In this video,
I will present the Limit-Comparison Test for improper integrals
and show how to use it with various examples.
In many cases,
it is too difficult to compute the value of an improper integral
directly from the definition.
And in those cases, our first goal is to see whether we can determine
whether the improper integral is convergent or divergent.
That means whether it is a well-defined number or not.
And the way we do that
is by comparing it to another integral that hopefully is easier,
one that we know whether it's convergent or divergent.
There are two main tools to do this, two theorems.
There is the Basic Comparison Test that we already know,
and then there is the Limit-Comparison Test,
which is another way of doing this,
and that's the theorem I will present in this video.
Here is the statement.
Let a be a real number, and I begin with two functions f and g.
I assume they're both positive and continuous from a to infinity.
I'm going to compare their integrals.
Then the hypothesis, I need to assume that the limit of the quotient,
as the variable goes to infinity, exists.
That means that it is a number.
Now since they are both positive, if the limit exists,
it will be greater than or equal to 0, and I'm going to exclude the case 0.
So I assume this limit exists (is a number, not infinity)
and is greater than 0.
Then in that case, the theorem says that
the improper integral of f, from a to infinity,
and of g, from a to infinity,
then they are either both convergent or both divergent
So if I know that one of them is convergent,
then the other one must be convergent as well.
In the next video, I will write a formal proof for this theorem.
But for now, let's see some examples of how to use it.
For my first example,
I am looking at the integral from 1 to infinity of (x² + 3x)/√(x⁵ + 1),
and I just want to know whether this is convergent, a number,
or divergent, infinity.
And this function is positive, so those are the only two options.
I don't need to know the exact value,
just whether it is convergent or divergent.
For simplicity I'm going to give a name to this function.
Now, this integral is improper because the domain goes all the way to infinity.
So for the purpose of convergence or divergence,
the only thing that matters is what happens when x is large.
Well when x is large, between x² and 3x, the dominant term is x²,
and in the bottom, x⁵ is the dominant term.
So I'm going to say that when x is large,
(and this is very vague for now, I'm just trying to guess what to compare it with)
so my guess is that when x is very large,
the only term that matters is x² in the top and x⁵/² in the bottom.
Or in other words, 1/x¹/².
And this is going to be my candidate; this is what I'm going to call it g(x).
So I want to try to compare the integral of f with the integral of g.
This one is definitely simpler.
Now to do this properly, I need to use the Limit-Comparison Test.
So what I need to do now is calculate the limit as x goes to infinity
of the quotient of these two functions.
And if that is a positive number,
then I'm set and I can use the Limit-Comparison Test.
Let's do that.
Okay, and now this is a standard limit at infinity.
I won't do all the details. I invite you to do it at your own pace.
Probably the best thing to do is reduce to a single fraction
and then factor out the biggest exponent,
but if you do that you will conclude that the limit is 1.
Perfect. So 1 is within what we need to use the Limit-Comparison Test.
So that means that the integral of g and the integral of f
will behave the same way,
will both be convergent or will both be divergent.
But luckily, we know what the integral of g is.
We know that the integral from 1 to infinity of g is simply...
and this is infinity because 1/2 is less than 1.
So now I put together the fact that the limit is 1
and the fact that this one is divergent,
and these two things tell me that by the Limit-Comparison Test,
the integral of f is also divergent, is also infinity.
And that is the answer to the original question.
In the second example,
I want to look at the integral from 1 to infinity of sin(1/x²) dx,
and I only want to know whether it is convergent.
Notice that when x > 1, 1/x² is between 0 and 1,
and therefore sine of that is positive.
So this is a positive function.
It can only be convergent, a number, or divergent to infinity.
Okay. This integral is improper at infinity;
the only thing that matters is what happens when x is large,
for the purpose of convergence.
And what I can notice is that as x approaches infinity, 1/x² approaches 0.
All right, so I have sine of a function that is approaching 0.
And this makes me think of an important limit that we know well.
If I calculate the limit of sin(t)/t as t approaches 0,
we know that this limit is 1. It's a very standard one.
So this suggests that since I have sine of a function that is approaching 0,
I could compare sine of this function with the function itself.
And that's what I'm going to do.
So I'm going to give a name to this function,
I'm going to call f(x) = sin(1/x²)
and then I'm gonna call g(x) = 1/x².
And now I want to compare the integral of g with the integral of f.
This one is certainly simple.
To do that properly, we know from the Limit-Comparison Test
I need to calculate the limit of the quotient as x approaches infinity.
Let's do that.
And now perhaps you are ready to just say this is 1 right away
because this is sine of a function that goes to 0 over the same function,
and otherwise this is 0/0 and you can use L'Hôpital.
And after using L'Hôpital, you will conclude that this limit is exactly 1.
Great. 1 is good for the purpose of the Limit-Comparison Test.
That means that the integral of f and integral of g behave the same way.
They're both convergent, or they're both divergent.
But we know what the integral of g is.
We know that this integral is convergent because 2 is greater than 1.
So finally, I put together the fact that the limit is 1
and the fact that this integral is convergent,
and then by the Limit-Comparison Test,
I can conclude that the other integral is also convergent.
And that is the answer to the original question.