MATH1120 Calculus II Solution to Supplementary Exercises on Improper Integrals Alex Fok November 2, 2013

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1 () Solution : MATH Calculus II Solution to Supplementary Eercises on Improper Integrals Ale Fok November, 3 b ( + )( + tan ) ( + )( + tan ) +tan b du u ln + tan b ( = ln + π ) (Let u = + tan. Then du = + ) Therefore the given improper integral converges. Solution : Note that the improper integral is of type I because there is an infinite integration it. Let f() = ( + )( + tan ) and g() =. Note that + f() g() and both are continuous for [, ). Since By the Direct Comparison Test, () Solution : b g() + tan b = π f() converges as well. 4 b 4 4 b 4 b b sin b sin b b sin b = π cos θ cos θ dθ ( Let = sin θ. Then = cos θdθ) dθ ( 4 = cos θ = cos θ because π θ π and cos θ for this range of θ) Therefore the given improper integral converges. Solution : Note that the integral is of type II because the integrand has an infinite c du discontinuity at =. In order to compare with the improper integral of the form u p, whose integrand has infinite discontinuity at u =, we make the substitution = u

2 (when =, u = ). 4 = 4( u)du 4 ( u) 4 = 4( u)du 4u 6u + 4u 3 u 4 Let f(u) = 4( u) 4u 6u + 4u 3 u 4. A comparison function in this type II case can be obtained by picking the lowest order terms of both the numerator and denominator, as the 4 lowest order terms matter the most when u. So g() can be chosen to be =. 4u u Both f(u) and g(u) are positive and continuous for u (, ], u 4( u) 4u 6u +4u 3 u 4 u u ( u) 4 6u + 4u u 3 = and du converges (because it is of type II and p = < ). By the Limit Comparison u Test, the original integral converges as well. (3) Solution : The given integral is of type I. Let f() =. A comparison function g() + can be obtained by picking the highest order terms of both the numerator and denominator. So g() =. Both f() and g() are positive and continuous for [, ), + + = + and diverges. By the Limit Comparison test, the original integral diverges as well. Solution :

3 3 b + + tan b tan tan b tan sec θdθ sec θ sec θdθ ln tan θ + sec θ ln b + + b ln( + 5) = ( sec θ = sec θ because π < θ < π and sec θ > for θ in that range) Therefore the given integral diverges. (4) Solution : The integral is of type II because the integrand has an infinite discontinuity at =. We can apply the trick similar to that used in () and make the substitution = + u. = du u + u Let f(u) =. g(u) can be chosen by picking the lowest order term of both the u + u numerator and the denominator. So g(u) =. Note that both f(u) and g(u) are u positive and continuous for u (, ], u u+u u u u u + u u + u = and u du converges. converges as well. Solution : By the Limit Comparison Test, the given improper integral

4 4 a + a + a + = a sec sec a sec sec a sec sec = ln( + 3) sec θdθ sec θ tan θdθ tan θ (Let = sec θ. Then = sec θ tan θdθ) sec θdθ (Because a is positive, < θ < π, and tan θ > ) Therefore the given integral converges. (5) See Quiz solution. (6) Solution : The improper integral is of type I. Let f() = g() can be chosen to be = 3. Note that both f() and g() are positive and continuous for [, ), 4 f() g() = 3 and converges. By the Limit Comparison Test, the given improper integral 4 converges as well. Solution : Resolve the integrand into partial fractions: = 3 (4 ) = A + B + C 4 3 = A(4 ) + B(4 ) + C = (4A + C) + (4B A) B

5 4A + C = A = 4B A = 3 = B =. B = C = 4 ( + 4 ) 4 b ( + 4 ) 4 (ln b + ln(4b ) + ln 3) b ln b 4b + + ln 3 = ln ln 3 5 = + ln 3 4 Remark.. While it is legitimate to split the interval of integration, one may not split the integrand of any improper integral. It is incorrect to conclude that the given improper integral in this question diverges because ( + 4 ) 4 + The LHS converges while the RHS diverges. 4 4 (7) Solution : The given improper integral is of mied type because it has two infinite integration its. So e + e = e + e + e + e When, e is the dominant term. Let f() = e, g() = + e e f() g() and both are continuous for [, ), = e. Since by the Direct Comparison Test, g() = e b e (e b + ) = f() converges as well. When, e is the dominant term. So we can let the comparison function for the interval of integration (, ] be h() = e = e. Since f() h() and both are

6 6 continuous for (, ], h() = a a e a ( ea ) = by the Direct Comparison Test, integral converges. Solution : f() converges as well. All in all, the given improper e + e = = e + e = u + u du u u + du = tan u + C = tan e + C e + e + (Let u = e. Then du = e, = du u ) e + e [tan e ] b + a [tan e ] a = π tan + tan = π So the given improper integral converges. (8) The integral is of type I. Let f() = + sin. Note that f() = + sin + = Let g() =. Since f() g() and both are continuous for [, ), and g() converges. By the Direct Comparison Test, the given improper integral converges. (9) See Quiz solution.

7 7 () The integral is of mied type because it has an infinite integration it and the integrand has an infinite discontinuity at =. ln = ln + a + a ln a + ln a ln ln + du u + b ln b ln ln du u (Let u = ln. Then du = ) a +(ln(ln ) ln(ln a)) + (ln(ln b) ln(ln )) = ln(ln ) () + ln(ln ) = DNE So the given improper integral diverges. () The integral is of type II with infinite discontinuity at =. Let f() = ln and g() =. Note that g() f(). In order to apply the version of the Direct ln Comparison Test which requires the non negativity of relevant functions, we shall consider instead f() and g(). We have g() f() and both are continuous for (, ]. Note that g() a + By the Direct Comparison Test, a ln a + [ ln(ln )] a = f() diverges. So does the given improper integral. () The integral is of mied type because there are two infinite integration its and the integrand has one infinite discontinuity at =. So ( + e ) = ( + e ) + ( + e ) + ( + e ) + ( + e ) Consider ( + e ). Let f() = ( + e ) and g() =. Both of them are positive and continuous for (, ], f() + g() + (+e ) + + e = and diverges. By the Limit Comparison Test, so does the given improper integral. ( + e diverges as well, and )

8 8 (3) Applying integration by parts twice, we have e cos = e sin e cos + C [ e e sin e ] b cos cos ( e sin e ) cos = (4) = (Here e b sin b = by Squeeze Theorem: e b e b sin b e b and ±e b =. Same reason for e b cos b = ) So the given improper integral converges. Remark.. We may not apply the version of the Comparison Tests as in the notes and the tetbook in this case because the integrand e cos is not always nonnegative for all [, ). Nonetheless, we may apply the following Squeeze-Theorem type argument: as e e cos e for [, ) and both does the given integral. dt is of type II. Let f(t) = t and continuous for t (, ], and t f(t) t + g(t) t + = dt diverges. By the Limit Comparison Test, t + t dt + t cos + + cos = e and e converge, so and g(t) =. Both f(t) and g(t) are positive t t dt diverges as well. Now dt (It is indeed of the form ) (By L Hopital s Rule)