Explanation Given information: 1 θ 2 − 1 Formula: ∫ a b 1 x d x = [ ln | x | ] a b ∫ a b 1 c x + d d x = [ 1 c ln | c x + d | ] a b Calculation: Test the below expression using the direct comparison test. 1 θ 2 − 1 = 1 ( θ − 1 ) ( θ + 1 ) Write the function f ( θ ) as below. f ( θ ) = 1 ( θ − 1 ) ( θ + 1 ) Write the function g ( θ ) as below. g ( θ ) = 1 θ The following term cannot be tested using this direct comparison test as the following condition exists as given below. f ( θ ) < g ( θ ) on [ 0 , 1 ] f ( θ ) > g ( θ ) on [ 1 , ∞ ] This does not satisfy the condition of continuity within the given limit [ 0 , ∞ ] and the condition 0 ≤ f ( θ ) ≤ g ( θ ) . The discontinuity exists in the function at the value of 1 for θ , where it will reach infinity. Hence, use the limit comparison test. Test the below expression using the limit comparison test. 1 θ 2 − 1 = 1 ( θ − 1 ) ( θ + 1 ) Apply also the following test for the convergence as below. lim θ → ∞ f ( θ ) g ( θ ) = lim θ → ∞ 1 ( θ 2 − 1 ) 1 θ = lim θ → ∞ θ ( θ 2 − 1 ) Apply l’Hospital’s rule as the term lim θ → ∞ θ ( θ 2 − 1 ) yields lim θ → ∞ ∞ ∞ . lim θ → ∞ f ( θ ) g ( θ ) = lim θ → ∞ 1 ( 2 θ ) = lim θ → ∞ 1 ∞ = 0 Therefore, the term is not within the range ( 0 , ∞ ) . The value should fall in between the range 0 < x < ∞ and the functions f ( θ ) and g ( θ ) are not continuous. Hence, the integral term f ( θ ) cannot be tested with this limit comparison test. Express the integral as f ( θ ) having discontinuity at θ equal to 1 as below. ∫ 0 ∞ d θ ( θ − 1 ) ( θ + 1 ) = lim b → 1 − ∫ 0 b d θ ( θ − 1 ) ( θ + 1 ) + lim b → ∞ ∫ 1 b d θ ( θ − 1 ) ( θ + 1 ) (1) Write the integrand term in the form of partial fraction as below. 1 θ 2 − 1 = 1 ( θ − 1 ) ( θ + 1 ) = A ( θ − 1 ) + B ( θ + 1 ) Simplify it as below

University Calculus: Early Transcendentals (4th Edition)

4th Edition

ISBN: 9780134995540

Author: Joel R. Hass, Christopher E. Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr.

Publisher: PEARSON

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Chapter 8.7, Problem 38E

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Whether an integral converges or diverges by testing it either using the direct comparison test, or the Limit comparison test by adopting any one of the appropriate method.

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Chapter 8.7, Problem 37E

Chapter 8.7, Problem 39E

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