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The formulas for integrals in the textbooks usually define indefinite integral up to a constant term. Yet the natural integration constant for antiderivative can be fixed from the following formula involving Fourier transform making the antiderivative no less definite than derivative:

$$f^{(-1)}(0)=\frac{i}{2\pi}\int_{-\infty}^{+\infty} \frac{1}{\omega} \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega $$

The following is the list of antiderivatives derived using this formula and its generalizations. Particularly, for any even function, its antiderivative is odd, thus if exists in $x=0$, its value is 0 there.

$$(a^x)^{(-1)}=\frac{a^x}{\ln a};\qquad f^{(-1)}(0)=\frac{1}{\ln a}$$ $$(\sin ax)^{(-1)}=-\frac1a \cos ax;\qquad f^{(-1)}(0)=-\frac{1}{a}$$ $$(\cos ax)^{(-1)}=\frac1a \sin ax;\qquad f^{(-1)}(0)=0$$ $$(\sinh ax)^{(-1)}=\frac 1a \cosh ax;\qquad f^{(-1)}(0)=\frac{1}{a}$$ $$(\cosh ax)^{(-1)}=\frac 1a \sinh ax;\qquad f^{(-1)}(0)=0$$ $$(\sin^3 ax)^{(-1)}=\frac{\cos (3 a x)-9 (\cos (a x))}{12 a};\qquad f^{(-1)}(0)=-\frac2{3a}$$ $$(x e^{-x^2})^{(-1)}=-\frac{e^{-x^2}}{2};\qquad f^{(-1)}(0)=-\frac12$$

etc.

So why the antiderivatives are given with arbitrary constants rather than these distinguished ones?

