I’ve got one last thing to wrap up in my coverage of integration. Well, I never talked about Brown v. Board of Ed., but most curricula these days don’t cover the technique “by court order” anymore.
Did I get the title of the post wrong? No. It turns out that I covered half of this topic two years ago as I prepared to push into infinite series. Back then, I dealt with what happened when we wanted to integrate over an infinite interval. Then I defined an infinite series to be what happened when we used a particular integrator in our Riemann-Stieltjes integral. This could also be useful in setting up multiple integrals over -dimensional intervals that extend infinitely far in some direction.
But there are other places that so-called “improper integrals” come up, like the example I worked through the other day. In this case, the fundamental theorem of calculus runs into trouble at the endpoints of our interval. Indeed, we ask for an antiderivative on a closed interval, but to get the endpoints we need to have the antiderivative defined on some slightly larger open interval, and it just isn’t.
So here’s what we do: let be defined on and integrable with respect to some integrator over the interval for all . Then we can define each integral
Just as before, we define the improper integral to be the one-sided limit
If this limit exists, we say that the improper integral converges. Otherwise we say it diverges. Similarly we can define the improper integral
taking the limit as approaches from the left.
Just as for the first kind of improper integral, we have analogues of the direct comparison and limit comparison tests, and of the idea of absolute convergence. Each of these is exactly the same as before, replacing limits as approaches with limits as approaches some finite point from the left or right.
We can also combine improper integrals like we did before to integrate over the whole real line. For example, we could define
As before, we must take these limits separately. Indeed, now we can even say more about what can go wrong, because these are examples of multivariable limits. We cannot take the limit as and together approach their limiting points along any particular path, but must consider them approaching along all paths.
In practice, it’s pretty clear what needs to be done, and when. If we have trouble evaluating an antiderivative at one endpoint or another, we replace the evaluation with an appropriate limit.
A neat little variation on differentiating an integral from last time combines it with the fundamental theorem of calculus. It’s especially interesting in the context of evaluating iterated integrals for irregular regions where the limits of integration may depend on other variables.
Let is a continuous function on the rectangle , and that is also continuous on . Also, let and be two differentiable functions on with images in . Define the function
Then the derivative exists and has the value
In fact, if we forget letting depend on at all, this is the source of some of my favorite questions on first-semester calculus finals.
Anyway, we define another function for the moment
for and in and in . Then .
The fundamental theorem of calculus tells us the first two partial derivatives of immediately, and for the third we can differentiate under the integral sign:
Then we can use the chain rule:
Another question about “partial integrals” is when we can interchange integral and differential operations. Just like we found for continuity of partial integrals, this involves interchanging two limit processes.
Specifically, let’s again assume that is a function on the rectangle , and that is of bounded variation on and that the integral
exists for every . Further, assume that the partial derivative exists and is continuous throughout . Then the derivative of exists and is given by
Similar results hold where and are vector values, and where derivatives in terms of are replaced outside the integral by partial derivatives in terms of its components.
So, if and we can calculate the difference quotient:
where is some number between and that exists by the differential mean value theorem. Now to find the derivative, we take the limit of the difference quotient
as we take to approach , the number gets squeezed towards as well. Since we assumed to be continuous, the limit in the integrand will equal , as asserted.
There are some remaining topics to clean up in the theory of the Riemann-Stieltjes integral. First up is a question that seems natural from the perspective of iterated integrals: what can be said about the continuity of the inner integrals?
More explicitly, let be a continuous function on the rectangle , and let be of bounded variation on . Define the function on by
Then is continuous on . Similar statements can be made about other “partial integrals”, where and are each vector variables and we use in place of the Stieltjes integrator .
Specifically, this is a statement about interchanging limit operations. The Riemann-Stieltjes integral involves taking the limit over the collection of tagged partitions of , while to ask if is continuous asks whether
As we knew back when we originally discussed integrators of bounded variation, we can write our integrator as the difference of two increasing functions. It’s no loss of generality, then, to assume that is increasing. We also remember that the Heine-Cantor theorem tells us that since is compact, is actually uniformly continuous.
Uniform continuity tells us that for every there is a (depending only on so that for every pair of points and , with we have .
So now let’s take two points and with and consider the difference
where we’ve used the integral mean value theorem. Clearly by choosing the right we can find a to make the right hand side as small as we want, proving the continuity of .
