Your calculus prof lied to you (probably) 

Ye, I love it. Absolute value is piecewise defined and +C can be a piecewise defined function that has derivative zero! So the two are equivalent.

Anytime the value of the locally constant function changes, the derivative at that point is undefined, so the only time that +C can jump to a different constant is when the original function is undefined (like things involving 1/x). is this correct?

@@aryanjoshi3342 If the definition of indefinite integral is extended in such a way that it includes cases with discontinuous domain, probably yes. Though generally I'd just avoid such cases and, if absolutely needed, consider each needed domain interval separately and use concepts such as 'derivative' and 'integral' only to such domain-restricted functions.

So if a student writes “+C+1” where your grading key has “+C”, you take off a point, right? By “you” here, I mean your grading software or just grader.

Oh absolutely. And to be clear, setting aside the clickbait title, I don't make a big deal of this point to my students either. I do like showing the example as I think it helps solidify the concepts, but for all practical purposes I use ln|x|+C like everyone else:D

@@DrTrefor I disagree with the idea that it's OK to wave hands as willy nilly as calculus books do. Why don't they at least point out that they're abusing the terminology and the notation, in favor of speed? The Higher level courses, like linear algebra, are harder to grade because no one wants to address these things before the damage is done. One of my engineering colleagues told me he had students who couldn't understand why he asked them to find the derivative of pressure with respect to temperature in the ideal gas law (Pv=NRT), because they said it doesn't depend on x (I think he actually quoted the student as saying "there's no x"). Calculus books wrongly identify expressions with functions and don't teach that that's an abuse of notation and terminology, and then almost never do they use variables other than x or t as independent variables, making it tougher to clean up that mess after it's entrenched in the students' minds. The claim then that (ax)'=a is false, but it's ubiquitous on quizzes and tests as the only correct answer. If a good student writes (ax)'=a'x+ax', the grading key says they're wrong when they're right. Without context the problem is misstated and does more harm than good.

@@DrTrefor I get this. In many situations that call for the indefinite integral you end up only needing 'an' antiderivative rather than all of them. Either because you have some initial value that lets you define what the constant C is, or because the problem eventually leads into a definite integral. So in a lot of cases the piecewise definition doesn't really bring anything useful. But even so it's still a good idea to understand when the shorthand doesn't actually include It's a bit of a convenient shorthand that is fine for most situations, but it's still a good idea to keep in mind what it's lacking. Additionally, correct me if I'm wrong, but you could likely correct for this at the end if you really wanted to be complete, by at the end adding a little piecewise definition for C depending on the which interval of the domain x is in. Sort of how like when taking a definite integral it's common to omit the +C entirely since you know it's going to cancel out, but it's still important to remember that that +C is still really there and what it means. Also, if I'm not mistaken you could correct for this at the end (or at any step) if you really wanted to, by adding a piecewise definition for C for each interval of x separated by a point of discontinuity.

Also, this issue would crop up for integrating lots of other functions: f(x)=1/x^2, f(x)=3/(x-5)^3, etc. And imagine asking for the "full" answer for the indefinite integral of tan(x).

@@justsomeboyprobablydressed9579 there are so many places where the notation seems slightly buggy. this is an interesting find. the first thing that came to mind is piecewise spline curves.

but if 1/x is undefined at x=0 and A δ(x) is only nonzero at x=0, shouldn't the A H(x) term remain in the form for the antiderivatives of 1/x, since 1/x should exactly equal 1/x + A δ(x)?

@@rarebeeph1783 A distribution isn't defined pointwise, so we cannot say that the distribution 1/x is undefined at x=0. Saying that δ = 0 on ℝ∖{0} is okay, however. Within the theory of distributions, what happens when we add A H(x) to the antiderivative, as is done in the video, is that we add terms A δ(x) to the derivative. Thus, our antiderivatives are not for 1/x but for 1/x + A δ(x). So, the only antiderivatives of (principal value) 1/x are ln |x| + B.

