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Consider the expression $$\frac 10 - \frac 10$$

This is an indeterminate expression, and even though it's of the form $a - a$, we can't say that it evaluates to $0$ because $a$ itself is undefined here.

This got me thinking about how a definite integral might be indeterminate. I have two questions:

1. Consider a function $f(x)$ that is continuous everywhere but at $x=x_1$, where it is undefined. Is $$ \int_{x_1}^{x_2} f(x) \, \mathrm dx$$ defined? If so, what is its value? I was thinking that if $f$ is undefined at $x_1$, then $F$ is undefined there also, and so by the definition of the integral, this would be undefined.

Yet, if I remember that there is no differentiation (here I mean "differentiation" in a non-mathematical sense) between $ \int_{x_1}^{x_2} f(x) \, \mathrm dx$ and $ \int_{x_1+\delta}^{x_2} f(x) \, \mathrm dx$, then I become inclined to think that $$ \int_{x_1}^{x_2} f(x) \, \mathrm dx = \lim_{h \rightarrow x_1} \int_{h}^{x_2} f(x) \, \mathrm dx$$

But this creates a problem, doesn't it? If a function that is undefined at certain points can still be integrated, then it's not very practical to talk about integrability because we can just break any otherwise non-integrable function up into several integrals of which we take the limit.

**Subquestion: ** Can we make any non-integrable range integrable by splitting the range on non-integrable points and taking the limit at each discontinuity? If so, does the handedness of the limit matter?


2. Consider a function $f(x)$ that is continuous everywhere but at $x=x_1$, where it is undefined. Is $$ \int_{x_1}^{x_1} f(x) \, \mathrm dx$$ defined? If so, what is its value?

From a geometric standpoint (thinking of a definite integral as an area), then it seems obvious that this should be zero regardless of whether or not $f$ is defined at $x_1$. But perhaps this is only "obvious" in the same way that $\frac 10 - \frac 10 = a - a$ would "obviously" be zero, which we know is not the case. Is evaluating this integral tantamount to evaluating $\frac 10 - \frac 10$?

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  • $\begingroup$ What definition of integral are you using? Lebesgue integration allows integration over open intervals. $\endgroup$ – ajotatxe Dec 9 '17 at 19:11
  • $\begingroup$ @ajotatxe I was thinking of the Riemann integral. I've never heard of the Lebesgue integral; I'll check that out. $\endgroup$ – Tiwa Aina Dec 9 '17 at 19:12
  • $\begingroup$ Be careful when stating that $\frac{1}{0}$ is "indeterminate." It is not. It is simply not defined. The expression $\lim_{x\to 0} \frac{1}{x}$ (for example) is often described as an indeterminate form of the type $\frac{1}{0}$, but the expression $\frac{1}{0}$ remains undefined. $\endgroup$ – Xander Henderson Dec 9 '17 at 19:12
  • $\begingroup$ consider the function $f(x)=\dfrac{d}{dx}|x|$, we have that $f(0)$ is undefined but $F(0)=0$, another example is $f(x)=\frac{1}{2\sqrt{x}}$. $\endgroup$ – Holo Dec 9 '17 at 19:23
  • $\begingroup$ if a function is continuous everywhere but $x_1$ and the right and left side of the limit of $x_1$ are equal then you can say that $\int_{x_1}^{x_1} f(x) \, \mathrm dx=\lim_{h\to x_1} \int_{h}^{h} f(x) \, \mathrm dx=0$ $\endgroup$ – Holo Dec 9 '17 at 19:30
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You are essentially asking how an improper (Riemann) integral can be defined. Typically, an integral $$ \int_{a}^{b} f(x)\,\mathrm{d}x $$ is said to be improper if either $a$ or $b$ is infinite, or if $\lim_{x\to a} f(x)$ or $\lim_{x\to b} f(x)$ is infinite. The Riemann integral is not exactly well-suited for asking these kinds of questions, but we can generally get away with considering the limit, as you have. That is, define the improper integral (in the case where $a=-\infty$ or the limit of $f$ at $a$ is infinite) by setting $$ \int_{a}^{b} f(x)\,\mathrm{d}x := \lim_{t\to a^{+}} \int_{t}^{b} f(x)\,\mathrm{d}x. $$ As long as $f$ is integrable on every interval of the form $[t,b]$ and the limit exists, then this is a reasonable approach, and one that (I think) is appropriate for your question.


