A question about infinity and calculus

[Dark Lord of the Bith]

Today in my Calc II class, the professor was talking about log and exponential functions, and near the end, she put some integrals up on the board for us to do, namely, integral 1/x dx from 1 to e, from -e to -1, and -1 to e. The first two are simple, and the third one is obviously improper, though we haven't actually learned about those in class yet. If you simply evaluate each of them, they all come out to be 1, and then she says the third one is meaningless because of the asymptope. All of this, of course, is rather elementary.

When I did the third problem, though, applying the limit as a approaches 0 so we don't quite deal with infinity, the final step of the solution is
Lim_a->0 ln|a| - ln|-1| + ln|e| - ln|a|
Couldn't you cancel the ln|a|'s, and be left with just 1? That's what you would do to limits in any other case.

I talked to her after class, and I asked about it using the graph, saying that the area from -1 to 0 and 0 to 1 were the same, regardless of whether they were infinite or not, since the function was symmetric, and thus you could subtract the infinities in this case. She tried to explain that you couldn't because not all infinities are the same, which is a trivial concept that I understood anyway. I said that in this case, they were, and she pretty much said that you just can't do it that way.

So I'm left unsatisfied. I understand that infinity is just a concept and not really a quantity, but...Why aren't the areas equal? -1 to 0 is the same distance as 0 to 1 and the function is symmetric about the origin, so the two regions should be equal in area. Or am I just going to have to settle for "It just doesn't work like that"?

[Alan Bolte]

This is where either you give up on being satisfied and realize that these sort of rules are there for a well-defined reason that you won't fully understand unless you take a few more years of classes, or you give up on higher math as a whole. At least, that's my interpretation of it. I gave up at a different point and decided I didn't actually want to be an engineer.

[Eframepilot]

The areas aren't equal. Your answer deals with the third problem, integrating from -1 to e, which is divided into nonequal intervals of -1 to 0 and 0 to e. The second interval is clearly larger.

[Alan Bolte]

Naw, I think he means that -1 to 0 and 0 to 1 are the same, and 1 to e is 1, so the answer is 1.

[Grog]

The problem with your solution is that you want the answer to be the same however you approach the limit, for example
Lim_a->0 ln|a| - ln|-1| + ln|e| - ln|a|
Should be the same as
Lim_a->0 ln|2a| - ln|-1| + ln|e| - ln|a|
For your integration problem at least there should be no good reason for why those would give different answers if the integral is defined.

[Steel]

Grog has the answer there, as the integral is not defined, you can actually make tha answer anything you want it to be by approaching the limit in various different ways. If you choose your a to be a = f(b) and have that function->0 then you can make the answer wildly different. Just because you can make a limit turn out to be what you want it to does not mean that it is that or even exists.

[Surlethe]

As I recall, \Int_0^c{1/x dx} blows up for all c>0, so you can do weird things adding and subtracting it. (Note that \lim_{a->0} \Int_a^1{1/x dx} = log|1| - log|a| = -log|a|, which increases without bound. Since adding or subtracting any finite amount will not change the divergence, you get the integral diverging from zero to any c>0.)

Part of the problem is that, since it does increase without bound, you can increase to infinity faster from one side than the other, as Grog pointed out, and this will lead to a discrepancy. This simply means the limit is not well-defined.

Another way of looking at it is that since your subdivisions of the interval (for the integral) are always finite (remember, you're tending to zero), you can subdivide the interval differently on both sides of the origin. Doing this, it's possible to contrive a way to always make one side always fit inside the other, so that you'll get a nonzero difference as a goes to zero (taking left-handed partitions on one side and right-handed partitions on the other would do the trick, I think).

[Kuroneko]

So I'm left unsatisfied. I understand that infinity is just a concept and not really a quantity, but...Why aren't the areas equal? -1 to 0 is the same distance as 0 to 1 and the function is symmetric about the origin, so the two regions should be equal in area. Or am I just going to have to settle for "It just doesn't work like that"?

Your reasoning would be correct if the problem asked you to compute the Cauchy principal value of 1/x on [-1,e] about 0, which is indeed 1. However, improper integrals are not defined that way because the Cauchy principal value in general fails to satisfy certain important properties, e.g., Int_[a,b] f(x) dx = Int[a,c] f(x) dx + Int[c,b] f(x) dx. Worse, the Cauchy principal value can be different depending on the choice of the point about which it is taken, so it is not even a consistent alternative for other types of integrals.

[Dark Lord of the Bith]

Your reasoning would be correct if the problem asked you to compute the Cauchy principal value of 1/x on [-1,e] about 0, which is indeed 1. However, improper integrals are not defined that way because the Cauchy principal value in general fails to satisfy certain important properties, e.g., Int_[a,b] f(x) dx = Int[a,c] f(x) dx + Int[c,b] f(x) dx. Worse, the Cauchy principal value can be different depending on the choice of the point about which it is taken, so it is not even a consistent alternative for other types of integrals.

Since we haven't done any improper integrals in class yet, all I really know about them is what I got from reading ahead last year, so I didn't know about this Cauchy principal value. Is it safe to assume that I'll learn more about this in class eventually?

I think I understand a little better now. Essentially, problems arise since the limit doesn't exist so the integrals diverge, right? The picture of the graph is just sort of misleading because the asymptope doesn't behave that way, and the nature of limits makes it so all the numbers involved aren't really exact quantities?