@@rarebeeph1783 A better way to think of it is in terms of measures. Instead of thinking of the dirac delta as a function of a real variable (because it isn't), you should think of it as a measure concentrated at zero. Moreover, the notation "δ(x)" for such a concept is wholly confusing, mathematically incorrect, and pedagogically vapid. One should denote it only as δ_0 ("delta subscript zero"), and realize that it maps measurable sets to real numbers. If the measurable set E includes the number 0, then δ_0 maps E to 1, and otherwise, it maps E to zero. That way, when one computes the integral, over the measurable set E, of the dirac delta times a real valued function f of one real variable, is either 0 or f(0), according to whether 0 is in the set E or not. For the function f given by the formula f(t)=1/t, 0 is not in the domain of f, so without a richer interpretation of integration as in Lebesgue integration or some other variants on Riemann integration and Improper Riemann integration, you cannot assign a meaning to the integral in question even in this case. A problem with the question at hand is that 0 is not in the domain of the function in question, so integrals of it against the dirac distribution don't really make sense. I guess one can (and perhaps does) extend the theory by extending the notion of these integrals to try to extend the integrand to the closure of its domain, but in this case, one still has the problem that the unboundedness of the function requires one to first extend the real number system to the extended real numbers so that both lazy 8 and its negative are members of the codomain (and perhaps of the domain). In this case, we still cannot make sense of f(0) in a way that is not seen as ad hoc, because the limit from the left and the limit from the right at 0 do not agree. One could instead use the Riemann sphere, I guess, which is the one point compactification of the complex plane, and then one does not include the negative of lazy 8 in the codomain or the domain, and one "defines" 1/0 to be lazy eight, which is the point of compactification, located at a pole of the Riemann Sphere, usually considered to be the "North Pole", diametrically opposite the pole at 0, usually considered to be the "South Pole". In this nice repair of the situation, one would be able to recover the property described for the dirac distribution for bounded functions, namely that the integral of "δ(x)/x" (abusing the notation in the sophomoric way in ODE textbooks), would be interpreted (`calculated'?) as f(0)=\infty.

Is that language of gods?

thats a dam good answer, it just looks at the discontiniouty in the most elegent way possible. I went about it in a dumber way with insisting on the complex derivative to make sense as well

I'd disagree with removing the + C because then you'll abuse = notation, but I'd be in favour of getting the C to explicitly mean "the set of all functions with derivative 0 over the domain". In this case, the piecewise C and D would quallify, because it's derivative is 0 across the entire domain (since the discontinuity isn't included in the domain) This is something we implicitly do through abuse of notation, but this qualifier would make it formal. In this case the + C would always make the equality valid

I would also be happy with this. It slightly changes the criticism, which is then that most calc textbooks don't put the added qualifiers on what an indefinite integral means that you did here.

And of course the "trick" works because |x| is itself piecewise, or with the alternate definition of sqrt(x^2), because sqrt is not a true inverse function because of x^2 isn't 1-1.

Eh. I think the term "indefinite integral" should be retired altogether, as it is more unhelpful and confusing and misleading, than it is useful.

Indefinite integral is just a confusing way of saying antiderivative. If it is defined differently, a indefinite integral of f should be all function F satisfy F(b)-F(a) equal integral of f from a to b in the domain

@@Noname-67 That is fine for continuous functions, but for other functions you have to use something more general than the Lebesgue integral if you want to recover the usual class of antiderivatives.

Yes! If we take the complex result restricted to the real line, we actually know the difference between the constants on the left and right, and it's imaginary.

(confession: I was a graduate student doing particle physics calculations before I figured this out)

The abuse of notation in asymptotic growth is much more egregious, though. It’s insane that someone would choose “=” to represent anything that’s not equality, or even an equivalence relation.

or, as another guy said here, since you don't use indef integrals on disconnected domains (to compute definite integrals), you give two separate antiderivatives on two connected parts

Awesome!

I'd say the main "practical" point is just reinforcing the normal idea that you can't just integrate between two values, you have to check the domain first to picture you don't have some improper integral. That point is well made in normal calculus exposition, but this helps reinforce it I think.

Not just unnecessary, wrong. d/dz ln|z|=Re(z)/|z|^2

Indeed!

This is not only true for functions with poles, but for any function that is continuous on a disconnected domain. The antiderivative of function f(x)=x, where we restrict the domain to be the nonzero real numbers will have the same problem.

@@tracyh5751 Better yet, as I indicated in another reply, it is even reasonable to work with totally disconnected domains, like the rational number field, in which the proper way to state the corresponding theorem is the following: Theorem: Let f and g be functions with the same derivative. Then f and g differ by some pseudoconstant function. A function c is locally constant at a point p provided that there's an open neighborhood of p on which c is constant, and we say that a function c is locally constant if it is locally constant at every point in its domain. On a connected domain, every locally constant function is constant. (There's a better result available, by the way, and I'll leave it as an exercise 🙂.) We call a function h a pseudo-constant function if its derivative is zero on its domain. Not all pseudo-constant functions are locally constant, although the converse holds. In fact, Charatonik (r.i.p.) and I studied the notion of an absolute derivative, which is closely related to the notion of a derivative (and in the cases considered in this forum, are computed by finding the absolute value of the derivative), and we constructed functions on totally disconnected spaces (like the rational numbers or a cantor set) that are everywhere differentiable, have zero derivative everywhere (and so are pseudo-constants), but are nowhere locally constant, meaning that if p is a point in the domain and U is an open neighborhood of p, then such a function is not constant on U. The heat of the result is that with a base of clopen sets, one carefully constructs a sequence of functions, each of which is locally constant at less points than the previous sequence member, but approximates the previous sequence member everywhere. This does serious damage to the uniqueness theorem for differential equations, unless one phrases the uniqueness theorem in terms of equivalence classes whose members only differ by a pseudo-constant.

