1

I understand that definite integrals can be defined as the limit of Riemann sums but I have yet to see a similar definition for the indefinite integral. I suppose $$ \int f(x) \, dx = \{F(x):F'=f \} $$ might do the trick but I'm unsure if this really a satisfactory definition. Alternatively, with the knowledge of the fundamental theorem of calculus, we might define $$ \int f(x) \, dx = \int_{a}^{x} f(t) \, dt + C $$ but this definition seems odd in that uses quite a deep result to define something which is usually introduced very early on in calculus classes. So is there a formal definition of an indefinite integral, and if so, what is it?

Bernard
  • 175,478
Joe
  • 19,636
  • 2
    An indefinite integral of $f$ is any function whose derivative is $f$. It is not uniquely defined. – Kavi Rama Murthy Jan 11 '21 at 23:17
  • as far as I know there is no formal definition of indefinite integration, this is because its not needed or useful but to teach badly mathematics to people. It use must be avoided completely – Masacroso Jan 11 '21 at 23:18
  • It is simply an anti-derivative, and the notation $\int f(x),\mathrm dx$ is justified by the 1st fundamental theorem of Calculus. – Bernard Jan 11 '21 at 23:19
  • @KaviRamaMurthy So is the first possible definition I gave, $\int f(x) , dx = {F(x):F'=f }$, satisfactory? – Joe Jan 11 '21 at 23:20
  • Yes, any element of your set is called 'an indefinite integral' and there is no such thing as 'the indefinite integral'. – Kavi Rama Murthy Jan 11 '21 at 23:22
  • If one wants something definite, one needs to specify a basepoint, e.g., $F(x)=\int_a^x f(x'),dx'$. (I sometimes notate that as $F(x)=\int_0 f(x),dx$ to avoid introducing an auxiliary variable.) – Semiclassical Jan 11 '21 at 23:27
  • @Semiclassical it wont work for functions with disconnected domain, like $f(x):=1/x$. The use of indefinite integral must be avoided completely, if you use it you will go to mathematical hell. – Masacroso Jan 11 '21 at 23:30
  • 1
    See here. – Noah Schweber Jan 11 '21 at 23:31
  • 1
    @KaviRamaMurthy : In first- and second-year calculus, that's what an indefinite integral is, but beyond that, one may encounter subtleties. – Michael Hardy Jan 11 '21 at 23:55
  • @Bernard $\qquad \uparrow \qquad$ – Michael Hardy Jan 11 '21 at 23:56
  • @MichaelHardy: Are you alluding to Denjoy-Perron-Kurzweil (not Kurt Weill ;o))-Henstock integral? – Bernard Jan 12 '21 at 00:16
  • @Bernard : Actually I had in mind somewhat less exotic stuff than that. More later, maybe. – Michael Hardy Jan 12 '21 at 00:55
  • @Masacroso I don't understand why you think indefinite integration should be avoided. Could you elaborate please? – Joe Jan 12 '21 at 08:17
  • Indefinite integral is just a set of antiderivatives, differing by a constant. – sergei ivanov Jan 12 '21 at 13:18
  • 2
    @Joe there is no coventional definition for $\int f(x)\mathop{}!d x$, and any definition that you choose will be bad in some: or it just apply to functions with connected domains, or it define the set of antiderivatives of $f$, in the latter then the notation is terrible as it induces to think that you can get any antiderivative integrating, and this is false for functions with disconnected domains. In any case the notation doesn't add something useful, just confusion. If you want to talk of antiderivatives then just use that word, you dont need any problematic notation – Masacroso Jan 12 '21 at 19:26
  • @Masacroso OK, thanks for clarifying. That makes a lot of sense. I have a feeling that antiderivatives should really just be thought of as computational tool that in some cases can you help compute (definite) integrals. I don't see many uses for them other than that, in contrast to integrals which can be used to find areas and volumes, among other things. Is my understanding correct? – Joe Jan 12 '21 at 19:43
  • @Joe antiderivatives are useful because they have all the information about area or volume packed in a function. They are important in differential geometry and other branches of mathematics – Masacroso Jan 12 '21 at 19:57
  • @Masacroso Right, so antiderivatives are useful; it's just the notation we use is very misleading. – Joe Jan 12 '21 at 19:58
  • @Joe the point is that you dont need a notation to say that $F$ is an antiderivative of $f$. An antiderivative of a function $f: A\to \mathbb{R}$, for some $A\subset \mathbb{R}$, is a function $F:A\to \mathbb{R}$ such that $F'=f$. Of course, when $A$ is an interval, you can show that all derivatives of $f$ are of the form $F(x)=C+\int_{a}^x f(t)\mathop{}!d t$ for some $C\in \mathbb{R}$ – Masacroso Jan 12 '21 at 20:04
  • @Masacroso Thanks again. You have been very helpful. – Joe Jan 12 '21 at 20:14

1 Answers1

5

You should stick to the definition of an indefinite integral.

Given a function $f:I\to\mathbb{R}$ defined on an open interval $I$, if $F:I\to\mathbb{R}$ is a function such that $F'(x)=f(x)$ for every $x\in I$, then we call $F$ an antiderivative of the function $f$. It is easy to check the two following two facts:

  • If $F$ is an antiderivative $f$ on the interval $I$, then so is $F+C$ for any constant $C$;

  • If $F$ and $G$ are two antiderivatives of $f$, i.e, $F'(x)=G'(x)=f(x)$ for all $x\in I$, then there exists (exercise!) a constant $C$ such that $F=G+C$.

We define the notion of the "indefinite integral" of the function $f$ (on the interval $I$) as a family of functions:

$$ \textrm{indefinite integral of } f=\{F:I\to\mathbb{R}\mid F'(x)=f(x)\ \textrm{for all } x\in I\} $$

Because of these two facts above, we write $$ \int f(x)\,dx=F(x)+C $$ for any antiderivative $F$ of $f$ (on the interval $I$).

amWhy
  • 209,954
  • 4
    the problem is that there is no canonical definition of indefinite integration, you cannot stick to something that doesn't exists. – Masacroso Jan 12 '21 at 19:27