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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?

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closed as off-topic by Eric Wofsey, abx, Andy Putman, Stefan Kohl, Lucia Aug 23 at 14:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Eric Wofsey, abx, Andy Putman, Stefan Kohl, Lucia
If this question can be reworded to fit the rules in the help center, please edit the question.

7  
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 at 8:46
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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 at 8:53
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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 at 9:24
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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 at 14:19
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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 at 17:30

1 Answer 1

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 $0$.

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3  
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 at 14:03
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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 at 14:10
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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 at 14:46
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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 at 15:04
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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 at 15:40

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