Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:

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1 Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl is improper when it doesn t meet the conditions given ove. Specificlly, one of two things could hppen: The intervl isn t of the form [, ] ecuse one or the other or oth of the endpoints of the intervl is infinite. f is not continuous on the intervl [, ] ecuse there s point on [, ] where f is not defined. We ll refer to these s Type A nd Type B improper integrls, respectively. There is one more cse, where oth of the prolems ove hppen for the sme integrl; we cll this Type C improper integrl. We ll mostly e concerned with Type A, so don t worry too much out Types B nd C. Solving Type A Improper Integrl We cn get round the prolem of hving ± s n upper or lower ound y first replcing it with vrile. Usully, this new vrile is,, or c, ut you cn use just out whtever letter you like. To see this, let s look t the integrl e x dx. Our first step is to replce the here, I ll use to get new integrl: e x dx. Strictly speking, this isn t exctly the integrl tht we strted with. But it does stisfy the criteri from the First Fundmentl Theorem, so we cn evlute it in the usul wy: e x dx = e x = e e = e +. At this point, we ve found the re under the curve e x on the intervl [, ]. We cn evlute the originl integrl y letting : e x dx e x dx

2 e + e +. We now clculte the limit y noticing tht e = ; s, the denomintor goes to. e This mens tht the frction goes to zero. This gives us the vlue of our integrl: e e x dx e + = + =. So the re under the curve e x from to is. In this cse, we sy tht the integrl converges to. Of course, there s nothing to sy tht we ll lwys get limit tht isn t infinite. For exmple, let s try to evlute x dx. First of ll, we hve n infinite lower ound insted of n infinite upper ound. We get our new integrl y replcing the lower ound I ll replce it with : x dx. Agin, we ve now rigged things so tht we hve n integrl tht stisfies the conditions of the First Fundmentl Theorem. Evluting our new integrl, we get x dx = 4 x4 = = Now we evlute our originl integrl y tking the limit s : x dx = x dx lim = 4 lim Notice tht since is negtive nd pproching, 4 is positive nd pproching + ; thus, our limit is. In this cse, we didn t get finite vlue for our limit. So we sy tht the integrl diverges. In oth of the integrls we just evluted, we hd either or s one of our ounds of integrtion. If the integrl hs oth nd s ounds, we cn still evlute it. Specificlly, suppose tht we wnt to evlute the integrl dx. Here, our first cse is to rek up x 2 + our intervl nd our integrl into two pieces. We cn pick relly ny point we wnt, ut it s dvisle to pick something tht s esy to compute. I ll try reking things prt t x =. This mens tht we cn tke x 2 + dx = 4 4 x 2 + dx + x 2 + dx 2

3 Notice tht this reking the intervl prt rule works even if the intervl is finite nd the integrl isn t improper. We re just using it here s tool to get nicer computtions. By using this, we hve tht our originl integrl is sum of two other ones. The two new integrls re of forms we ve lredy seen, so we cn evlute them y replcing the infinities in the ounds; let s do this s shown elow: x 2 + dx + x 2 + dx Agin, you cn cll the new ounds whtever you like, ut it s importnt here tht you use different letters in the two integrls so tht you don t incorrectly cncel things lter on. We evlute these two integrls y reclling tht rctnx is n ntiderivtive of x 2 + dx + x 2 + dx = rctnx + rctnx = rctn rctn + rctn rctn. The first thing to notice is tht rctn =. The second thing is to recll tht lim x rctnx = π nd tht lim 2 x rctnx = π. Don t worry if you don t rememer this fct out limits 2 of rctn; this sort of thing won t come up on the finl exm. In ny cse, we now tke limits to evlute our originl integrl: x 2 + dx = = rctn lim x 2 + dx + x 2 + dx x 2 + dx + lim x 2 + dx rctn + lim rctn rctn = π + π 2 2 = π. It turns out tht this integrl converges, to π. Now, it s entirely possile tht the integrl won t converge in ny nice wy. The wy to decide whether or not n integrl of the to vriety diverges is to see whether or not either of the pieces diverge once you ve roken it prt. If t lest one of the two new integrls diverges, then your originl one diverges. x 2 + : Solving Type B Improper Integrl Integrls of Type B re those which hve finite intervl, ut ren t continuous everywhere on this intervl. Specificilly, there is point in the intervl somewhere for which the function isn t defined. This is wht I ve een clling the prolem point. If the prolem point is t one or the other of the endpoints, we my still evlute the integrl y first inserting new ound.

4 4 As n exmple, let s suppose we wnt to evlute dx. The prolem point here is t x 2 x =, since is not defined t this point. We get round this y replcing with s ound x 2 of integrtion: x dx. 2 Techniclly, is smll positive numer. Think of this s strting lte when constructing our intervl. The point, though, is tht we now hve n integrl tht meets the conditions of the First Fundmentl Theorem, nd we cn evlute our new integrl in the usul wy: x dx = 2 x = +. We cn now recover our originl integrl y tking limit: lim + = + lim + + Notice tht we chose to e positive numer; so when we let pproch zero, we re only tking right-hnd limit. It s now esy to see tht the limit ove diverges to. Consequently, our originl integrl diverges. Another sitution tht could rise with Type B integrls is if the prolem point isn t t the endpoints of our intervl, ut is somewhere in etween. To see this, we consider the exmple of dx. We mke use of the sme reking prt trick tht we did for certin integrls of x Type 2/ A: dx = dx + dx. x2/ x2/ x2/ However, the choice of rek prt point mtters here. We need to choose the prolem point in this integrl it s x = s the point t which we rek prt the integrl. Notice tht we now hve two integrls of Type B; wht s more, we hve lredy discussed how to evlute these types. Crrying out the process, we replce the prolem endpoint in ech integrl: dx + x2/ dx. x2/ We finlly hve two integrls tht meet the conditions of the First Fundmentl Theorem. Let s evlute them: dx + dx x2/ x2/ = x / + x / = / / + / / = / + + / = / / + 6. And lstly, we tke limits s nd s + : dx dx x2/ x2/ + lim + dx. x2/

5 5 / lim + / + 6. But oth of these limits re zero, since they oth involve positive powers of nd. Consequently, 6 is ll tht s left. And thus, Finl Comments dx = 6. x2/ Notice tht we hven t discussed how to test the convergence of improper integrls other thn to mke n ttempt t evluting integrls. Wht ends up hppening with ll of the ove exmples is tht either we find vlue, or find tht the integrl diverges.

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