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4.10 Improper Integrals

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4.10 Improper Integrals 4.10. IMPROPER INTEGRALS 237 4.10 Improper Integrals 4.10.1 Introduction b In Calculus II, students defined the integral a f (x) dx over a finite interval [a, b]. The function f was assumed to be continuous, or at least bounded, otherwise the integral was not guaranteed to exist. UnderR these conditions, assuming an b antiderivative of f could be found, a f (x) dx always existed, and was a number. In this section, we investigate what happens when these conditions are not met. R Definition 337 (Improper Integral) An integral is an improper integral if either the interval of integration is not finite (improper integral of type 1) or if the function to integrate is not continuous (not bounded that is becomes infinite) in the interval of integration (improper integral of type 2). 1 x Example 338 e dx is an improper integral of type 1 since the upper limit 0 of integration isZ infinite. 1 dx 1 Example 339 is an improper integral of type 2 because is not con- 0 x x tinuous at 0. Z dx Example 340 1 is an improper integral of type 1 since the upper limit 0 x 1 1 of integration isR infinite. It is also an improper integral of type 2 because x 1 is not continuous at 1 and 1 is in the interval of integration. 2 dx 1 Example 341 2 is an improper integral of type 2 because 2 is 2 x 1 x 1 not continuous atZ 1 and 1. Example 342 tan xdx is an improper integral of type 2 because tan x is Z0 not continuous at . 2 We now look how to handle each type of improper integral. 4.10.2 Improper Integrals of Type 1 These are easy to identify. Simply look at the interval of integration. If either the lower limit of integration, the upper limit of integration or both are not finite, it will be an improper integral of type 1. Definition 343 (improper integral of type 1) Improper integrals of type 1 are evaluated as follows: 238 CHAPTER 4. INTEGRALS t 1. If f (x) dx exists for all t a, then we define ≥ Za t 1 f (x) dx = lim f (x) dx t Za !1Za 1 provided the limit exists as a finite number. In this case, f (x) dx is Za 1 said to be convergent (or to converge). Otherwise, f (x) dx is said a to be divergent (or to diverge). Z b 2. If f (x) dx exists for all t b, then we define ≤ Zt b b f (x) dx = lim f (x) dx t t Z1 !1Z b provided the limit exists as a finite number. In this case, f (x) dx is b Z1 said to be convergent (or to converge). Otherwise, f (x) dx is said to be divergent (or to diverge). Z1 b 1 3. If both f (x) dx and f (x) dx converge, then we define a Z Z1 a 1 1 f (x) dx = f (x) dx + f (x) dx a Z1 Z1 Z The integrals on the right are evaluated as shown in 1. and 2. 4.10.3 Improper Integrals of Type 2 These are more diffi cult to identify. One needs to look at the interval of in- tegration, and determine if the integrand is continuous or not in that interval. Things to look for are fractions for which the denominator becomes 0 in the in- terval of integration. Keep in mind that some functions do not contain fractions explicitly like tan x, sec x. Definition 344 (improper integral of type 2) Improper integrals of type 2 are evaluated as follows: 1. if f is continuous on [a, b) and not continuous at b then we define b t f (x) dx = lim f (x) dx t b Za ! Za 4.10. IMPROPER INTEGRALS 239 b provided the limit exists as a finite number. In this case, f (x) dx is said a b Z to be convergent (or to converge). Otherwise, f (x) dx is said to be a divergent (or to diverge). Z 2. if f is continuous on (a, b] and not continuous at a then we define b b f (x) dx = lim f (x) dx t a+ Za ! Zt b provided the limit exists as a finite number. In this case, f (x) dx is said a b Z to be convergent (or to converge). Otherwise, f (x) dx is said to be a divergent (or to diverge). Z c 3. If f is not continuous at c where a < c < b and both f (x) dx and a b Z f (x) dx converge then we define Zc b c b f (x) dx = f (x) dx + f (x) dx Za Za Zc The integrals on the right are evaluated as shown in 1. and 2. We now look at some examples. 