Mat104 Fall 2002, Improper Integrals From Old Exams

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1 Mat4 Fall 22, Improper Integrals From Old Eams For the following integrals, state whether they are convergent or divergent, and give your reasons. () (2) (3) (4) (5) converges. Break it up as The first of these is proper and finite. The second behaves like the integral of / 3 on [, ) and thus converges. + converges. As, goes to much more slowly than does. (Think about the graphs.) Therefore when is very close to, the denominator +. So this integral will behave like the integral of / on [, ], and this integral converges. + converges. As goes to, the integrand behaves like 3 = 3. 5/2 diverges. Break it up into two integrals The 3 + first integral is proper and finite. The second can be compared to the integral of / on [, ) which diverges. ln converges to. Here we can compute directly since integration by parts tells us that ln = ln + C. Evaluating at the = endpoint gives ln =. For the other endpoint we have to take the limit as goes to. For this we need L Hôpital s rule. ln lim ln = lim / = lim / / = lim =. 2 So evaluating at the = endpoint gives. (6) diverges. The only difficulty is that the denominator is when =. There are e a couple of approaches we could take. The easiest is to use the Taylor series for e. Then we know that e = + higher powers of and as goes to zero, the higher powers of will vanish much more rapidly. So this function behaves essentially like / when is close to. Since / diverges, this integral will also. Alternatively, we could compute the integral, making the substitution u = e and then use partial fractions. (7) converges. The only difficulty is that we have an infinite endpoint. The integrand is asymptotic to / as goes to infinity. Since / converges, this integral will as well. (To compare these we should break up the integral. First integrate from to, which gives a finite value. Then integrate further from out to. This gives a finite value as well by comparison to /.) 3 (8) diverges. Again the only problem is that we have an infinite endpoint. As ln + 4 goes to infinity, 4 grows much faster than ln. Thus the integrand will be asymptotic to 3 / 4 = / as goes to infinity.

2 2 (9) 3 + converges. Break it up into an integral from to plus the integral from to. When is close to, the integrand will behave like / since 3 goes to much more rapidly than does. Since the integral of / on [, ] converges, so will /( 3 + ). As goes to infinity, grows much more slowly than 3, so /( 3 + ) / 3 when is very large. Since the integral of / 3 on [, ) is converges, so will the integral of /( 3 + ) on [, ). () diverges. Here the easiest method is to use the Taylor series for cos. It tells cos us that cos = /2 + higher powers of. Since the higher powers of die out more rapidly when is close to, /( cos ) behaves like 2/ as goes to. Therefore the given integral will behave like the integral of 2/ on [, ] and this integral diverges. () (2) (3) (4) (5) (6) (7) e cos converges. We could use integration by parts twice to compute the integral and then take limits. On the other hand, since e dies out more rapidly than any power of, we can conclude that e < once gets big enough, say, when > (Check it graphically). So e cos < cos < Since the integral of / on [, ) converges, so will the integral of e cos on [, ). Since there is no problem with our function on [, ], the given integral converges. e 2 diverges. We have to split it up and think about what happens as we approach and what happens as we approach infinity separately. To think about what is happening at the endpoint, we notice that the numerator goes to. So e 2 / / as goes to zero. Since the integral of / on [, ] diverges, so will the integral of e 2 /. (Remark: The integral of this function on [, ) will converge again because the eponential dies out very very rapidly.) + converges. The only problem is that we have an infinite endpoint SInce the integrand is asymptotic to / 3 the integral will converge diverges. The only problem is that we have an infinite endpoint. The integrand is asymptotic to / so the integral diverges. e diverges. The only issue is what happens at. Since the numerator approaches this function will behave like / as goes to zero. sin converges. Again the only issue is what happens as we approach. Since sin when is close to zero, we see that the integrand behaves like / =. + converges. As goes to zero, dominates. (the other term dies out much faster) So this integral behaves like / near zero.