We first defined the Jacobian determinant as measuring the factor by which a transformation scales infinitesimal pieces of -dimensional volume. Now, with the change of variables formula and the mean value theorem in hand, we can pull out a macroscopic result.
for some , where .
Simply take in the change of variables formula
The mean value theorem now tells us that
for some between the maximum and minimum of on . But then since is connected, we know that there is some so that , as we asserted.
Today we finish up the proof of the change of variables formula for multiple integrals:
So far, we’ve shown that we can chop up into a collection of nonoverlapping regions and into their preimages . Further, within each we can factor , where fixes each component of except the last, and fixes that one. If we can show the formula holds for each such region, then it will hold for arbitrary (compact, Jordan measurable) .
From here we’ll just drop the subscripts to simplify our notation, since we’re just concerned with one of these regions at a time. We’ve got and its preimage . We’ll also define , so that . For each real we define
Then , since preserves the last component of the vector. We also define
The lowest and highest points in along the th coordinate direction. Now we can again define and set up the iterated integral
We can apply the inductive hypothesis to the inner integral using , which only involves the first coordinates anyway. If we also rename to , this gives
Which effectively integrates as runs over . But now we see that lies within the projection of , as we defined when we first discussed iterated integrals. We want to swap the order of integration here, so we have to rewrite the limits. To this end, we write , , and define
which runs over the part of above some fixed point in . Then we can reverse the order of integration to write
Now we can perform the one-dimensional change of variables on the inner integral and swap out the variables through to write
But now we recognize the product of the two Jacobian determinants as the Jacobian of the composition:
and so we can recombine the iterated integral into the -dimensional integral
Finally, since we can replace the composition. We can also replace by since there’s no chance anymore for any evaluation of to go outside . We’re left with
Essentially, what we’ve shown is that we can always arrange to first change variables (which we handle by induction) and then by the last variable. The overall scaling factor is the -dimensional scaling factor from the first transformation times the one-dimensional scaling factor from the second, and this works because of how the Jacobian works with compositions.
This establishes the change of variables formula for regions within which we can write as the composition of two functions, one of which fixes all but the last coordinate, and the other of which fixes that one. Since we established that we can always cut up a compact, Jordan measurable set into a finite number of such pieces, this establishes the change of variables formula in general.
Okay, let’s get to actually proving the change of variables formula for multiple integrals. To be explicit:
Let be a continuously differentiable function defined on an open region . Further, assume that is injective and that the Jacobian determinant is everywhere nonzero on . The inverse function theorem tells us that we can define a continuously differentiable inverse on all of the image .
Further, let be a Jordan measurable subset of , and let be defined and continuous on . Then we have the change of variables formula
We will proceed by induction on the dimension . For , this is exactly the one-dimensional change of variables formula, which we already know to be true. And so we’ll assume that it holds in all -dimensional cases, and prove that it then holds for -dimensional cases as well.
Since the Jacobian determinant is nonzero, we cannot have for all ; at least one of the components must have a nonzero partial derivative with respect to at any given point . Let’s say, to be definite, that . We will now (locally) factor into the composite of two functions and , which will each have their own useful properties. First, we will define
This is clearly continuously differentiable, and it’s even injective on some neighborhood of , by our assumption that . Further, the Jacobian determinant is exactly the partial derivative , and so the inverse function theorem tells us that in some neighborhood of we have a local inverse , with in some neighborhood of . We can now define
for , and define
Then for each in a small enough neighborhood of we have . The first function leaves all components of fixed except for , while the second function leaves fixed. Of course, if we used a different partial derivative , we could do the same thing, replacing instead with in , and so on.
Now if is a Jordan measurable compact subset of , then its inverse image will also be compact since is continuous. For every point in , we can find some neighborhood — which we can take to be an -dimensional interval — and a factorization of into two functions as above. As we move around, these neighborhoods form an open cover of . And since is compact, we can take an open subcover.
That is, we can cover by a finite collection of open intervals , and within each one we can write , where the function leaves all the components of fixed except for the last, while leaves that last one fixed. By subdividing these intervals, we can assume that they’re nonoverlapping. Of course, if we subdivided into open sets we’d miss the shared boundary between two subintervals. So we’ll include that boundary in each subinterval and still have nonoverlapping intervals .