Anixx
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    Does that constant prescription method give constants consistently for $\int f'g=fg-\int fg'$? If $g(x)=f(x+a)$, do we have $g^{(-1)}(x)=f^{(-1)}(x+a)$? – Joonas Ilmavirta Aug 23 '14 at 08:46
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    Are they really "distinguished" to a calculus student? The major application of finding anti-derivatives in a standard calculus text is to calculate definite integrals. In this application, they all work the same. Moreover, your definition uses mathematics that is particularly difficult to define to calculus students (i.e. complex exponentiation). – PVAL Aug 23 '14 at 08:46
  • @Joonas Ilmavirta yes – Anixx Aug 23 '14 at 08:47
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    I agree with PVAL: the definition is so complicated that bringing it up in calculus does more harm than good. The definitions seems to have two nested integrals neither of which converges absolutely, so it would be a great source of confusion. And keeping a general constant emphasizes that differentiation can only be inverted up to a constant. – Joonas Ilmavirta Aug 23 '14 at 08:53
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    In what sense is the proposed "definition" a definition? For example, if $f(x)=1$? And what properties can then be proved for it? – Lutz Mattner Aug 23 '14 at 09:02
  • @Joonas Ilmavirta but they could just include it in the tables, saying "we already calculated it for you". I also did not see any computer algebra system that would strictly produce integrals with the natural constant. – Anixx Aug 23 '14 at 09:13
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    @Lutz Mattner in that case the calculation involves Dirac Delta function and comes (up to a factor) to $\int_{-\infty}^{+\infty}\frac{\delta(x)}{x}dx$ Since Dirac Delta is even, it being devided by $x$ would be odd so one can reasonably assume the symmetric integral of an odd function to be zero. So $(1)^{(-1)}=x$. This also comes to zero for all natural powers of $x$. – Anixx Aug 23 '14 at 09:16
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    @Anixx, I don't think saying "we did for you" is good. The students should be able to reproduce or verify any integration result given to them in a calculus class or book. To me at least (as a student) those natural constants would have seemed quite magical, and I would have been dissatisfied with the lack of explanation. Calculus (and mathematics at large) looks quite arbitrary to many students as is, and I wouldn't want to add to it without a clear gain. – Joonas Ilmavirta Aug 23 '14 at 09:24
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    Why on earth would anyone want to do this? It solves no problem, is confusing, and is entirely arbitrary (I could write down any number of other conventions that are equally "natural" and give different answers). No offense, but this is a totally absurd suggestion. It's also not really a research-level math question, so I have voted to close. – Andy Putman Aug 23 '14 at 14:19
  • @Andy Putman it is not arbitrary. With if exponent satisfies $f(x)^{(-1)}=f(x)$ and sine satisfies $f(x)^{(-4)}=f(x)$, for instance. – Anixx Aug 23 '14 at 14:37
  • @Anixx: Your question is not research-level, so it's not appropriate here. Please post a question on math.stackexchange, and I suggest simply asking there why undetermined constants of integration are important in math, without offering your proposal by Fourier transforms as part of the question. – KConrad Aug 23 '14 at 14:44
  • @KConrad the Fourier transform is what makes it non-elementary, but I concede that this question may be too opinion-based. I will try to pose a more concrete question. – Anixx Aug 23 '14 at 14:46
  • Oh, rats! I think this question has a mathematically interesting answer, and I wanted to post it last night, but now I can't because the question is on hold. I'll post a short version here in the comments, and maybe it can be turned into a proper answer if the question is reopened. – Vectornaut Aug 23 '14 at 17:13
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    @Anixx, I think these "natural integration constants" come from a surprising fact about differentiation: the derivative operator on $L^1(\mathbb{R})$ is injective! If $f$ is an $L^1$ function, it has at most one $L^1$ antiderivative; if that antiderivative exists and is regular enough, your formula for $f^{(-1)}(0)$ gives its value at zero. The reason this natural choice of antiderivative doesn't show up in intro calculus classes is that these classes take place not in $L^1(\mathbb{R})$, but in $C^1(\mathbb{R})$, where the derivative operator is not injective. – Vectornaut Aug 23 '14 at 17:30
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    @Vectornaut: in more concrete terms, what you're saying is that the only constant function in $L^1({\mathbf R})$ is $0$. Of course there are many reasons people might be interested in functions defined on closed and bounded intervals, where the result you describe breaks down. – KConrad Aug 23 '14 at 17:34
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    @KConrad: Yes, but the OP appears to be interested in functions on $\mathbb{R}$. More generally, I think that the question the OP really wanted to ask is a question I also had when I first saw the Fourier transform: "Wait a minute! If differentiation in Fourier-land is multiplication by $\omega$, then the derivative operator is injective! But I learned in intro calculus that the derivative operator is not injective! What's going on?" – Vectornaut Aug 23 '14 at 17:44
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    @KConrad: Of course, the OP knows best what kind of answer they're actually looking for, but if they do want one along the lines I described above, maybe the negative reactions to this question could be ameliorated by removing the references to calculus textbooks, and making it clear that this is a question about the existence of a "natural antiderivative" in Fourier analysis. – Vectornaut Aug 23 '14 at 17:45
  • @Vectornaut it seems people here tend to close questions they dont know how to answer. One of my questions was similarly closed, then another user said he has an answer and now the question has 34 upvotes. – Anixx Aug 23 '14 at 17:51
  • @Anixx, is my guess about the motivation for your question correct? Do you think the answer I suggested is useful? – Vectornaut Aug 23 '14 at 17:59
  • @Vectornaut yes, I think it is very useful, at least, very interesting. – Anixx Aug 23 '14 at 18:00
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    @KConrad, Anixx: Yay! In that case, how would you feel about editing the question to remove the references to calculus textbooks? Like I said, I think people are currently reacting negatively to the question because they're reading it as a question about calculus teaching, rather than a question about Fourier analysis. – Vectornaut Aug 23 '14 at 18:07
  • @Vectornaut actually I have another, non-Furier-analysis related formula that gives the same values. So the question was intended as abstract one, why these natural integration constants are disrespected in integral tables (whether in textbooks or in computer algebra systems or whatever). I think your answer may shed a light on this. – Anixx Aug 23 '14 at 18:10
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    Anixx: I'm tempted to say that if your other formula gives the same values, then it's really the Fourier-analysis-related formula in disguise, so this is still a question about Fourier analysis. Nonetheless, if what you really want is a discussion about whether integration constants are useful in general, then I suppose it's best to leave the question as it is. Ultimately, my advice is to write your question so that you get answers which are satisfying to you. If you want answers like KConrad's below, you should write it one way; if you want answers like mine, you should write it another way. – Vectornaut Aug 23 '14 at 18:23
  • @Vectornaut the other formula is here: http://mathoverflow.net/questions/179202/prove-that-these-two-definitions-of-natural-integration-constant-coincide-when It would be wonderful to see any connection of that formula with Furier analysis :-) I would welcome the answers like yours. I voted to reopen. – Anixx Aug 23 '14 at 18:27

1 Answers1

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A very important role for the undetermined constants in indefinite integrals, in fact perhaps their first really essential role, is in solving basic differential equations. The constants of integration are used to find the solution satisfying the initial conditions for the differential equation. Changing the initial conditions will change the solution, which mathematically corresponds to changing the undetermined constants in indefinite integration. You'd make the whole subject of differential equations more awkward by avoiding the undetermined constants. In fact, it's hard to imagine how anyone could teach or learn basic differential equations without those undetermined constants from integration. (Could the OP write about Fourier transforms but not have studied differential equations?)