The passing to the complex plane, with the zero removed or a ray removed, introduces a restriction that doesn’t exist on the reals with the zero removed. The continuity on this larger connected set. That is why you get a result that when restricted to the reals gives you a smaller set of solutions.

You should be careful doing this if you haven’t taken complex analysis. As the other comment stated, your perspective added restrictions that weren’t previously there, therefore you get a smaller solution set. In order to use ln in the complex plane, you need a lot to get there first

Just take them as the separate functions on the intervals they’re defined on, assuming they’re broken into intervals. Piecewise isn’t hard, it’s more mental struggle than actual math capability struggle. Just try to shake that thought process

I mean, depending on your analysis course, ln(x) is sometimes just defined as ln(x) = int_1^x 1/t dt for x in the domain (0, infinity), so there's no more justification really needed to show that the antiderivative of 1/x is ln|x| (or some transformation of ln|x|). From there, you can define its inverse exp(x) and find all the relevant properties of both.

@@andrewkarsten5268 I think you have just described me perfectly. It’s more of my low self-confidence in my mathematical abilities stemming from my K-12 education where I would fail most of my math classes every year and make them up in my summer. And here I am, just recently graduating with a mathematics degree..

@@ericy1817 I think you may be correct because I think that is how my professor defined, if I remember correctly. The good news is that I think I passed my analysis course. I’m still waiting for the final grades to be posted, but I think I did sufficiently well to pass with a passing grade. Honestly, even though I more than likely passed, I feel like I didn’t master the material the level I would have liked. I felt like I barely scrapped by because I didn’t have sufficient time to study or learn the material since I learn at such a slow pace given my learning disabilities. I love math, even though I consider myself to be terrible and slow at it.

Geogebra!

Exactly! 1/x is just the simplest example

in 1 dimension, yes. But we can just call them open intervals (open connected has a sort of 2D or more connotation)

Or even without discontinuities. Take for example f(x) = |x|, which is everywhere continuous, though not differentiable at x=0. The derivative of |x| is x/|x| for any non-zero x, but this is also the derivative of any piece-wise defined function such as f(x) = |x|+C1 if x>0 and f(x) = |x|+C2 if x

It works, absolutely.

negative numbers (but not 0) have a logarithm in the complex numbers, but the complication is that they have infinitely many logarithms: if z is a logarithm of -x then so is z + 2i pi*k for any integer k. one says that the complex logarithm is a multi-valued function with infinitely many branches

Do you know of any examples of functions with vertical asymptotes and convergent integrals on both sides, whose most obvious choice for an antiderivative has a jump discontinuity? I'm trying to find a function that motivates the need to split the integral at the vertical asymptote and carry out two improper integrals. Every example I come up with, usually has continuity that is salvaged when taking the anti-derivative, and you can get away with ignoring the vertical asymptote, and still get the correct answer.

you're spot on

Exactly. The specific choice of 1/x was just as an example, any function whose domain is a disjoint union of intervals has the same issue.

@@DrTrefor ... Wow, interesting. I was about to ask how you deal with calculation of areas that include the point of discontinuity (if the area remains convergent) but then I realized that you need to break the integral in the same intervals at each point of discontinuity anyway. So for example if you want the are under 1/x^2 between -2 and 3 you have to find the area between -2 and 0 plus the area between 0 and 3; you cannot do it in one step. So using different arbitrary integration constants for each segment would not affect the result.

Exactly

Just powerpoint!

Geogebra! It’s free and awesome

@@DrTrefor thank you, maybe you also know program for xyz axes for drowing?

Exactly! 1/x is just a template function here

Pretty much anything that has infinity is going to require you to split it there, otherwise the integral would likely just be infinity. But, if you do split it, you get a couple of improper integrals that may be finite, depending upon the specifics..

In this case, the function wouldn’t be differentiable at 0

@@DrTrefor yeah, so the value at 0 wouldn't affect the derivative. right?

ln(z) = ln(|z|exp(iarg(z)))= ln|z| + iarg(z) Absolute value becomes complex modulus and the phase factor is added to the constant. If you can have two different constants on the real line for the positive and negative domains, how do you ensure it's continuous when extending to the complex plane?