Thanks everyone.

[drachefly]

I've taken a fair amount of calculus and never encountered the term 'Cauchy principle value'.

At first, I thought, "What happens if you take the analytic continuation of the integral of 1/x?"
Well, ln(z) has an infinite number of alternate values (branch cut) at every point except 0, where it isn't defined. So, that doesn't get you anwhere useful either. Except it does give you only discrete options separated by 2?i.

Now, I would point out that in many particular practical cases of using such an integral, it is clear that you should let the two sides cancel; but this will not always be the case.

[Kuroneko]

Since we haven't done any improper integrals in class yet, all I really know about them is what I got from reading ahead last year, so I didn't know about this Cauchy principal value. Is it safe to assume that I'll learn more about this in class eventually?

In introductory calculus, probably not. In introductory complex analysis, it is much more likely. I mention it only because your idea does actually have a useful and well-defined meaning elsewhere in mathematics.

I've taken a fair amount of calculus and never encountered the term 'Cauchy principle value'.

Which is a shame. The general form of Cauchy's integral formula is f(z_0) = (PV)Int_C[ f(z)/(z-z0) ], where C is an arbitrary contour enclosing region D and phi is the sector of D at z_0, where the Cauchy principal value is taken about the poles along C, if any. Other theorems, such as the residue theorem and the Sokhotski-Plemelj formulae, can likewise be naturally stated using the CPV.

Now, I would point out that in many particular practical cases of using such an integral, it is clear that you should let the two sides cancel; but this will not always be the case.

Of course, it's one thing to fudge a calculation because it makes sense, and quite another to rigorously define exactly in what manner one is fudging it--the CPV does exactly that.

[Dark Lord of the Bith]

Since we haven't done any improper integrals in class yet, all I really know about them is what I got from reading ahead last year, so I didn't know about this Cauchy principal value. Is it safe to assume that I'll learn more about this in class eventually?

In introductory calculus, probably not. In introductory complex analysis, it is much more likely. I mention it only because your idea does actually have a useful and well-defined meaning elsewhere in mathematics.

Then I'll probably run into it eventually, since I'm planning on getting at least a M.A. in math. In fact, my college offers an introduction to complex analysis that I could take next fall, since they only list Calc III as a prerequisite. Would you recommend that course, or are there others that I should take first? And what other recommendations would you have for courses that I should consider?

[Surlethe]

Then I'll probably run into it eventually, since I'm planning on getting at least a M.A. in math. In fact, my college offers an introduction to complex analysis that I could take next fall, since they only list Calc III as a prerequisite. Would you recommend that course, or are there others that I should take first? And what other recommendations would you have for courses that I should consider?

Your university probably has at least the requirements for a bachelor's degree mapped out; but courses you should definitely consider include discrete mathematics; linear algebra; abstract algebra; and real analysis. Also, if you're interested and have the inclination (and you probably do, since you intend to get an M.A.), try reading about math on the side to broaden your horizons.

[drachefly]

I've taken a fair amount of calculus and never encountered the term 'Cauchy principle value'.

Which is a shame. The general form of Cauchy's integral formula is f(z_0) = (PV)Int_C[ f(z)/(z-z0) ], where C is an arbitrary contour enclosing region D and phi is the sector of D at z_0, where the Cauchy principal value is taken about the poles along C, if any.

I don't understand this notation. It may be an artifact of coercing it into a flat text stream, but I suspect it's more of an artifact of my taking complex analysis but _not_ real analysis.

Of course, it's one thing to fudge a calculation because it makes sense, and quite another to rigorously define exactly in what manner one is fudging it--the CPV does exactly that.

In the cases I'm thinking of it isn't fudging. Sure, if you're handed an integral with no context, it'd be fudging. But when you're using the math to solve some exterior problem, it's possible you can use a more fundamental argument than the equation itself to establish the symmetry, then take the integral of only the asymmetric part.

[Kuroneko]

I don't understand this notation. It may be an artifact of coercing it into a flat text stream, but I suspect it's more of an artifact of my taking complex analysis but _not_ real analysis.

Look up the Cauchy integral formula, which expresses (i.2pi)f(z_0) as a certain integral around a countour containing z_0. By the Cauchy integral theorem, the integral itself is zero if z_0 is outside the contour, which can be equated with (i.0)f(z_0). If z_0 is allowed to be on the contour, the integral turns out to be (i.phi)f(z_0), where phi is found in the manner described above--the measure of the sector is 0 for z_0 outside the contour and 2pi for z_0 inside, pi if the contour is smooth around z_0, and anywhere for sharp turns. Except for one little detail--the integral itself may in fact be divergent in the strict sense, but the theorem holds for Cauchy principal value regardless.

In the cases I'm thinking of it isn't fudging. Sure, if you're handed an integral with no context, it'd be fudging. But when you're using the math to solve some exterior problem, it's possible you can use a more fundamental argument than the equation itself to establish the symmetry, then take the integral of only the asymmetric part.

Of course, if you've rigorously defined your handling of the integral beforehand, it's not fudging anymore either, but simply another method. The apparent disagreement seems to be a fairly silly question of nomenclature--perhaps the physicist might say "we're only interested in the asymmetric part about c" where the mathematician might say "it's a principal value integral about c."

[drachefly]

Okay, so I did run into the concept, I just didn't hear the name.