In this video I will study the most important family of improper integrals:
the integral from 1 to infinity of 1/xᵖ with respect to x.
This is not a single integral, it's a family of them.
For each real number p I get a different one.
Some are convergent, and some are divergent.
I want to know which ones.
This family is very important not because it is complicated
(the calculation will be relatively simple),
but because it comes up again and again.
Later, we will learn how to determine whether other integrals
are convergent or divergent by comparing them to this one.
At this moment, I invite you to pause the video
and try to answer the question before continuing watching.
This is an improper integral because the domain goes all the way to infinity.
So like in the previous examples, we set it up as a limit.
First I integrate between 1 and a number b, the same function,
and then I take the limit as b approaches infinity.
So first integrate, and then take the limit.
In order to integrate, I need to find an antiderivative for this function.
And I'm going to break that into two cases:
when p = 1, I will have 1/x, and we know an antiderivative is the logarithm.
But in all other cases,
I just have a power whose antiderivative is also a power.
So I will do this second case first, when p is not 1.
The antiderivative of x⁻ᵖ is going to be x⁻ᵖ⁺¹/(-p+1).
That works because I omitted the case p = 1.
And then I need to evaluate it when x = b and when x = 1.
Let's do that.
Now to calculate the limit, I'm going to have to consider a couple of cases.
I'm taking the limit as b approaches infinity,
so for the purpose of the limit,
b is the variable and P is just a constant.
This is a fixed number, and so is this.
This is the only part that depends on b.
When b approaches infinity,
it matters whether this exponent is positive or negative,
so I'm going to break it into two cases:
when -p+1 is positive and when -p+1 is negative.
if -p+1 is positive, I have b approaching infinity raised to a positive exponent;
that also approaches infinity;
divided by a positive number is still infinity;
and the constant doesn't matter.
So in this case,
I'm going to get that the limit does not exist because it is infinity.
But in the second case, when the exponent is negative,
b approaching infinity raised to a negative exponent,
that's going to go to 0.
So this term is 0 and doesn't matter, and I'm left only with this term.
And I enter the minus in the fraction and I get that this is the number 1/(p-1).
But more importantly, in the first case the improper integral is divergent,
and in the second case the improper integral is convergent.
And perhaps it's simpler to summarize the first case by saying p < 1,
that's an equivalent inequality,
and the second case by saying that p > 1.
So we have only one case left. What happens when p equals exactly 1?
Now I go back here and when p = 1, I have the function 1/x.
The antiderivative is lnx, which I have to evaluate at b,
minus evaluate at 1.
And then do all the steps; as b approaches infinity,
I'm going to have lnb that also approaches infinity.
So in this case, I also get infinity, and the integral is also divergent.
So in these two cases it's divergent, and in this case only it is convergent.
Here is a summary of the calculation we just did.
The improper integral from 1 to infinity of 1/xᵖ
is convergent if and only if p > 1.
If you ever need the specific value, you can repeat the calculation we just did.
And it is divergent, not just divergent but specifically to infinity,
if and only if p < 1 or equal to 1, and put those two cases together.
This summary is important and it's actually worth memorizing
because it appears a lot.
Once you learn to use comparison test,
both for improper integrals and for series,
you will be using this a lot it.
It is the most ubiquitous family of improper integrals.