Improper Integrals

Definition.
An improper integral,definedby
? ? M
f(x)dx = lim f(x)dx
aM?? a
issaidtoconvergeifthelimitexists(divergesifthelimitdoesnotexist).
?
e?kxdx =1/k (k> 0)
Example 1.
0
? M
M
e?kxdx =(?1/k)e?kx
= (1/k)(1 ? e?kM)
0 0
Takingthelimitas M ??,wefind e?kM ? 0 and
?
e?kxdx =1/k
0
Werewritethiscalculationmoreinformallyasfollows,
0
?
e?kxdx =(?1/k)e?kx
?
0
= (1/k)(1 ? e?k?)=1/k (since k> 0)
?
e?kxdx =1/k has an easier formula than the Note that the integral over the infinite interval
? M 0
correspondingfiniteintegral e?kxdx = (1/k)(1?e?kM). As a practical matter, for large M,the
0
term e?kM isnegligible,soeventhesimplerformula 1/k servesasagoodapproximationtothefinite
integral. Infinite integrals are often easier than finite ones, just as infinitesimals and derivatives are
easierthandifferencequotients.
Application: Replace x by t =timeinsecondsinExample1.
R = rateofdecay =numberofatomsthatdecaypersecondattime 0.
Atlatertimes t> 0 thedecayrateis Re?kt (smallerbyanexponentialfactor e?kt)
Eventually (over time 0 ? t< ?) every atom decays. So the total number of atoms N is calculatedusingtheformulawefoundinExample1,
?
Re?ktdt = R/k
N =
0
Thehalflife H ofaradioactiveelementisthetime H atwhichthedecayrateishalfwhatitwasat the start. Thus
e?kH =1/2=??kH = ln(1/2) =? k = (ln 2)/H
1
?
?????
Lecture 35 18.01 Fall 2006
Hence
R = Nk = N(ln 2)/H
Let us illustrate with Polonium 210, which has been in the news lately. The half life is 138 days
or
H = (138days)(24hr/day)(602sec/hr) = (138)(24)(60)2seconds
Using this value of H, we find that one gram of Polonium 210 emits (1 gram)(6 × 1023/210
atoms/gram)(ln2)/H = 1.661014 decays/sec ? 4500 curies
At 5.3 MeV per decay, Polonium gives off 140 watts of radioactive energy per gram (white hot).
Polonium emits alpha rays, which are blocked by skin but when ingested are 20 times more dangerous
than gamma and X-rays. The lethal dose, when ingested, is about 10?7 grams.
?
Example 2. dx/(1 + x2) = ?/2.
0
We calculate,
? M
M
dx = tan?1 x = tan?1M ? ?/2 1 + x2
0
0
as M ? ?. (If ? = tan?1M then ? ? ?/2 as M ? ?. See Figures 1 and 2.)
x
y = tan(x)
M
?
x = ?/2
x = -?/2
.
Figure 1: Graph of the tangent function, M = tan ?.
2
?