It is quite misleading to avoid facing the plain fact that on an interval of the real line, $f'(x) = g'(x) \Longleftrightarrow f(x) = g(x) + C$ for some constant $C$, or even more basically $f'(x) = 0 \Longleftrightarrow f(x) = C$. These undetermined constants in integration are an essential feature of differentiation, just as much as the fact that a system of linear equations $A{\mathbf x} = {\mathbf 0}$ can have a nonzero solution (or, in the language of abstract algebra, that homomorphisms can have nontrivial kernels).

An $n$th order constant coefficient linear differential equation generally has an $n$-dimensional solution space (e.g., $y'' + y = 0$ has solution space $\{a\sin x + b\cos x : a, b \in {\mathbf R}\}$, which is important in physics). These $n$ dimensions, intuitively, come from integrating $n$ times to pass from the differential equation back to its solutions, because each integration introduces an undetermined constant, so an $n$th order differential equation will have $n$ undetermined constants for its solutions (hence an $n$-dimensional solution space). If you want to have an intuition for higher-order differential equations then you want to have the language of undetermined constants available.

Finally, it seems to me that the OP is suggesting (indirectly) that all antiderivatives be fixed by specifying their value at $x = 0$, but the value of most functions at $0$ are no more special or important in general than their value at $x = 1$ or at other numbers, so specifying an indefinite integral by its value at 0 is not in general going to make anything simpler for the purpose of applications. And what would you do for functions like $f(x) = 1/x$, which aren't even defined at $x = 0$? If you want to discuss antiderivatives of functions on an interval that does not contain $0$, it doesn't make sense to specify an antiderivative in terms of its value at $0$.

