# Discontinuities when integrating

1. Nov 22, 2013

### MathewsMD

Is there a proof that shows why indefinite integrals cannot be assessed when there are an infinite number of discontinuities but definite integrals are can only be assessed when there is no discontinuities? Why does the fact whether there is one or infinite make a difference? Any mathematic proofs would be greatly appreciated! :)

2. Nov 23, 2013

### AlexChandler

Can you reference the results you are talking about? They do not seem correct as stated... for example you can certainly take a definite integral of a step function, which has a discontinuity.

3. Nov 23, 2013

### MathewsMD

From my Calculus class, we were told that you cannot take the definite integral a function with at least one vertical asymptote in the specified domain you are integrating. We were also told that that you cannot integrate over the entire domain of the function if there are infinite vertical asymptotes. I just don't understand why this is true and have yet to see any mathematical proofs in class or online. Any help would be great!

4. Nov 23, 2013

### WannabeNewton

Your statement in post #1 is incorrect. Any bounded function on a closed interval whose set of discontinuities has Lebesgue measure zero is Riemann integrable. In particular, since any countable set is of Lebesgue measure zero, a bounded function on a closed interval with countably many discontinuities is Riemann integrable.

The difference in post #3 is that you are talking about unbounded functions.

5. Nov 23, 2013

### gopher_p

If there is a vertical asymptote, then the function is unbounded. It can be shown that the unique limit of Riemann sums does not exist in this case, and so the Riemann integral also does not exist.

6. Nov 23, 2013

### AlexChandler

If the OP was familiar with measure theory, I don't think the question would have been asked in the first place.

7. Nov 23, 2013

### MathewsMD

Okay. So why must there be infinite discontinuities and not just one for the indefinite integral to not be integrable? I'm assuming that you were speaking about the definite integral, right?

Also, if you're taking the indefinite integral of sin2x/sinx (as an example)

= ∫ sin2x/sinx
= ∫ 2sinxcosx/sinx
= ∫ 2cosx
= 2sinx + C

But there are an infinite number of discontinuities at x=∏k where k is any integer since the function 2sinxcosx/sinx is not defined for those values on (-∞, ∞). But lim x→∏k sin2x/sinx = 2 preventing any discontinuities. So do we not consider there to be any discontinuities in this case?

Any explanations you could provide would be amazing. If you could also please provide links to redirect me for further reading that would be great as well. Thanks!

8. Nov 23, 2013

### WannabeNewton

There is no measure theory required for this. The stated theorem is a basic result of calculus. You don't need to know measure theory to know what a set of Lebesgue measure zero is.

9. Nov 23, 2013

### gopher_p

The domain of the indefinite integral/antiderivative of a function is assumed to be the same as the domain of the function, although this is not usually stated overtly. The Fundamental Theorem of Calculus assures us that functions which are Riemann integrable have antiderivatives. Since sin(2x)/sinx is Riemann integrable on any finite interval by the Lebesgue integrability criterion mentioned by WannabeNewton, it has antiderivative. That antiderivative is 2sinx+C for x≠kπ.

10. Nov 23, 2013

### MathewsMD

But it's not integrable on x = ∏k....doesn't this mean it has no indefinite integral, and only a definite integral when ∏k is not in between or included in the upper or lower limits?

11. Nov 23, 2013

### gopher_p

Sorry. Since sin(2x)/sin(x) is not defined at k∏, it would not be properly Riemann integrable on any interval containing an integer multiple of ∏. However if we were to extend the function to be defined at k∏ for all k, even if the extension weren't continuous the resulting function would be properly Riemann integrable on any finite interval.

If you are taking a standard introductory calculus course (i.e. one geared towards a typical engineering student), then it's likely your text includes a theorem stating something along the lines of "If f is defined and bounded on [a,b] and continuous at all but finitely many points of [a,b], then f is Riemann integrable on [a,b]". While this is a true statement, it is not the whole truth; there are functions with infinitely many discontinuities in a finite interval which are Riemann integrable.

12. Nov 23, 2013

### MathewsMD

Thanks for the clarification!

13. Nov 23, 2013

### MathewsMD

By the way, do you mind naming some examples? The ones I've found online say that a function with an infinite number of vertical asymptotes are not Riemann integrable...If there is a single hole at a value for x and no actual value for f(x) at x versus just a discontinuity like in a Heaviside step function, is it Riemann integrable?

14. Nov 24, 2013

### gopher_p

Let $f$ be defined on $\mathbb{R}$ by $f(x)=0$ if $x$ is not rational and $f(x)=\frac{1}{q}$ if $x=\frac{p}{q}$ is rational and $\frac{p}{q}$ is reduced. It can be shown that this function is continuous at each irrational point and discontinuous at each rational point. It can also be shown that $f$ is Riemann integrable on every interval $[a,b]$ and $\int_a^bf(x)\ dx=0$.

It sounds like you may be confusing discontinuities with asymptotes. While it is true that an asymptote does correspond to a discontinuity, not all discontinuities involve asymptotes. An asymptote occurs when there is an infinite limit; $\lim_{x\rightarrow a^-}f(x)=\pm\infty$ or $\lim_{x\rightarrow a^+}f(x)=\pm\infty$.

If there is just a hole, then the function is not Riemann integrable simply because the function is not defined there. It's kinda just a technicality because we can always just define the function to be whatever we want at that point, i.e. just make something up, and it really has no effect on the new function's integrability.

