# Remark weve seen before that ş 1 1x d xdiverges if

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Remark: we’ve seen before that ş 1 0 1 x d x diverges. If we imagine the solid that would result from choosing a = 0, it would have a scant volume of π cubic units, but a silhouette (side view) of infinite area. S-18: Our goal is to decide when this integral diverges, and where it converges. We will leave q as a variable, and antidifferentiate. In order to antidifferentiate without knowing q , we’ll need different cases. The integrand is x ´ 5 q , so when ´ 5 q ‰ ´ 1, we use the power 446
rule (that is, ş x n d x = x n + 1 n + 1 ) to antidifferentiate. Note x ( ´ 5 q )+ 1 = x 1 ´ 5 q = 1 x 5 q ´ 1 . ż t 1 1 x 5 q d x = \$ & % h x 1 ´ 5 q 1 ´ 5 q i t 1 with 1 ´ 5 q ą 0 if q ă 1 5 h log x i t 1 if q = 1 5 h 1 ( 1 ´ 5 q ) x 5 q ´ 1 i t 1 with 5 q ´ 1 ą 0 if q ą 1 5 = \$ & % 1 1 ´ 5 q ( t 1 ´ 5 q ´ 1 ) with 1 ´ 5 q ą 0 if q ă 1 5 log t if q = 1 5 1 5 q ´ 1 ( 1 ´ 1 t 5 q ´ 1 ) with 5 q ´ 1 ą 0 if q ą 1 5 . Therefore, ż 8 1 1 x 5 q d x = lim t Ñ8 ż t 1 1 x 5 q d x = \$ & % 1 1 ´ 5 q lim t Ñ8 t 1 ´ 5 q ´ 1 = 8 if q ă 1 5 lim t Ñ8 log t = 8 if q = 1 5 1 5 q ´ 1 1 ´ lim t Ñ8 1 t 5 q ´ 1 = 1 5 q ´ 1 if q ą 1 5 . The first two cases are divergent, and so the largest such value is q = 1 5 . (Alternatively, we might recognize this as a “ p -integral” with p = 5 q , and recall that the p -integral diverges precisely when p ď 1.) S-19: This integrand is a nice candidate for the substitution u = x 2 + 1, 1 2 d u = x d x . Remember when we use substitution on a definite integral, we also need to adjust the limits of integration. ż 8 0 x ( x 2 + 1 ) p d x = lim t Ñ8 ż t 0 x ( x 2 + 1 ) p d x = lim t Ñ8 1 2 ż t 2 + 1 1 1 u p d u = lim t Ñ8 1 2 ż t 2 + 1 1 u ´ p d u = \$ & % 1 2 lim t Ñ8 u 1 ´ p 1 ´ p t 2 + 1 1 if p 1 1 2 lim t Ñ8 h log | u | i t 2 + 1 1 if p = 1 = \$ & % 1 2 lim t Ñ8 1 1 ´ p h ( t 2 + 1 ) 1 ´ p ´ 1 i if p 1 1 2 lim t Ñ8 h log ( t 2 + 1 ) i = 8 if p = 1 447
At this point, we can see that the integral diverges when p = 1. When p 1, we have the limit lim t Ñ8 1/2 1 ´ p h ( t 2 + 1 ) 1 ´ p ´ 1 i = 1/2 1 ´ p lim t Ñ8 ( t 2 + 1 ) 1 ´ p ´ 1/2 1 ´ p Since t 2 + 1 Ñ 8 , this limit converges exactly when the exponent 1 ´ p is negative; that is, it converges when p ą 1, and diverges when p ă 1. So, the integral in the question converges when p ą 1. S-20: First, we notice there is only one “source of impropriety:” the domain of integration is infinite. (The integrand has a singularity at t = 1, but this is not in the domain of integration, so it’s not a problem for us.) We should try to get some intuition about whether the integral converges or diverges. When t Ñ 8 , notice the integrand “looks like” the function 1 t 4 . We know ş 8 1 1 t 4 d t converges, because it’s a p -integral with p = 4 ą 1 (see Example 1.12.8 in the CLP–II text). So, our integral probably converges as well. If we were only asked show it converges, we could use a comparison test, but we’re asked more than that. Since we guess the integral converges, we’ll need to evaluate it. The integrand is a rational function, and there’s no obvious substitution, so we use partial fractions. 1 t 4 ´ 1 = 1 ( t 2 + 1 )( t 2 ´ 1 ) = 1 ( t 2 + 1 )( t + 1 )( t ´ 1 ) = At + B t 2 + 1 + C t + 1 + D t ´ 1 Multiply by the original denominator.