Then we can define . Since is injective, these regions will also be nonoverlapping. And they’ll cover , just as covered . So we can define and write
Our proof will thus be complete if we can show that the change of variables formula holds for these regions, and we will pick up with this final step next time.
In the one-variable Riemann and Riemann-Stieltjes integrals, we had a “change of variables” formula. This let us replace our variable of integration by a function of a new variable, and we got the same answer. This was useful because the form of the resulting integrand might have been simpler to work with in terms of using the fundamental theorem of calculus.
In multiple variables we’ll have a similar formula, but it will have an additional use. Not only might it be used to simplify the integrand, but it might simplify the region of integration itself! Of course, there might also be a trade-off between these two considerations, as many students in multivariable calculus classes might remember. A substitution which simplifies the region of integration might make antidifferentiating the integrand (in any of the resulting variables) impractical, while another substitution which simplifies the integrand might make the region a nightmare to work with.
The formula in one variable looked something like this:
where (along with the induced transformation ) is a continuously differentiable function on with and . Notice that could extend out beyond , but if it went above it would have to come back down again, covering the same region twice with opposite signs. This is related to the signed volumes we talked about, where (in one dimension) an interval can be traversed (integrated over) from left to right or from right to left.
The picture gets a little simpler when we assume that is strictly monotonic. That is, either is strictly increasing, , and ; or is strictly decreasing, , and (traversing in the opposite direction). In the first case, we can write our change of variables relation as
while in the second case, reversing the direction of integration entails adding a negative sign.
but in this case, the derivative is strictly negative. We can combine it with this new sign, and rather elegantly write both cases as
In all of these cases, we know that the inverse function exists because of the inverse function theorem. Here the Jacobian determinant is simply the derivative , which we’re assuming is everywhere nonzero.
In essence, the idea was that measures the factor by which stretches intervals near . That is, the tiny bit of one-dimensional volume gets stretched into the tiny bit of (unsigned) one-dimensional volume . And this works because at a very small scale, little changes in transform almost linearly.
So in higher-dimensional spaces, we will assume that transforms small enough changes in almost linearly — is differentiable — and that the Jacobian determinant is everywhere nonzero, so we can invert the transformation. This gives us hope that we can write something like
Since is invertible, integrating as ranges over is the same as letting range over , so the region of integration lines up, as does the integrand. All that’s left is to figure out how we should replace .
Now this is not the differential of a variable . When it shows up in a multiple integral, it’s a tiny little bit of -dimensional volume. And we measure the scaling of -dimensional volumes with the Jacobian determinant! The same sign considerations as before tell us that either the Jacobian determinant is always positive or always negative, and in either case we can write
or, using our more Leibniz-style notation
We will start proving this formula next time.
My description of how to evaluate a multiple integral over some region other than an -dimensional interval by using iterated integrals might not have been the clearest, so I’m hoping an example will help illustrate what I mean. Let’s calculate the Jordan content of a three-dimensional sphere of radius , centered at the origin. This will also help cement the fact that Jordan content is closely related to what we mean by “volume”.
So, we have our sphere
We can put it inside the interval , so we know that the Jordan content is
Now we want to peel off an integral from the inside. Let’s first integrate over the variable . Projecting the sphere onto the plane we’re left with the circle
and we need to write as the region between the graphs of two functions on this projection. And indeed, we can write
Thus we can write
Next we’ll integrate with respect to . Projecting onto the line we find . And then we find
which lets us peel off another integral from the inside
Finally, we can start evaluating these integrals. The innermost integral goes immediately, using the fundamental theorem of calculus. An antiderivative of with respect to is , and so we find
So we put this into our integral, already in progress:
Here’s where things start to get difficult. I’ll just tell you (and you can verify) that an antiderivative of with respect to is
So the fundamental theorem of calculus tells us
Technically, this is an improper integral (which we haven’t really discussed), so we need to take some limits. The denominators are limits as some dummy variable approaches zero from above. Thus we continue evaluating our integral
This one is easy: an antiderivative of is , and so
And this is exactly the well-known formula for the volume of a sphere of radius . There are easier ways to get at this formula, of course, but this one manages to illustrate the technique of iterated integrals with variable limits of integration on the inner integrals.