3 3 (8) (9) ( ) 2/3 converges. Compute directly ( ) 3 diverges. The only issue is that we have an infinite endpoint. 3 As goes to infinity, the highest powers of will dominate. So the integrand will behave like /( 3/2 3/2 ) = / 3 = /. (2) (2) (22) (23) (24) (25) (26) (27) (28) (29) tan diverges. Compute directly, using the substitution y = cos. e converges. The only issue is the infinite endpoint. When is large the e 2 + integrand will behave like e /e 2 = /e. Compute directly or use the fact that e grows faster than any power of so /e dies out faster than any power of. 2 sin converges. Compare to /2. sin + 4 converges. Compare to /( + 4 ) and then to / 4. sin + converges. When is small the numerator will be well-approimated by 4 and the denominator will be well-approimated by. So the integrand behaves like / when goes to zero. diverges. This is the same as integrating /( ) which behaves like integrating /. diverges. Compute directly or use the p-test converges. Near the integrand behaves like 2/3 /2/3 which gives a convergent integral on [, ]. When is large the integrand behaves like / 4 which gives a convergent integral on [, ). 3 e converges. Compute directly (a pain) or use the fact that the eponential dies out faster than any power of, say faster than 5. This allows you to compare the integral to that of / which gives convergence. ln converges. Since ln grows more slowly than any power of we can say + 2 that ln /( + ) < /( + ) when is large enough. Since /( + ) / 3/2 we get convergence at the infinite endpoint, the only possible problem. (3) diverges. This integral has problems at both endpoints. This means we have ln to split the integration, say integrating first from to 2 and then integrating again from 2

4 4 (3) (32) (33) (34) to. To understand what is happening at = we could make the substitution = u+ ln = du (u ) 2 ln( + u) and then use a known Taylor series to understand this integral. Since ln( + u) = u u 2 /2 + u 3 /3... we see that the denominator ( 2u + u 2 )(u u 2 /2 + u 3 /3... ) is of the form u + higher powers of u. So the integrand behaves like /u as u goes to zero, and therefore this integral (as well as the original integral) diverges. While we re here let me say that the other integral, as runs from 2 to converges. To see this, observe that ln > and thus /( ln ) < /. By comparison 2 ln converges. sin converges. The only problem is the = endpoint. When is small, sin, so this integral behaves like that of / and converges. e ( + e 2 ) diverges. Multiplying out the integrand we get e + e The second integral here is finite, and the first is infinite since e goes to infinity as does. ln converges. Compute directly using integration by parts and take the limit using L Hôpital s Rule. converges. The only problem is the infinite endpoint. The integrand is asymptotic to / 3 as goes to 2 3 infinity. (35) (36) (37) (38) (39) + cos diverges. The only problem here is that denominator vanishes at =. Since the numerator approaches 2 as goes to, the integrand behaves like 2/ when goes to. ln cos + converges. Here the only problem is the infinite endpoint. ln cos + ln + < since ln grows more slowly than any power of. + 3/2 ( + ) converges. When is close to zero, then dominates. That is as. When is very large, then = 3/2 dominates + 4 converges. as ( + ) ( + ) 3/ converges. When is close to, 3 dominates. When is very large, dominates.

5 5 Other problems involving improper integrals () Find the arc length of the curve given by = e t cos t and y = e t sin t for t <. = e t sin t e t cos t dy = e t cos t e t sin t ( ) 2 + ( ) 2 dy = = 2e 2t So the arc length is given by the improper integral 2e t = 2. (2) Find te t or show that it diverges. Use integration by parts to show that te t e t and then (3) Evaluate e te t =. te t = arcsin(ln ). Make the substitution w = ln and the integral becomes e /2 arcsin(ln ) = arcsin(w) dw Using integration by parts with u = arcsin w and dv = dw we find that arcsin w dw = w arcsin w + w 2 and the definite integral works out to be π (4) Evaluate +. Here we get lim t arctan t arctan = π 2 π 4 = π 4.

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