# Where say b i will give a real physexample of this at

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. Where, say, b→∞. I will give a real physicalexample of this at the next lecture. This integral will have a domain whichis an infinite interval. It could be [a,∞) or (–∞,b] or even (–∞,∞). Suchintegrals also occur very frequently in statistics (and therefore they"infiltrate" almost all experimental sciences!).A completely naive interpretation of the areawould declare that since thelength of the base is infinite, the total area must somehow be forced to beinfinite. More subtly, the height in certain cases will decrease fast enough so that the total integral can be thought of as finite.There may also be a sort of defect (?) in the range of f. For simplicity, consider the situation where,although f is "nice" for x>a, as x→a+, f(x) gets larger and largerand larger: in fact, we might need totry to "integrate" or compute the value of ∫abf(x)dx even if f tends to ∞ as x→a.Again, a first look could convince you that such "areas" need to be infinite, also, because the height isinfinite. But, actually, the way f grows as x→a+is what matters. It is possible to imagine that the growth off is so controlled that the total approximating areas don't tend to ∞. And maybe, in such situations, weshould be able to compute the integral.I will first consider the "defect in the domain" case.Toy example #1Look at y=1/x2. The integral ∫1B(1/x2)dx (I'm using B as an abbreviation for BIG) can be computeddirectly:1B(1/x2)dx=–1/x|1B=1–(1/B) (be careful of the signs!)Now if B→∞, certainly 1–(1/B)→1. Then we will declare that the improper integral ∫1(1/x2)dx converges and its value is 1.Toy example #2Look at y=1/x. Consider the analogous integral ∫1B(1/x)dx (again think of B as a BIGnumber). We compute it:1B(1/x)dx=ln(x)|1B=ln(B)–ln(1)=ln(B)–0I wanted also to consider here the behavior as B→∞, but some students seemed to be confused (this isconfusing!). What does happen to ln(B)Diary outline for Math 152, spring 2009: second section38 of 417/26/2009 8:17 AM
when B gets large? If you only have a loose idea of the graph in your head, well the log curve might not look like it is increasing too fast. Well, itactually is not increasing very fast, but it isincreasing. Look: ln(10) is about 2.3, so ln(102)=2ln(10) is about 4.6 (that's 2·2.3), and ln(103)=3ln(10)is about 6.9 (that's 3·2.3), etc. Here etc. means I can get values of ln as large as I want by taking ln's of large enough powers of 10 (hey, ln of105,000is bigger than ... 10,000: so there!). So the values of ∫1B(1/x)dx do notapproach a specific number as B→∞. Therefore we say: theimproper integral ∫1(1/x)dx diverges.The distinction between converges (approaches a specific finite limit) and diverges (does not approach a specific finite limit) is the one that isimportant in applications and that motivates the distinct use of the two words.Geometric constrastI love pictures. I like computation, but I can barely tolerate (!) "algebra". But I introduced the actual definition of {con|di}vergent improper