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Comparison Test for Improper Integrals
Now that we’ve seen how to actually compute improper integrals we need to address
one more topic about them. Often we aren’t concerned with the actual value of these
integrals. Instead we might only be interested in whether the integral is convergent or
divergent. Also, there will be some integrals that we simply won’t be able to integrate
and yet we would still like to know if they converge or diverge.
To deal with this we’ve got a test for convergence or divergence that we can use to
help us answer the question of convergence for an improper integral.
We will give this test only for a subcase of the infinite interval integral, however
versions of the test exist for the other subcases of the infinite interval integrals as
well as integrals with discontinuous integrands.
Comparison Test
If
on the interval
then,
1.
If
converges then so does
.
2.
If
diverges then so does
.
Note that if you think in terms of area the Comparison Test makes a lot of sense.
If
is larger than
then the area under
must also be larger than the area under
.
So, if the area under the larger function is finite (
i.e.
converges) then the area under the smaller function must also be finite
(
i.e.
converges). Likewise, if the area under the
smaller function is infinite (
i.e.
diverges) then the
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View Full Documentarea under the larger function must also be infinite (
i.e.
diverges).
Be careful not to misuse this test. If the smaller function converges there is no reason
to believe that the larger will also converge (after all infinity is larger than a finite
number…) and if the larger function diverges there is no reason to believe that the
smaller function will also diverge.
Let’s work a couple of example using the comparison test. Note that all we’ll be able
to do is determine the convergence of the integral. We won’t be able to determine the
value of the integrals and so won’t even bother with that.
Example 1
Determine if the following integral is convergent or divergent.
Solution
Let’s take a second and think about how the Comparison Test works. If this integral is convergent
then we’ll need to find a larger function that also converges on the same interval. Likewise, if this
integral is divergent then we’ll need to find a smaller function that also diverges.
So, it seems like it would be nice to have some idea as to whether the integral converges or
diverges ahead of time so we will know whether we will need to look for a larger (and convergent)
function or a smaller (and divergent) function.
To get the guess for this function let’s notice that the numerator is nice and bounded and simply
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 Fall '08
 prellis
 Improper Integrals, Integrals

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