In this video, I will explain how to work with doubly improper integrals.
These are integrals that are improper in more than one way,
most of the time at both ends of the interval of integration.
I will begin with an example.
I want to calculate the integral from minus infinity to infinity
of the function x with respect to x.
Here is what the graph of the function looks like.
This integral is improper for two reasons:
the domain goes all the way to minus infinity, and all the way to infinity.
Which means we first have to figure out how to define this.
It may make sense or maybe it's not even well-defined.
Geometrically, what I'm doing is looking at this area
between the graph of the function and the x-axis,
but the area above the x-axis counts as positive
and the area below the x-axis counts as negative.
So is it tempting to say that by symmetry the answer will be 0?
This function is odd, so this region is exactly equal to this region.
So they'll cancel and we'll come up with 0.
I want to argue that that is an incorrect explanation,
and that is because the area above the x-axis is infinity,
the area below the x-axis is going to give us minus infinity,
and I cannot cancel plus with minus infinity.
But let me do this in a bit more detail.
We know that we can't integrate between minus infinity and infinity directly,
we can only integrate a function that's continuous,
for example, on an interval that includes both endpoints.
So perhaps we can set this up by integrating between -R and R,
where R is a positive real number, the same function.
So integrating between here and here
and then taking the limit as R goes to infinity.
This seems like a reasonable interpretation.
I integrate between here and here and then I make R grow bigger and bigger.
So what happens if I calculate this limit?
An antiderivative of x is x²/2, and then I evaluate at the endpoints,
so I get R²/2 - (-R²)/2, and this quantity here is identically 0.
So the limit is 0.
Okay, so, this calculation argues for 0.
It is true that the value of this limit is 0,
but the question is whether it is okay to choose the value of this limit
and assign it as the value of the integral.
Here is an argument why I shouldn't do this.
In this first calculation, I've chosen to center my domain at 0,
but my domain is all the real numbers, and there's nothing special about 0.
I could have chosen to center it at 1 instead.
So instead of integrating between -R and R,
why not integrate between 1 - R and 1 + R?
It should be equally good.
Then integrate the function, and then take the limit.
That seems like as much a valid interpretation as the previous one.
So I won't do the details, it's an easy computation.
I invite you to do it,
and you'll see that in this case the answer comes out to be infinity.
So which one is it?
The answer is none of them.
Any good interpretation, any good definition for this quantity,
should be one that agrees in this calculation and in this one.
And the fact that both of them give me different answers is an indication
that I should not define this to be a number at all,
that this is not well defined, that this is divergent.
And once again, that is because I cannot cancel plus infinity with minus infinity.
If you ever allow infinity and minus infinity to cancel with each other,
then bad things happen.
So perhaps this argument doesn't convince you.
Here's another way to look at the same problem.
Before I try to compute this integral, I'm going to perform a substitution.
I'm going to change x into u + 1 with a new variable u,
and so dx will be du.
So what happens when I do that? I get ∫(u + 1)du.
And now as X approaches minus infinity, so does u,
and as x approaches infinity, so does u.
And now I can break this into the sum of two integrals using linearity.
And look what happens now:
this piece here is exactly the same as the piece I started with,
so this is again I.
Whereas this integral here, this one is unambiguous.
This one is definitely infinity.
It's the integral of the constant function 1
from minus infinity to infinity. Just picture the graph.
So I end up with the conclusion that I must be equal to I plus infinity.
Now this again is an argument for why it's better to say that I is not a number.
If I try to make this integral convergent and assign it the value of a number,
in particular 0, I get a contradiction here.
If we ever try to cancel infinity with minus infinity,
which is what I was doing at the beginning,
we have to give up all the things we take for granted and that are very useful,
like linearity of the integral or the substitution rule.
It is better just to say that this integral is divergent
and it is not a number.
Okay, but how do we come up to that conclusion in a rigorous systematic way?
Here is the original integral again.
As we said, it is improper at minus infinity, and it is improper at infinity.
What I'm going to do is break it into two pieces,
each one of them improper at only one end.
For example, I can break this at 0,
and I have the sum of these two integrals.
And now each one of these two pieces is an improper integral.
It may still be a number or not.
But each one of them, I know exactly how to study.
There are no cancellations possible.
So I'm going to write each one of them separately as a limit.
The first one can be written as this limit,
whereas the second one can be written as this limit.
and now for the integral (the original one) to be convergent,
I need both limits to exist separately.
If they exist separately (don't mix them yet),
then everything will be alright.
Then I can try to do cancellations and use properties,
and I won't draw any contradictions.
Unfortunately, in this case that doesn't happen,
because the first limit is going to be minus infinity,
and the second limit is going to be plus infinity.
Which means in this case, the original integral is simply divergent.
And we stop there.
Any attempt to define this integral to be any particular number
will end up in nice properties being broken and contradictions.
Here is the general strategy.
Let's say I want to calculate an integral, I'm going to call it A,
which is multiply improper.
This means that it is improper, it has problems, in more than one way,
perhaps both ends of integration like the previous one,
or perhaps there's a problem at an intermediate point
in the middle of the domain.
What we are always going to do is first break the domain into pieces,
so that the integral over each piece is improper at one point and one point only,
because then each one of those pieces I know how to study it independently,
how to set it up as a limit,
and determine whether it is convergent or divergent.
Those are all of the same kind.
Then if all the pieces are convergent separately,
then the original improper integral is also convergent,
and then you can use all the properties
you expect to be able to use for integrals,
and you never get contradictions, and everything is fine.
But if at least one of the pieces is divergent,
then we say the original integral is also divergent,
and then it is NOT a number.
Any attempt to define it as a number will probably lead to trouble.
You may wonder how we choose how to break the domain,
and the answer is that it doesn't matter.
This is a little exercise:
the outcome won't depend on how you break the original domain
as long as you break it into pieces such that
each piece is only improper at one endpoint.
For example, let's say I want to integrate from 0 to infinity
this function, 1/√x(1+x²)dx.
This is improper at infinity, because they always are,
but also at 0, because the function is undefined at 0
and I have a vertical aymptote.
The way to deal with that is to separate it into two parts:
the integral from 0 to 1 and the integral from 1 to infinity.
Now each one of them is improper in a simple way,
they're like previous examples, and I can study them separately.
If they're both convergent, then the full integral is convergent.
If at least one of them is divergent, then the full integral is divergent.
Or here is an example that looks different:
the integral from -1 to 1 of 1/x² dx.
This one is improper, but it's a bit hidden.
It's improper at 0
because the function is not defined at 0 and has a vertical asymptote at 0.
The way to deal with it is to split this integral into two pieces,
the integral from -1 to 0 and integral from 0 to 1,
because now each one of them is improper in a familiar way,
I know how to deal with it,
I know how to set it up as a limit and how to study it.
If both of them are convergent separately, the full thing is convergent.
But if at least one of them is divergent, then the full thing is also divergent.