{Integral of smaller function }}~\textrm{converges} \]

EXAMPLE 9 Choosing the Right Comparison

Does \(\displaystyle\int_1^{\infty} \dfrac{dx}{\sqrt x + e^{3x}}\) converge?

Solution Since \(\sqrt{x} \geq 0\), we have \(\sqrt{x} + e^{3x} \geq e^{3x}\) and therefore \[\frac{1}{\sqrt{x} + e^{3x}} \leq \frac{1}{e^{3x}} \]

Furthermore, \[ \int^\infty_1 \dfrac{dx}{e^{3x}} = \left.\lim_{R\to\infty} -\frac13 e^{-3x}\right|^{R}_1 = \lim_{R\to\infty} \frac13 \big(e^{-3} -e^{-3R}\big) = \frac{1}{3} e^{-3}\quad\textrm{(converges)}\]

Our integral converges by the Comparison Test: \[ \underbrace{\int_1^{\infty} \dfrac{dx}{e^{3x}}}_{\textrm{Integral of larger function}}\quad \textrm{converges}\quad \Rightarrow\quad \underbrace{\int_1^{\infty}\dfrac{dx}{\sqrt x + e^{3x}}}_{\textrm{Integral of smaller function}} \quad\textrm{also converges} \]

Had we not been thinking, we might have tried to use the inequality \[ \frac{1}{\sqrt x + e^{3x}} \le \frac{1}{\sqrt x} \]

However,\(\int_1^{\infty} \frac{dx}{\sqrt x}\) diverges (\(p\)-integral with \(p<1\)), and this says nothing about our smaller integral (Figure 7.16).

Figure 7.16: The divergence of a larger integral says nothing about the smaller integral.

EXAMPLE 10 Endpoint Discontinuity

Does \(J = \displaystyle\int_0^{0.5} \dfrac{dx}{x^{8} + x^{2}}\) converge?

Solution This integral has a discontinuity at \(x=0\). We might try the comparison \[ x^{8} + x^{2} > x^{2} \quad \Rightarrow \quad \frac1{x^{8} + x^{2}} < \frac1{x^{2}} \]

However, the \(p\)-integral \(\int_0^{0.5} \frac{dx}{x^{2}}\) diverges, so this says nothing about our integral \(J\), which is smaller. But notice that if \(0 < x <0.5\), then \(x^8<x^2\), and therefore \[ x^{8} + x^{2} <2 x^{2} \quad \Rightarrow \quad \frac1{x^{8} + x^{2}} > \frac{1}{2x^{2}} \]

Since \(\int_0^{0.5} \frac{dx}{2x^{2}}\) diverges, the larger integral \(J\) also diverges.