@@cryora it's not continuous on real plane though. The value of ln(0) is undefined, so the fact that constants are different doesn't matter

@@YuriyNasretdinov I mean continuous on the complex plane except z=0. If you have two different constants for x0 on the real line, there must be intermediate values when you go off the real line and into the imaginary axis.

Only when the integral converges as you approach the asymptote, does it even make sense to talk about a definite integral crossing a vertical asymptote. The integral of 1/x diverges on both sides, so you can't integrate it across the singularity. For instance, the integral of 1/(x-1)^(1/3), from 0 to 9. The indefinite integral is 3/2*(x - 1)^(2/3) + C. Or more accurately per this video, it should be 3/2*(x - 1)^(2/3) + C + D*H(x-1), where H is the Heaviside step function, and D is another arbitrary constant. Keeping it simple and using the conventional answer for the indefinite integral, you can evaluate it across the two bounds. You will get the same answer as you would if you split the integral in two segments, and add the two segments together. The reason is that the real valued versions of (x-1)^(2/3) will connect to each other with continuity on both sides of x=1, and form a cusp at x=1, instead of a vertical asymptote.

Go bearcats!

Yup!

The problem here is that the derivative is not defined at x=n in your definition, so it can't be an anti-derivative on all of R.

The problem with a piecewise approach to functions with only one interval in the domain is that the two pieces will have to be continuous where the two intervals meet. But in order for that to happen, the constants for the two pieces will have to be the same. This is kind of what breaks down with the antiderivative of 1/x. There is no need to force continuity at x = 0, because the function 1/x isn't even defined there.

@@tracyh5751 Thanks for the explanation! I didn’t know a piecewise function had to be continuous with a single interval.That would answer my question well, but how come that’s the case? I mean if I really wanted to make such a function like the f(x) I wrote, would that just be not allowed?

@@sunruiheng4255 The problem is at the endpoints. When we say that a function is differentiable on its domain, we mean something specific in terms of limits, and the way those limits are computed is what calculus books refer to as "two sided" limits. In hither level real analysis courses, one may (and that they do, by the way) define "derivative from the left" only in terms of a limit from the left, and similarly on the right, fit functions of one real variable. Then one of these can be used on your example, because the relevant result ("differentiability implies continuity") can be weakened to "left differentiability implies left continuity", and then (a variant of) your system of constants can be pasted together over the entire domain in such a way as to constitute one "left pseudo-constant", meaning that the function is left differentiable everywhere (but perhaps not right differentiable everywhere), so that adding it to the identity function didn't change the left derivative anywhere, i.e. yielding another function that's a (left) anti-derivative of the constant function 1. (Actually, I didn't check carefully whether one needs to weaken to left differentiability to fix your example or right differentiability, so please adjust accordingly.) This direction of the discussion has a natural extension to multi-variable functions, in which directional derivatives are considered, but that's a good deal trickier, until you start learning measure theory, and then you only concern yourself with differences between functions on sets of positive measure. For the example in the video, the function domain certainly has positive measure, so the pseudo-constant need only have zero derivative on a set whose complement has zero measure, and the example Dr. Bazett described certainly fits the bill, but your example isn't quite there because your attempt to build a pseudo-constant (without you calling it that...) doesn't yield a function that's almost everywhere (except on a set of measure zero) constant. A lot of variation is available, as long as you're willing to weaken some things here and there and be quite careful, which this video points out is apparently not quite the goal of calculus text authors.

@@sunruiheng4255 I believe the reasoning would be that your derivative function is continuous at the boundary point between the 2 intervals, this would require the antiderivative to also be continuous there as well. If in your example you made C1 and C2 be different then the function would not be continuous at n, and as a result the derivative of this piecewise function would not be defined at n, so the derivative in that case would be f'(x) = 1; x /= n I think this is what Dr. Trefor Bazett meant in his response. Yes, you defined x=n in your antiderivative, but he said that x=n would not be defined in the derivative (due to the discontinuity of the antiderivative). And additionally yes, you could make a piecewise function like the f(x) in your initial post if you wanted, there's nothing wrong with that. The only issue here is that the derivative of such a function would not be f'(x) = 1, and therefore it's not an antiderivative of f'(x) = 1.

Yup!

While it’s true you can’t do a proper integral over 0, with absolute values you can still integrate from say -2 ti -1

I’m that case it wouldn’t be differentiable at 0

exactly! Any time the domain is disconnected like this it works

But then the anti derivative won't be continuous at the point where you change constants so you won't be able to take a derivative there. This doesn't matter for integrating 1/x or any 1/x^k (k >= 1) as they're not defined at zero so do not need to be differentiable there. So as long as we have a hole to take advantage of, we can put in a split and make it so that the resulting non-differentiable point lines up with the original undefined point.

@@sjth2497 oh right. Thanks for the correction.