In this video I will show some examples of how to use the Basic Comparison Test
to determine whether an improper integral is convergent or divergent.
I assume you have watched the previous video in which I introduced the theorem.
A very quick summary of the main two results in that video.
First, for improper integrals of positive functions only,
ONLY for positive functions, there are only two options:
the improper integral must be either divergent to infinity or convergent.
That's it.
And that's why we use a notation of saying that less than infinity
means convergent.
Not infinity means convergent, only for positive functions.
Second, the theorem said the following:
If I have two functions and they are continuous and positive
and I assume that g >= f,
then if the improper integral of the smaller one is divergent,
then the improper integral of the bigger one must be divergent as well,
and if the improper integral of the bigger one is convergent,
then the improper integral of the smaller one must be convergent as well.
For the details of these two results, watch the previous video.
How am I going to use this?
I'm going to take new improper integrals
and I want to determine whether they are convergent or divergent.
I'm going to use this theorem to compare them with other integrals
I already know are convergent or divergent,
and luckily we have a very useful family of functions to compare with.
Specifically, we already know that the integral from 1 to infinity of 1/xᵖ dx
is convergent exactly when p > 1,
and the same integral is divergent exactly when p <= 1.
I will include a link in the description to the video where I prove this.
So let's take more improper integrals and compare them to these ones
and try to quickly conclude whether they are convergent or divergent.
In this first example I want to determine
whether the integral from 1 to infinity of sin²x/x² dx is convergent or divergent.
We could try to use the definition,
but that would require computing an antiderivative for this function,
and that is a bit difficult.
If we only need to know whether this is convergent (a number) or divergent,
then we can use the basic comparison and it will be faster.
The difficult part is sin², but the sine is always bounded, we know that.
We can say that for any x >= 1, and actually for any real number,
sin²x is between 1 and 0,
and therefore sin²x/x² is always between 1/x² and 0.
And now I'm comparing this function with another one,
and for this one I know the outcome of the improper integral.
We know that the integral from 1 to infinity of 1/x² dx is convergent,
because 2 is greater than 1.
So this is now ready to use the Basic Comparison Test.
We know that the integral of the big function is convergent
(less than infinity),
so the integral of the smaller function must also be convergent (also a number).
So the integral from 1 to infinity of sin²x/x² dx is convegent.
That's it! That was easy.
Let's do another one.
This second example looks quite similar,
but there is a reason I want to explore it.
It's the integral from 1 to infinity of sin²x/x dx.
I can begin like I did in the previous example
and simply say that for any x >= 1,
I can bound this function between 1/x and 0.
And just like before, I know the outcome for the improper integral for 1/x.
We know that the integral from 1 to x of dx/x is infinity.
So I can now use the Basic Comparison Test,
and in this case the improper integral of the bigger function is infinity,
is divergent, so... this doesn't help,
because if the bigger one is divergent,
then the smaller one could be divergent or convergent.
So as it is, the Basic Comparison Test doesn't help.
Okay. Ultimately,
this improper integral is going to be either convergent or divergent.
(Spoiler: it is divergent.)
But given what I know so far, I don't know what the answer is yet.
So I have to say "try something else, I don't know".
Well, let's look at one final example where the Basic Comparison Test does help.
The third example is more interesting.
I want to determine whether the integral from 1 to infinity of ln(x)/x² dx
is convergent or divergent.
In this case, the complicated part is this logarithm.
And logarithm is not bounded, but I know something about it.
If I want to compare this with powers, I know from the Big Theorem
that logarithm grows slower than any power.
Specifically, I can say that logarithm is much smaller than x.
Now be careful: this doesn't mean that logarithm is smaller than x
for all values of x.
The actual definition of this is that lim_(x → ∞)[ln(x)/x] is 0.
So eventually ln(x) is smaller than x, but maybe not for all values of x.
Or more precisely, what this means is that for large values of x,
since this is approaching 0,
for example, I can say that for large values of x,
ln(x)/x < 1, and therefore ln(x) < x.
But again, only for large values of x.
However, I want to argue that this is enough.
I'm trying to determine whether this improper integral
is convergent or divergent.
That is only going to depend on what happens for large values of x.
So for example, let's say that this inequality is only true whenever x > 10.
Well from 1 to 10, this is perfectly well-defined,
and whether it's convergent or divergent
only depends on what happens from 10 to infinity.
So as a general rule, whenever I'm only studying convergence or divergence,
if the conditions I have are only true for large values of x,
for x's starting at a certain point,
that's good enough, and the theorems are also true.
So I'm going to use this comparison and try to use the Basic Comparison Test.
Since ln(x) < x, I can now say that ln(x)/x² < x/x²,
and of course it's also non-negative, and this is equal to 1/x.
Great — I'm comparing this with a function whose improper integral I know,
because we know that the integral from 1 to infinity of 1/x dx is infinity.
And now we use the Basic Comparison Test.
And it looks like if we are stuck again; but not quite.
Okay, if I do it directly, we're stuck,
because in this case the integral of the bigger function is infinity, divergent,
so the integral of the smaller function could be divergent or convergent.
This doesn't give me enough information.
But I'm going to change things slightly.
The big theorem doesn't only tell me that logarithm is much smaller than x,
it tells me that every logarithm is much smaller than every power.
So I'm going to use a power that helps me more.
Instead of x, I'm going to put here x¹/².
And so this limit is going to be 0 when I put x¹/²,
and therefore I can also put x¹/² here now.
What large x means is different; maybe this is only true whenever x > 1000.
And so here I can put x¹/², and here I end up with x³/².
And that's better because if I put x³/² there,
now the integral is convergent, not divergent.
And now the integral of the bigger function is convergent,
so by the Basic Comparison Test,
I can conclude that the integral of the smaller function
is also convergent.