4JJGB` EkBhcImproper integralskBhc4-An example that leads to an improper integral+kBhcConsider a rocket launched from the surface of the earth. How much work is needed to move the rocket from the surface of the earth to a certain height? The force the rocket moves against is the force of gravity,kBhcfGrapkBhcxkBhc  = - carF>kBhc m kkBhcxopxEkBhc2, kBhc where kBhcxkBhcE> is the distance from the center of the earth in (say) miles, kBhcmkBhc' is the mass of the rocket, and kBhckkBhc is a constant. Let us assume that the rocket has a weight of 100,000 lbs at the surface of the earth, which is roughly 4000 miles from the center of the earth. This allows us to determine kBhc k mkBhc.-kBhc -100000 = carFMkBhc m kGrapkBhc 4000opxEkBhc2.kBhc m k = 1.6 10opxEkBhc 12kBhc lb - mileopxEkBhc2/kBhc]VGiven all this, how much work does it take to lift the rocket up to, say 400 miles up?0kBhc W = etnIcarF>kBhc m kkBhcxopxEkBhc2kBhcxkBhc 4400kBhc 4000*kBhcHow much work does it take to lift the rocket up to a point in space where the Earths gravity is essentially negligible? Certainly if we go all the way out to infinity, the gravitational attraction of the Earth will be negligible.kBhc W = etnI}carF>kBhc m kkBhcxopxEkBhc2kBhcxkBhckBhc 4000kBhcOHHow do we do this integral? The naive way to procede is to pretend that kBhckBhcXQ is a number and do this integral the same way we would do any definite integral.kBhc W = - carF&kBhc m kkBhcxkBhc  |sbuSkBhc 4000opxEkBhckBhcmfThe more proper way to do this integral is to side-step the issue of the infinity by means of a limit.kBhc W = tmiLkBhc A kBhc GrapetnI}carF>kBhc m kkBhcxopxEkBhc2kBhcxkBhcAkBhc 4000kBhc  = tmiLkBhc A kBhc GrapkBhc - carF&kBhc m kkBhcxkBhc  |sbuSkBhc 4000opxEkBhcA kBhc = tmiLkBhc A kBhc GrapzkBhc - carF&kBhc m kkBhcAkBhc  + carF)kBhc m kkBhc 4000kBhc  = carF)kBhc m kkBhc 4000 kBhcThe integral with an kBhckBhc| in it is an example of an improper integral. The best way to understand improper integrals is to look at a proper integral: etnI\kBhcfGrapkBhcxkBhcxkBhcbkBhca kBhcTo make this proper: kBhc Both kBhcakBhc  and kBhcbkBhc" have to be finite numbers.kBhc f(x)kBhc1* also has to be continuous over the range kBhc [a,b]kBhc.kBhc~How do you deal with an improper integral? Use some sort of limit to put off the problem until you are done with the integral.kBhcHere are some examples: etnI\kBhceopxEkBhcxkBhcxkBhckBhc0kBhc  = tmiLkBhc A kBhc GraphetnI\kBhceopxEkBhcxkBhcxkBhcAkBhc0kBhc  = tmiLkBhc A kBhc GraphkBhceopxEkBhcxkBhc|sbuSkBhc0opxEkBhcAkBhc  = tmiLkBhc A Grap;kBhceopxEkBhcAkBhc  - 1kBhc  = etnIxcarF<kBhc1kBhcxopxEkBhc3kBhcxkBhc1kBhc0kBhc  = tmiLkBhc A 0kBhc etnIxcarF<kBhc1kBhcxopxEkBhc3kBhcxkBhc1kBhcAkBhc  = tmiLkBhc A 0Grap2kBhc -1/2 xopxEkBhc -2kBhc|sbuSkBhcAopxEkBhc1kBhc  = tmiLkBhc A 0Grap\kBhc-1/2 + carF>kBhc1kBhc 2 AopxEkBhc2kBhc  = +etnIpcarF4kBhc1trqSkBhcxkBhcxkBhc1kBhc0kBhc  = tmiLkBhc A 0kBhc etnIpcarF4kBhc1trqSkBhcxkBhcxkBhc1kBhcAkBhc  = tmiLkBhc A 0Grap0kBhc2trqSkBhcxkBhc|sbuSkBhcAopxEkBhc1kBhc  = tmiLkBhc A 0Grap@kBhc 2 - 2trqSkBhcAkBhc kBhc  = 2kBhc81If the limit is finite, we say that the