More generally, the "right" approach might be measure-theoretic. Instead of working with the Riemann integral, we can work with the Lebesgue integral. An entire discussion of the definition of the Lebesgue integral is far beyond the scope of a single answer on MSE, so let me get to the punchline:

In the Lebesgue theory, we integrate over sets, rather than intervals. In notation, $$ \int_{A} f(x)\,\mathrm{d}\mu(x) $$ is the integral of the function $f$ over the set $A$, with respect to something called a measure $\mu$. The measure isn't really playing a role in the current discussion, so don't worry about too much—for now, you might think of it as a bit of decoration that distinguishes a Riemann integral from a Lebesgue integral.

There are certain sets that are very small (called null sets), which can be thrown away without changing the value of the integral. Singleton points, countable sets (i.e. the integers) and the usual middle-third Cantor set are examples of such "small sets". In other words, if $N$ is a null set, then $$ \int_{N} f(x)\,\mathrm{d}\mu(x) = 0, \qquad\text{and}\qquad \int_{A\setminus N} f(x)\,\mathrm{d}\mu(x) = \int_{A} f(x)\,\mathrm{d}\mu(x). $$ Moreover, it turns out that if a function is (properly) Riemann integrable, then it is also Lebesgue integrable, and the two methods of integration give the same number.

So suppose that we want to evaluate $$ \int_{a}^{b} f(x)\,\mathrm{d}x $$ as a Riemann integral, but $f$ is not defined at $a$. Assuming that a $f$ is not too pathologically behaved, we can instead interpret this as a Lebesgue integral, i.e. $$ \int_{a}^{b} f(x)\,\mathrm{d}x = \int_{[a,b]} f(x)\,\mathrm{d}\mu(x). $$ But the singleton set $\{a\}$ is a null set, and so we have $$ \int_{[a,b]} f(x)\,\mathrm{d}\mu(x) = \int_{(a,b]} f(x)\,\mathrm{d}\mu(x). $$


In short, depending on the nature of $f$, it might be appropriate to regard the integral as in improper Riemann integral and take limits, or to regard it as a Lebesgue integral and apply that theory, instead.

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  • $\begingroup$ You might want to be more specific and say "the Cantor middle thirds set", because there are generalizations of the Cantor construction --- where the "middles" that one inductively removes have a carefully chosen sequence of smaller and smaller proportions --- which produce "fat" Cantor sets having positive Lebesgue measure. On Wikipedia they go by the name of Smith-Volterra-Cantor sets. $\endgroup$ – Lee Mosher Dec 9 '17 at 19:54
  • $\begingroup$ @LeeMosher Indeed. I should have been more careful---I even have such a non-standard Cantor set tattooed to my arm. ;) I'll fix that. Thanks. $\endgroup$ – Xander Henderson Dec 9 '17 at 20:14
  • $\begingroup$ Hah! Interesting! It must have hurt. $\endgroup$ – Lee Mosher Dec 9 '17 at 21:14
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1) Such integrals are defined improper integrals and they are defined as limit around the point where the function is not defined. Only for simplicity we write:

$$\int_{x_1}^{x_2} f(x) \, \mathrm dx $$ but as you noted the true definition is that:

$$\int_{x_1}^{x_2} f(x) \, \mathrm dx := \lim_{h \rightarrow x_1^+} \int_{h}^{x_2} f(x) \, \mathrm dx$$

2) In this case the integral does'n make sense (how ever you can also define it by limit and set its value to 0)

https://en.wikipedia.org/wiki/Improper_integral

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  • $\begingroup$ you have to be sure that the left and right side limits are equal $\endgroup$ – Holo Dec 9 '17 at 19:33
  • $\begingroup$ @Holo It's set as a definition, I update to make it more clear. Thanks. $\endgroup$ – gimusi Dec 9 '17 at 19:35
  • $\begingroup$ @Xander Henderson You are right. Thanks! $\endgroup$ – gimusi Dec 9 '17 at 20:21

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