KConrad
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  • "evaluation at x=0 is no more special or important in general than evaluation at x=1" - This formula provides the way for evaluating the constant at any point. There is also a general formula for antiderivative: $$f^{(-1)}(x)=\frac{i}{2\pi}\int_{-\infty}^{+\infty} \frac{e^{\omega x}}{\omega} \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt , d\omega $$, you can take x to be 1 or any other so to fix the constant, say, if you have a pole in zero. – Anixx Aug 23 '14 at 13:45
  • "Finally, the OP is suggesting that all antiderivatives be fixed by specifying their value at x=0," - no, I did not suggest so. See the formula above, you can take any x. " And what would you do for functions like f(x)=1/x" - for instance shift them by 1, find the antiderivative and shift back. – Anixx Aug 23 '14 at 13:57
  • Giving formula for $x=0$ does not impair generality in any way because one can easily using this formula and shift derive the constant for a function that is not defined in zero. I wonder how you do not see it. – Anixx Aug 23 '14 at 14:00
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    If you're allowing $x$ to be 0 or 1 or "any other" number to fix the constant, then you really are not fixing the constant! Besides, it seems all that you are after here is a "theory of fixing a constant", which is not their purpose in math. What is any actual advantage in math to specifying only one choice of a constant in integration? There are real uses of these undetermined constants, as in the theory of differential equations, and to avoid having undetermined constants sure seems to me like it hurts far more than it helps. – KConrad Aug 23 '14 at 14:03
  • For instance to fix the constant for the antiderivative of function $1/x$ one can evaluate the antiderivative of $1/(x-1)$ and see that it would be 0 in x=0. So antiderivative of $1/x$ would be 0 in x=1. – Anixx Aug 23 '14 at 14:03
  • "then you really are not fixing the constant!" - why? If you fix the constant at one x, it gets fixed for the whole function. – Anixx Aug 23 '14 at 14:04
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    Let $f(x) = 2x+3$. The antiderivative vanishing at $x = 0$ is $x^2 + 3x$. The antiderivative vanishing at $x = 1$ is $x^2 + 3x - 4$. The antiderivative vanishing at $x = a$ changes as you change $a$. If you know an antiderivative takes a particular value at a particular point then that antiderivative is determined elsewhere if it is defined on an interval, but changing the point of evaluation or the preferred value will change the antiderivative by a constant, which is the entire issue you are trying to avoid in the first place, so it isn't going away after all. Your strategy is ill-defined. – KConrad Aug 23 '14 at 14:10
  • I've already answered your original equation by pointing out the very important role of undetermined constants in differential equations. If you want to make further comments on my answer, please address that part of it. – KConrad Aug 23 '14 at 14:20
  • I do not get you. Why do you want the antiderivative vanishing? For this function the antiderivative vanishes only in x=0, in x=1 it is 4, no matter what method you employ. Look at the formula in the first reply. The antiderivative is defined in any point and does not differ whether you use one x ar another one to fix the constant. Do youy see it? – Anixx Aug 23 '14 at 14:26
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    I asked you to address the role of the undetermined integration constants in solving differential equations if you want to make further comments on my answer, since you are ignoring it. Does that application not provide an explanation to you for why we want the flexibility of the undetermined constants in integration? Differential equations are the first topic in analysis where those undetermined constants are fundamentally important (they certainly are not important for computing areas and volumes). – KConrad Aug 23 '14 at 14:31
  • My question was about tables of integrals. You can put there integrals in arbitrary form or with properly fixed constant (so that consecutive integration of sine would return to it again) thus it to satisfy $f^{(-4)}(x)=f(x)$. With arbitrary constants sine does not satisfy this equation and exponent does not satisfy $f(x)^{(-1)}=f(x)$. – Anixx Aug 23 '14 at 14:33
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    What a bizarre idea. Of course, no one need integrals of piece-wise defined functions, or functions defined only on a finite interval. – Oleg Eroshkin Aug 23 '14 at 14:37
  • Кстати, по-английски фраза «в точке» = at a point, не in a point (напр., "at x = 0" вместо "in x = 0"). – KConrad Aug 23 '14 at 14:38
  • @OlegEroshkin: why would "no one need" integrals of functions defined only a finite interval? Studying differential or integral operators on intervals $[a,b]$ is a basic topic in functional analysis, for instance. Or does the term "finite interval" mean something to you other than closed bounded intervals? – KConrad Aug 23 '14 at 14:41
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    @KConrad Sorry, that was a sarcastic remark. I found this construction absolutely absurd. Require functions to be defined on entire real line just to avoid choice of constant. Never mind that the "definition" make sense only for "nice" functions. – Oleg Eroshkin Aug 23 '14 at 14:46
  • @Oleg Eroshkin what do mean under "nice"? The definition can be equally well be used for piecewise-defined functions (where the Fourier transform is appropriate). It is even applicable to Dirac Delta's integration (its antiderivative would have 0 in zero and $\pm 1/2$ elsewhere). – Anixx Aug 23 '14 at 14:49
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    @Anixx And what it gives for $xe^{x^2}$? If function is defined only on an interval, does the constant depends on how you extend the function on entire line? – Oleg Eroshkin Aug 23 '14 at 15:04
  • @Oleg Eroshkin second - yes. First - I do not know the Fourier transform for that function, and I even did not claim the formula always converges. But there are other generalizations of the same idea, I will post a related question soon. – Anixx Aug 23 '14 at 15:08
  • @Oleg Eroshkin and regarding your example function, the value of antiderivative in x=0 would be 1/2. This can be easily obtained from the value for the function $x e^{-x^2}$ from the table in the question by noticing that $-x^2=(i x)^2$ and making a substitution $t=ix$. – Anixx Aug 23 '14 at 15:17
  • @KConrad can you remove the last paragraph from your answer as I already pointed that you wrong in attributing me the opinions you're citing there and that even it that were true, it would not impair the generality. – Anixx Aug 23 '14 at 15:33
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    I don't want to remove that paragraph since then some of the comments on my answer will no longer make sense. I have edited the start of that paragraph to make it clearer that it is my interpretation of what you are essentially doing. In any case, I do wish you had at least once addressed the part of my answer dealing with the application of undetermined integration constants in the solution to differential equations rather than ignore it so completely. – KConrad Aug 23 '14 at 15:40
  • Can you please remove the paragraph entirely because I already responded to you what to do with 1/x and showed you with 5-th grade school considerations that sticking to x=0 does not affect the generality. Regarding differential equations, I have nothing to say because my question did not mention them in any context. – Anixx Aug 23 '14 at 15:43
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    You didn't show how your Fourier method leads to an antiderivative for $1/x$ (what is it?). Concerning the role of differential equations, your question asks for a reason that "antiderivatives are given with arbitrary constants rather than these distinguished ones" and differential equations provide a context that answers that question. Have you solved differential equations and used the undetermined constants to find the unique solution fitting some initial conditions, or used the general constants to find a formula for the general solution of the equation? – KConrad Aug 23 '14 at 17:39
  • equation $x^2=2$ has two solutions. This does not prevent us from understanding only one number as $\sqrt{2}$. Regarding $1/x$ the matter is complicated. It can turn out that there is no natural integration constant. The Furier method implemented in Mathematica gives $\ln |x|+\gamma$ as the natural integral. – Anixx Sep 01 '14 at 12:58