integral kBhc convergeskBhc70, if it is not finite, we say that the integral kBhcdivergeskBhc.)kBhcaZThe final example shown here strikes some people as rather odd. The integrand blows up to kBhc +kBhc  as kBhcxkBhc approaches 0, so you might expect at first that the area under the curve is also infinite. This is actually not the case, because the function does not blow up fast enough to make the area infinite. Note that in the example preceding this one the functiokBhcRKn does blow up quickly enough to make the area under the curve be infinite.kBhc:3The next example shows a related thing going on at kBhckBhc.etnI]kBhceopxEkBhc -xkBhcxkBhckBhc0kBhc  = tmiLkBhc A etnI]kBhceopxEkBhc -xkBhcxkBhcAkBhc0kBhc  = tmiLkBhc A Grap/kBhc - eopxEkBhc -xkBhc|sbuSkBhc0opxEkBhcAkBhc  = tmiLkBhc A Grap>kBhc - eopxEkBhc -AkBhc  + 1kBhc  = 1kBhcIf you look at a plot of the integrand, you can see that it decays quickly at infinity, thus making it possible for the area under the curve to be finite.tlP2$@x@ykBhceopxEkBhc -x@2@4@6 @8$@10?0.2?0.4433333?0.6?0.8?1333333?1.2ffffff?1.4?1.6?1.8?2$$@$@)6/aG>M ?;G?WO3C?A~?'V7T?;-O?;_{?Bq#?"+h?o}?7Ga?ԕd?G?km?:h?ۺm?-?3cf@< ?t@kTq?gJg!i@Kktܝ?8-$@e+"f_?JF @p /=?>+e @a??]R_p @.#?kwa@9> D?etnIycarF<kBhc1kBhcxopxEkBhc3kBhcxkBhc0kBhc -1?kBhcand hence diverges. Well, it turns out ultimately that the integral does diverge, but we have to be a little more careful with our reasoning. The correct way to handle this example is to use partial fractions to isolate the trouble spot.@etnIcarFKkBhc1kBhcxopxEkBhc3kBhc  - 1kBhcxkBhc1kBhc0kBhc  = etnIcarFwkBhc1GrapkBhc x - 1Grap?kBhcxopxEkBhc2kBhc + x + 1kBhcxkBhc1kBhc0kBhc  = etnIcarFXkBhc -x/3 - 2/3kBhcxopxEkBhc2kBhc + x + 1kBhc  + carF*kBhc 1/3kBhc x - 1kBhcxkBhc1kBhc0BkBhcThe first of these two integrals is a proper integral over the interval in question, while the second integral matches the pattern in feR :(1)kBhc above with kBhc p = 1kBhc, and hence diverges.(kBhc)"Discontinuity at an interior pointDkBhcSometimes the function we are integrating fails to be continuous at some interior point in the interval over which we are integrating. The correct thing to do in that case is to isolate the point of discontinuity and then use limits. Consider this examplekBhc.EetnIcarFKkBhc1kBhcxopxEkBhc2kBhc  - 4kBhcxkBhc3kBhc0kBhc  = etnIcarF*kBhc 1/4kBhc x - 2kBhc  - carF*kBhc 1/4kBhc x + 2kBhcxkBhc3kBhc0kBhc  = etnIfcarF*kBhc 1/4kBhc x - 2kBhcxkBhc3kBhc0kBhc  - etnIfcarF*kBhc 1/4kBhc x + 2kBhcxkBhc3kBhc0FkBhcLEThe integrand in the first integral on the right is discontinuous at kBhc x = 2kBhc4-, which lies in the interior of the interval kBhc [0,3]kBhc.'. We use limits to handle this problem:HtmiL.kBhc A2opxEkBhc-kBhc etnIfcarF*kBhc 1/4kBhc x - 2kBhcxkBhcAkBhc0kBhc  + tmiL.kBhc A2opxEkBhc+kBhc etnIfcarF*kBhc 1/4kBhc x - 2kBhcxkBhc3kBhcAkBhc = tmiL.kBhc A2opxEkBhc-kBhc carF$kBhc1kBhc4kBhc  lnVsbAkBhc x - 2kBhc  |sbuSkBhc0opxEkBhcAkBhc  + tmiL.kBhc A2opxEkBhc+kBhc carF$kBhc1kBhc4kBhc  lnVsbAkBhc x - 2kBhc  |sbuSkBhcAopxEkBhc3kBhcWPBoth of these limits diverge, so we say that the original integral is undefined.