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Remark: we’ve seen before thatş101xdxdiverges. If we imagine the solid that wouldresult from choosinga=0, it would have a scant volume ofπcubic units, but asilhouette (side view) of infinite area.S-18:Our goal is to decide when this integral diverges, and where it converges. We willleaveqas a variable, and antidifferentiate. In order to antidifferentiate without knowingq, we’ll need different cases. The integrand isx´5q, so when´5q‰ ´1, we use the power446
rule (that is,şxndx=xn+1n+1) to antidifferentiate. Notex(´5q)+1=x1´5q=1x5q´1.żt11x5qdx=$’’’’’’&’’’’’’%hx1´5q1´5qit1with 1´5qą0ifqă15hlogxit1ifq=15h1(1´5q)x5q´1it1with 5q´1ą0ifqą15=$’&’%11´5q(t1´5q´1)with 1´5qą0ifqă15logtifq=1515q´1(1´1t5q´1)with 5q´1ą0ifqą15.Therefore,ż811x5qdx=limtÑ8żt11x5qdx=$’’’’’’&’’’’’’%11´5qlimtÑ8t1´5q´1=8ifqă15limtÑ8logt=8ifq=1515q´11´limtÑ81t5q´1=15q´1ifqą15.The first two cases are divergent, and so the largest such value isq=15. (Alternatively,we might recognize this as a “p-integral” withp=5q, and recall that thep-integraldiverges precisely whenpď1.)S-19:This integrand is a nice candidate for the substitutionu=x2+1,12du=xdx.Remember when we use substitution on a definite integral, we also need to adjust thelimits of integration.ż80x(x2+1)pdx=limtÑ8żt0x(x2+1)pdx=limtÑ812żt2+111updu=limtÑ812żt2+11u´pdu=$’’’&’’’%12limtÑ8u1´p1´pt2+11ifp‰112limtÑ8hlog|u|it2+11ifp=1=$’&’%12limtÑ811´ph(t2+1)1´p´1iifp‰112limtÑ8hlog(t2+1)i=8ifp=1447
At this point, we can see that the integral diverges whenp=1. Whenp‰1, we have thelimitlimtÑ81/21´ph(t2+1)1´p´1i=1/21´plimtÑ8(t2+1)1´p´1/21´pSincet2+1Ñ 8, this limit converges exactly when the exponent 1´pis negative; thatis, it converges whenpą1, and diverges whenpă1.So, the integral in the question converges whenpą1.S-20:•First, we notice there is only one “source of impropriety:” the domain of integrationis infinite. (The integrand has a singularity att=1, but this is not in the domain ofintegration, so it’s not a problem for us.)•We should try to get some intuition about whether the integral converges ordiverges. WhentÑ 8, notice the integrand “looks like” the function1t4. We knowş811t4dtconverges, because it’s ap-integral withp=4ą1 (see Example 1.12.8 inthe CLP–II text). So, our integral probably converges as well. If we were only askedshow 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 arational function, and there’s no obvious substitution, so we use partial fractions.1t4´1=1(t2+1)(t2´1)=1(t2+1)(t+1)(t´1)=At+Bt2+1+Ct+1+Dt´1Multiply by the original denominator.