4.10.4 Examples Evaluating an improper integral is really two problems. It is an integral • problem and a limit problem. It is best to do them separately. When breaking down an improper integral to evaluate it, make sure that • each integral is improper at only one place, that place should be either the lower limit of integration, or the upper limit of integration. 1 dx Example 345 2 1 x Z t dx This is an improper integral of type 1. We evaluate it by finding lim 2 . t 1 x t !1Z dx 1 1 1 dx First, = 1 and lim 1 = 1 hence = 1. x2 t t t x2 Z1 !1 Z1 240 CHAPTER 4. INTEGRALS 1 dx Example 346 1 x Z t dx This is an improper integral of type 1. We evaluate it by finding lim t 1 x t !1Z dx 1 dx First, = ln t and lim ln t = hence diverges. x t 1 x Z1 !1 Z1 dx 1 Example 347 2 1 1 + x This is an improperR integral of type 1. Since both limits of integration are infinite, we break it into two integrals. 0 1 dx dx 1 dx 2 = 2 + 2 1 + x 1 + x 0 1 + x Z1 Z1 Z 1 0 dx dx 1 Note that since the function 2 is even, we have 2 = 0 2 ; 1 + x 1 1 + x 1 + x dx we only need to do 1 . R R 0 1 + x2 R t 1 dx dx = lim 1 + x2 t 1 + x2 Z0 !1 Z0 and t dx 1 t 2 = tan x 0 0 1 + x Z 1 1 = tan t tan 0 1 = tan t Thus 1 dx 1 = lim tan t 1 + x2 t 0 !1 Z = 2 It follows that 1 dx = + 1 + x2 2 2 Z1 = Example 348 2 sec xdx 0 This is an improper integral of type 2 because sec x is not continuous at . We R 2 t evaluate it by finding lim sec xdx. t ! 2 Z0 4.10. IMPROPER INTEGRALS 241 First, t sec xdx = ln sec x + tan x t j jj0 Z0 = ln sec t + tan t j j 2 and lim (ln sec t + tan t ) = hence sec xdx diverges. t j j 1 ! 2 Z0 Example 349 sec2 xdx 0 This is an improper integral of type 2, sec2 x is not continuous at . Thus, R 2 2 sec2 xdx = sec2 xdx + sec2 xdx 0 0 Z Z Z 2 2 2 First, we evaluate 0 sec xdx. t R 2 sec2 xdx = lim sec2 xdx t Z0 ! 2 Z0 t sec2 xdx = tan t tan 0 Z0 = tan t Therefore, 2 sec2 xdx = lim (tan t) t Z0 ! 2 = 1 2 2 2 It follows that 0 sec xdx diverges, therefore, 0 sec xdx also diverges. Remark 350 RIf we had failed to see that theR above integral is improper, and had evaluated it using the Fundamental Theorem of Calculus, we would have obtained a completely different (and wrong) answer. sec2 xdx = tan tan 0 Z0 = 0 (this is not correct) 1 dx Example 351 x2 This integral isZ improper1 for several reasons. First, the interval of integration is not finite. The integrand is also not continuous at 0. To evaluate it, we break it so that each integral is improper at only one place, that place being one of the 1 0 1 1 dx dx dx dx 1 dx limits of integration, as follows: 2 = 2 + 2 + 2 + 2 . x x 1 x 0 x 1 x We then evaluate each improperZ integral1 . TheZ1 readerZ will verifyZ that it diverges.Z 242 CHAPTER 4. INTEGRALS Several important results about improper integrals are worth remembering, they will be used with infinite series. The proof of these results is left as an exercise. 1 dx Theorem 352 p converges if p > 1, it diverges otherwise. 1 x Proof. This is anZ improper integral of type 1 since the upper limit of integration t dx is infinite. Thus, we need to evaluate lim . When p = 1, we have already t xp !1 1 seen that the integral diverges. Let us assumeZ that p = 1. First, we evaluate the integral. 6 t dx t = x pdx xp Z1 Z1 x1 p t = 1 p 1 1 p t 1 = 1 p 1 p 1 p The sign of 1 p is important. When 1 p > 0 that is when p < 1, t is in the t1 p 1 numerator. Therefore, lim = thus the integral diverges. t 1 p 1 p !1 1 1 p When 1 p < 0 that is when p > 1, t is really in the denominator so that 1p t 1 1 1 dx lim = and therefore converges.
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