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Tuesday, 26 March 2013

LOCAL BEHAVIOR of FUNCTIONS PART 3


Recall that the definition of an integral tex2html_wrap_inline87 requires the function f(x) to be bounded on the bounded interval [a,b] (where a and b are two real numbers). It is natural then to wonder what happens to this definition if
1
the function f(x) becomes unbounded (we call this case Type I);
2
the interval [a,b] becomes unbounded (that is tex2html_wrap_inline101 or tex2html_wrap_inline103 )(we call this case Type II).
In both cases, we say that the integral tex2html_wrap_inline87 is improper.Case Type I: Consider the function f(x) defined on the interval [a,b] (where a and b are real numbers). We have two cases f(x) becomes unbounded around a or unbounded around b (see the images below)

and

For the sake of illustration, we considered a positive function. The integral tex2html_wrap_inline87 represents the area of the region bounded by the graph of f(x), the x-axis and the lines x=a and x=b. Assume f(x) is unbounded at a. Then the trick behind evaluating the area is to compute the area of the region bounded by the graph of f(x), the x-axis and the lines x=c and x=b. Then we let c get closer and closer to a (check the figure below)

Hence we have
displaymath143
Note that the integral tex2html_wrap_inline145 is well defined. In other words, it is not an improper integral.
If the function is unbounded at b, then we will have
displaymath149

Remark. What happened if the function f(x) is unbounded at more than one point on the interval [a,b]?? Very easy, first you need to study f(x) on [a,b] and find out where the function is unbounded. Let us say that f(x) is unbounded at tex2html_wrap_inline161 and tex2html_wrap_inline163 for example, with tex2html_wrap_inline165 . Then you must choose a number tex2html_wrap_inline167 between tex2html_wrap_inline161 and tex2html_wrap_inline163 (that is tex2html_wrap_inline173 ) and then write
displaymath175
Then you must evaluate every single integral to obtain the integral tex2html_wrap_inline87 . Note that the single integrals do not present a bad behavior other than at the end points (and not for both of them).
Example. Consider the function tex2html_wrap_inline179 defined on [0,1]. It is easy to see that f(x) is unbounded at x = 0 and tex2html_wrap_inline187 . Therefore, in order to study the integral
displaymath189
we will write
displaymath191
and then study every single integral alone.
Case Type II: Consider the function f(x) defined on the interval tex2html_wrap_inline195 or tex2html_wrap_inline197 . In other words, the domain is unbounded not the function (see the figures below).

and

The same as for the Type I, we considered a positive function just for the sake of illustrating what we are doing. The following picture gives a clear idea about what we will do (using the area approach)

So we have
displaymath199
and
displaymath201

Example. Consider the function tex2html_wrap_inline203 defined on tex2html_wrap_inline205 . We have
displaymath207
On the other hand, we have
displaymath209
Hence we have
displaymath211

It may happen that the function f(x) may have Type I and Type II behaviors at the same time. For example, the integral
displaymath215
is one of them. As we did before, we must always split the integral into a sum of integrals with one improper behavior (whether Type I or Type II) at the end points. So for example, we have
displaymath217
The number 1 may be replaced by any number between 0 and tex2html_wrap_inline219 since the function tex2html_wrap_inline221 has a Type I behavior at 0 only and of course a Type II behavior at tex2html_wrap_inline219 .
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