tgt

## Wednesday, 27 March 2013

### Tests of Convergence

It is very easy to see that a simple improper integral may be very hard to decide whether it is convergent or divergent. For example, the improper integral

is hard to study since it is very difficult to find an antiderivative of the function  . The tests of convergence are very useful tools in handling such improper integrals. Unfortunately some improper integrals fails to fall under the scope of these tests but we will not deal with them here.
Recall the p-Test: Regardless of the value of the number p, the improper integral

is always divergent. Moreover, we have
is convergent if and only if p <1
is convergent if and only if p >1
Note that one may generalize this test to include the following improper integrals

The conclusion is similar to the above one. Indeed we have
is convergent if and only if p <1
is convergent if and only if p <1
Comparison Test Let f(x) and g(x) be two functions defined on [a,b] such that

for any  . Then we have
If  is convergent, then  is convergent.
If  is divergent, then  is divergent.
Example. Decide on the convergence or divergence of

The p-Test implies that the improper integral  is convergent. Hence the Comparison test implies that the improper integral

is convergent.
We should appreciate the beauty of these tests. Without them it would have been almost impossible to decide on the convergence of this integral.
Before we get into the limit test, we need to recall the following:
we will say and write  when  if and only if

Limit test Let f(x) and g(x) be two positive functions defined on [a,b]. Assume that both functions exhibit an improper behavior at a and  when  , then we have
is convergent if and only if  is convergent.
This statement is still valid whether a is a finite number or infinite or if the improper behavior is at b.
Example. Establish the convergence or divergence of

Answer. Clearly this integral is improper since the domain is unbounded (Type II). Moreover since the function  is unbounded at 0, then we also have an improper behavior at 0. First we must split the integral and write

First let us take care of the integral  . Since

when  , and (because of the p-test) the integral

is convergent, we deduce from the limit test that

is convergent. Next we investigate the integral  . Since

when  , and (because of the p-test) the integral

is convergent, we deduce from the limit test that

is convergent. Therefore, the improper integral

is convergent.
Remark. One may notice that in the above example, we only used the limit test combined with the p-test. But we should keep in mind that it is not the case in general. The next example shows how the use of other tests is more than useful.
Example. Establish the convergence or divergence of

Answer. Again it is easy to see that we have an improper behavior at both 0 and  . Hence we must split the integral and write

The integral  is easy to take care of since we have

and because  is convergent (by the p-test), the basic comparison test implies that

is convergent. Next we take care of the integral  . Here we use the limit test. Indeed, since  when  , then we have

Because  is divergent (by the p-test), then the limit test implies that the integral

is divergent. Conclusion the improper integral

is divergent.
Remark. One may argue that the above example is in fact not a good one to illustrate the use of different tests. Since if we have showed first that the integral

is divergent via the limit test, then we do not need to take care of the other integral and conclude to the divergence of the given integral. A very good point. Now consider the improper integral

and show that in this case the integral is convergent. Let us point out that the trigonometric functions are very bad when it comes to look at what is happening at  . Hence the limit test is absolutely not appropriate to use...
Example. Establish the convergence or divergence of

Answer. This is clearly not an improper integral of Type II. Let us check if it is of Type I. First notice that  . Hence the function is unbounded at x=1 and x=3 (you must check it by taking the limit.. left as an exercise). Since 3 is between 2 and 4, we deduce that the integral is improper and the only bad point is 3. Hence we must split the integral to get

Let us take care of the integral . It is easy to see that when  , then we have

The p-test implies that the integral

is convergent. Hence by the limit test we conclude that the integral

is convergent. Using the same arguments, we can show that the integral

is also convergent. Therefore the integral

is convergent.
Note that all the tests so far are valid only for positive functions. One may then wonder what happens to improper integrals involving non positive functions