



When
we have talked about definite integrals,



the
limits of integration 


(the
numbers on the top and bottom of the funny S) 


have
always been real numbers, like p
or 12 or whatever. 


But
what if we wanted to use something like infinity (¥)
for the upper limit 





You
might think that any time we did that, 


the
answer that you would get would be infinity. 





Sometimes
you do, but not always. 





For
example, if you walked half way across a room 


and
then walked half of the remaining way 


and
then walked half of the distance that was left. 


If
you keep on walking half the remaining distance each time 


you
can take as many turns as you want 


but
you will never get to the other side of the room. 





This
is like integrating your step turns from one to infinity. 


The
total distance you walk will not be infinity. 


It
would keep getting closer to the distance across the room 


but
never quite get there. 





The
deal is, if the distance you move on each step 


gets
smaller each step "at a fast enough rate" 


the
distance traveled will approach some number. 





If
the distance you move on each step 


does
not get smaller "at a fast enough rate" 


the
distance traveled will approach infinity. 





Hey,
I bet you could write that as some kind of limit problem! 





Well,
now that you mention it ... 





Say
we have some function like ... 











The
graph of this looks like ... 











and
suppose that we want to find the area 


between
the graph line and the X axis 


from
the point where X = 1 to X = ¥. 











To
begin with, the problem would look like ... 











Now
we could just integrate and see what we get ... 











Now
that negative one over two times infinity term is pretty weird. 


Working
with it in official math takes a bit of work, 


but
it's obvious to most people that ... 











So
our area is equal to that (1/2) term. 


The
term simplifies to 1/2, 


but
the official, get all the points on the exam solution, 


looks
like this ... 











You
probably noticed that the graph has a vertical asymptote 


at
the point where X = 0. 





What
of we wanted to see if we could calculate the area 


from
X = 0 to X = 1? 











We
got a number for the area under the graph 


from
X = 1 to X = ¥
even though the graph area goes forever. 


Maybe
this unbounded area from X = 0 to X = 1 


will
give us a finite (some real number value) area. 


Let's
try ... 











Nope,
the area calculates as infinity. 





Interesting,
the 2 areas look about the same 


but
one has an infinite are and the other has a finite area. 





Let's
try another one ... 











Let's
calculate the area of this one from X = 0 to X = 1. 


That's
the part that blew up in the last example. 











Wow!
It worked that time. 


Now
let's try the X = 1 to X = ¥
part. 











Hmmm, 


On
this one, where the exponent was  1/2 


the
part where X went to infinity blew up. 


On
the one where the exponent was 2 


the
part where X went from zero blew up. 





I
wonder if we could generalize and say 


that
for exponents between 1 and 0 


we
get an infinite area when we integrate 


from
X equals some number to X equals infinity. 


And
a finite area when we integrate 


from
X equals zero to X equals some finite number. 





And
if it also always holds that when the exponent is less than 1 


the
reverse is true. 











YES! 


THIS
IS EXACTLY THE WAY IT WORKS! 





I
wonder what happens at the case between these two. 


That
is, at ... 





F(X)
= X^{1}












Let's
try it (both parts) 











Oh
well, the area of both parts is infinite. 





Two
comments. 


First,
the lim on the right used "b" instead of "a" to
avoid confusion. 


Second,
if you can't see how the limit of ln a as a approaches 0 is
infinity, 


calculate
ln 0.5, then ln 0.005, then ln 0.00005. 





copyright 2005 Bruce Kirkpatrick 
