Explanation Given information: ln | x | Formula: ∫ a b 1 x n d x = [ x − n + 1 − n + 1 ] a b d d x ( ln x ) = [ 1 x ] d d x ( ln ( − x ) ) = [ 1 x ] ∫ u ( x ) v ′ ( x ) d x = u ( x ) v ( x ) − ∫ v ( x ) u ′ ( x ) d x (1) Calculation: Test the below expression using the direct comparison test. ln | x | Write the function f ( x ) as below. f ( x ) = ln | x | Write the function g ( x ) as below. g ( x ) = 1 x The following term cannot be tested using this direct comparison test as the following condition exists as given below. f ( x ) > g ( x ) on [ − 1 , 0 ] f ( x ) < g ( x ) on [ 0 , 1 ] This does not satisfy the condition of continuity within the given limit [ − 1 , 1 ] and the condition 0 ≤ f ( x ) ≤ g ( x ) . The discontinuity exists in the function at the value of 0 for x , where it will reach infinity. Hence, use the limit comparison test. Hence, use the limit comparison test. Test the below expression using the limit comparison test. ln | x | Apply also the following test for the convergence as below. lim x → ∞ f ( x ) g ( x ) = lim x → ∞ ln | x | 1 x = lim x → ∞ x ln | x | = ∞ Therefore, the term is not within the range ( 0 , ∞ ) . Hence, the integral term f ( x ) cannot be tested with this limit comparison test. Express the integral f ( x ) as below. ∫ − 1 1 ln | x | d x = ∫ − 1 0 ln ( − x ) d x + ∫ 0 1 ln ( x ) d x (2) Consider the term ∫ 0 1 ln ( x ) d x from the equation (2). Let us take the term ln x as u ( x ) . u ( x ) = ln x Differentiate the term u with respect to x . u ′ ( x ) = 1 x d x Let us take the term 1 as v ′ ( x ) . Integrate the term with respect to x . ∫ v ′ ( x ) = ∫ d x v ( x ) = x Substitute ln x for u ( x ) , 1 x d x for u ′ ( x ) , x for v ( x ) and 1 for v ′ ( x ) in equation (1)

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Chapter 8.7, Problem 49E

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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 48E

Chapter 8.7, Problem 50E

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