AlexChandler said: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.
MathewsMD said: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!
WannabeNewton said: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.
gopher_p said: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.
AlexChandler said:If the OP was familiar with measure theory, I don't think the question would have been asked in the first place.
MathewsMD said: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!
gopher_p said: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.
MathewsMD said: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?
gopher_p said: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.
gopher_p said: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.
MathewsMD said: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?
gopher_p said: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##.
gopher_p said: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.?
MathewsMD said: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.
Discontinuities when integrating refer to points on a function where there is a sudden or abrupt change in its value. This can result in a break in the continuity of the function and can pose challenges when trying to integrate the function.
Discontinuities can be caused by a variety of factors, such as a sudden change in the slope of a function, a sharp turn or corner, a jump in the function, or a point where the function is undefined. These changes in the function can result in a break in its continuity and lead to difficulties when integrating.
To identify discontinuities when integrating, you can graph the function and look for any sudden changes or breaks in the graph. You can also look for points where the function is undefined or where there is a sharp turn or jump in the graph. Another way to identify discontinuities is by examining the limits of the function at certain points.
Discontinuities can make the process of integration more challenging, as they can result in breaks in the function's continuity. This means that the function may not have a definite value at certain points, making it difficult to find the area under the curve. Discontinuities may also require the use of different techniques, such as piecewise integration, to properly integrate the function.
In some cases, discontinuities can be avoided when integrating by using different techniques, such as substitution or integration by parts. However, some functions may have inherent discontinuities that cannot be avoided. In these cases, it is important to properly identify and handle the discontinuities in order to accurately integrate the function.