Timeline for Why calculus textbooks do not include the natural integration constants in the tables of integrals?
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29 events
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| Sep 1, 2014 at 12:58 | comment | added | Anixx | 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. | |
| Aug 23, 2014 at 17:39 | comment | added | KConrad | 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? | |
| Aug 23, 2014 at 15:45 | review | Suggested edits | |||
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| Aug 23, 2014 at 15:43 | comment | added | Anixx | 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. | |
| Aug 23, 2014 at 15:40 | comment | added | KConrad | 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. | |
| Aug 23, 2014 at 15:38 | history | edited | KConrad | CC BY-SA 3.0 |
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| Aug 23, 2014 at 15:33 | comment | added | Anixx | @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. | |
| Aug 23, 2014 at 15:17 | comment | added | Anixx | @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 $t=ix$. | |
| Aug 23, 2014 at 15:08 | comment | added | Anixx | @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. | |
| Aug 23, 2014 at 15:04 | comment | added | Oleg Eroshkin | @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? | |
| Aug 23, 2014 at 14:49 | comment | added | Anixx | @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). | |
| Aug 23, 2014 at 14:46 | comment | added | Oleg Eroshkin | @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. | |
| Aug 23, 2014 at 14:41 | comment | added | KConrad | @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? | |
| Aug 23, 2014 at 14:38 | comment | added | KConrad | Кстати, по-английски фраза «в точке» = at a point, не in a point (напр., "at x = 0" вместо "in x = 0"). | |
| Aug 23, 2014 at 14:37 | comment | added | Oleg Eroshkin | What a bizarre idea. Of course, no one need integrals of piece-wise defined functions, or functions defined only on a finite interval. | |
| Aug 23, 2014 at 14:33 | comment | added | Anixx | 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 $f(x)^{(-1)}=f(x)$. | |
| Aug 23, 2014 at 14:31 | comment | added | KConrad | 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). | |
| Aug 23, 2014 at 14:26 | comment | added | Anixx | 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? | |
| Aug 23, 2014 at 14:20 | comment | added | KConrad | 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. | |
| Aug 23, 2014 at 14:10 | comment | added | KConrad | 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. | |
| Aug 23, 2014 at 14:04 | comment | added | Anixx | "then you really are not fixing the constant!" - why? If you fix the constant at one x, it gets fixed for the whole function. | |
| Aug 23, 2014 at 14:03 | comment | added | Anixx | 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. | |
| Aug 23, 2014 at 14:03 | comment | added | KConrad | 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. | |
| Aug 23, 2014 at 14:00 | comment | added | Anixx | 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. | |
| Aug 23, 2014 at 13:57 | comment | added | Anixx | "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. | |
| Aug 23, 2014 at 13:54 | history | edited | KConrad | CC BY-SA 3.0 |
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| Aug 23, 2014 at 13:45 | comment | added | Anixx | "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. | |
| Aug 23, 2014 at 13:41 | history | edited | KConrad | CC BY-SA 3.0 |
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| Aug 23, 2014 at 13:26 | history | answered | KConrad | CC BY-SA 3.0 |