For example, $\sin(\frac{1}{x})$ isn't defined at $0$, and so it is not Riemann integrable on any interval including $0$. However if we were to extend that function to a new function $f$ defined by $f(x)=\sin(\frac{1}{x})$ if $x\neq 0$ and $f(0)=C$, then no matter what $C$ is, $f$ will be Riemann integrable on every interval. Conversely, no matter what $C$ is, the function $g$ defined by $g(x)=\frac{1}{x}$ for $x\neq0$ and $g(0)=C$ will not be integrable on any interval containing $0$.

15. Nov 24, 2013

### MathewsMD

So if I'm understanding this correctly:

if f(x) = sin(1/x) for x ≠ 0 and f(x) = C for x = 0

then the function is Riemann integrable but not integrable since C could be any value that changes the actual area of the graph, and thus the result.

So if a function is not defined for a point x in its domain (whether it is a vertical asymptote or just a general value for x for which f(x) is undefined) then we cannot take the definite integral of that function on an interval containing x. I still don't understand the reasoning behind why an indefinite integral can still be taken, though, but not if and only if there are an infinite number of discontinuities for which f(x) is undefined.

16. Nov 24, 2013

### MathewsMD

You lost me at the very end. Why is g(x) no longer integrable if it is defined for x = 0? You could technically say f(x) = sin(g(x)), no? Then would it be integrable or not?

17. Nov 24, 2013

### gopher_p

Could you maybe tell me exactly what you were told, paraphrasing as little as possible, regarding integrability/existence of antiderivatives and infinite "bad points? It seems that much of your confusion revolves around this in some way.

Also, just so I have an idea of where you are coming form, what is the nature of the calculus course that you are taking; i.e. what is the name of the course, what is your major, what is the "typical' major of the other students in your class, etc.?

18. Nov 24, 2013

### MathewsMD

It is first year Calculus 1000. It is the first semester and next semester will be Calculus II. I am undecided right now but leaning towards Medical Physics, and there are a lot of different students (science and business namely).

I asked my professor to clarify when a function is not integrable and the response from her was (with as little paraphrasing as possible): the definite integral cannot be calculated if there is at least one discontinuity in the interval of the upper and lower limits. The indefinite integral cannot be taken if there are infinite discontinuities in the graph.

My main concerns are:

1) Does discontinuity specifically mean where there is a break in the graph at some x value, BUT it is still defined for the x value? Or, does it mean where there is a break in the graph at some x value, BUT it is NOT defined for the x value (ex. vertical asymptote)? Or is it both?
2) Why must there be infinite discontinuities and not at least one for indefinite integral to be not be considered integrable? I keep hearing that we only consider it for the domain that the function is defined for, but wouldn't we require a new calculation for an exact answer for this? For definite integrals, if there is one discontinuity, it is not integrable. I don't understand why this doesn't hold true for indefinite integrals. I guess I have yet to see (or at least fully understand) any proofs that can specifically answer my question at the level of math I know currently.

Last edited: Nov 24, 2013
19. Nov 24, 2013

### gopher_p

I would characterize both of those statements as false. I've already given an example of a function with lots of discontinuities for which every definite integral can be calculated. $f(x)=\sec^2x$ has infinitely many discontinuities, and we can characterize all of its antiderivatives. If you replace the word "discontinuity" with "asymptote" then the first statement becomes true, but the second one is still false.

Giving your instructor the benefit of the doubt that she really knows the material and understood the question that you were asking, I would say that either (a) she was giving you the "freshman calculus" version of the answer (i.e. the one that allows you to to "correctly" answer the questions presented to you in the class) without stipulating that she's lying to you just a little bit or (b) you misheard or misinterpreted her response.

(1) You should have a working definition for "$f$ is continuous at $a$"; something along the lines of "$f$ is continuous at $a$ if and only if $\lim_{x\rightarrow a}f(x)=f(a)$". There are three things stipulated here. (a) That $\lim_{x\rightarrow a}f(x)$ exists, (b) that $f(a)$ makes sense (i.e. $f$ is defined at $a$, and (c) that these two numbers are equal. "$f$ is discontinuous at $a$" just means that one or more of these three conditions fails.

"break in the graph at some x value, BUT it is still defined for the x value" is when (c) (and possibly (a)) fails. "break in the graph at some x value, BUT it is NOT defined for the x value" is when (b) (and possibly (a)) fails and (c) doesn't even make sense. So they are both examples of types of discontinuities. An asymptote is a specific (and relatively nice) case when (a) fails. There are more exotic cases of discontinuity that really don't properly correspond to any kind of picture that can be drawn, and these cases are usually considered to be the most interesting.

(2) Again, I would characterize your understanding here to be false. More importantly, I really don't think it's all that necessary for you to be worrying about that right now. It's great that you're curious, but in my opinion you should focus on mastering the basics first (especially since finals are coming up soon). Regardless of your academic path, the material in a first year calculus course like the one that you're taking is meant to serve a springboard to learning about the more interesting stuff like what we're talking about here. Understanding things like limits, continuity, derivatives, Riemann sums, definite integrals, etc. beyond the computations is key towards grasping the more interesting and exotic ideas, because most of the truly interesting stuff (in my opinion) is beyond the computations.

For what it's worth, I think it would be very difficult to even state the full versions of the theorems that address your concerns in terms that you you could understand at your current level. That's not a knock on your abilities, just a statement on where you are along the path.

20. Nov 24, 2013

### MathewsMD

Okay! Thank you so much for your help!