Improper Integrals The integrals we have studied so far represent signed areas of bounded regions.

There are two ways an integral can be improper: (1) The interval of integration may be infinite. (2) The integrand may tend to infinity. We deal first with improper integrals over infinite intervals. One or both endpoints may be infinite: Improper Integrals However, areas of unbounded regions also arise in applications and are represented by improper integrals. Bell-shaped curve. The region extends infinitely far in both directions, but the total area is finite.

How can an unbounded region have finite area? To answer this question, we must specify what we mean by the area of an unbounded region. Consider the area under the graph of f (x) = e −x over the finite interval [0,R]: this area approaches a finite value : It seems reasonable to take this limit as the definition of the area under the graph over the infinite interval Thus, the unbounded region has area 1.

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 is defined as the following limit (if it exists): We say that the improper integral converges if the limit exists (and is finite) and that it diverges if the limit does not exist. Similarly, we define A doubly infinite improper integral is defined as a sum (provided that both integrals on the right converge):

Step 1. Integrate over a finite interval [2, R]. Step 2. Compute the limit as R → ∞.

CONCEPTUAL INSIGHT If you compare the unbounded shaded regions in the last two examples, 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 Our calculations show that x −3 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 For a > 0,

Improper integrals arise in applications when it makes sense to treat certain large quantities as if they were infinite. For example, an object launched with escape velocity never falls back to earth but rather, travels “infinitely far” into space. Escape Velocity The earth exerts a gravitational force of magnitude F (r) = GM e m/r 2 on an object of mass m at distance r from the center of the earth.distance (a) Find the work required to move the object infinitely far from the earth. REMINDER M e ≈ 5.98 · kg, r e ≈ 6.37 · 10 6 m The universal gravitational constant is… G ≈ 6.67 · 10 −11 N-m 2 /kg 2 A newton is… 1 kg-m/s 2 A joule is… 1 N-m. The work required to move an object from the earth’s surface (r = r e ) to a distance R from the center is

Escape Velocity The earth exerts a gravitational force of magnitude F (r) = GM e m/r 2 on an object of mass m at distance r from the center of the earth.distance (b) Calculate the escape velocity υ esc on the earth’s surface. The work required to move the object infinitely far from the earth. By the principle of Conservation of Energy, an object launched with velocity υ 0 will escape the earth’s gravitational field if its kinetic energy is at least as large as the work required to move the object to infinity—that is, if

Perpetual Annuity An investment pays a dividend continuously at a rate of $6000/year. Compute the present value of the income stream if the interest rate is 4% and the dividends continue forever. it was shared that the present value (PV) after T years at interest rate r = 0.04 is Over an infinite time interval, Although an infinite number of dollars are paid out during the infinite time interval, their total present value is finite.

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

An integral over a finite interval [a, b] is improper if the integrand becomes infinite at one or both of the endpoints of the interval. In this case, the region in question is unbounded in the vertical direction. For example, Infinite Discontinuities at the Endpoints is improper because the integrand f (x) = x −1/2 tends to as x → 0+. Improper integrals of this type are defined as one-sided limits.

DEFINITION Integrands with Infinite Discontinuities If f (x) is continuous on [a, b) but discontinuous at x = b, we define Similarly, if f (x) is continuous on (a, b] but discontinuous at x = a, In both cases, we say that the improper integral converges if the limit exists and that it diverges otherwise.

The integral is improper because the integrand has an infinite discontinuity at x = 0.

THEOREM 2 The p-Integral over [0,a] For a > 0, Theorem 2 is valid for all exponents p. However, the integral is not improper if p < 0.

In Section 9.1, we will compute the length of a curve as an integral. It turns out that the improper integral in our next example represents the length of one-quarter of a unit circle. Thus, we can expect its value to beSection 9.1

Sometimes we are interested in determining whether an improper integral converges, even if we cannot find its exact value. For instance, the integral cannot be evaluated explicitly. However, if x ≥ 1, then In other words, the graph of y = e −x /x lies underneath the graph of y = e −x for x ≥ 1. Therefore

Since the larger integral converges, we can expect that the smaller integral also converges (and that its value is some positive number less than e −1 ). This type of conclusion is stated in the next theorem. THEOREM 3 Comparison Test for Improper Integrals Assume that f (x) ≥ g (x) ≥ 0 for x ≥ a. 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. 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.

Notice that if 0 < x < 0.5, then x 8 < x 2, and therefore