Timeline for Why calculus textbooks do not include the natural integration constants in the tables of integrals? [closed]
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34 events
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| Sep 5, 2017 at 19:29 | review | Reopen votes | |||
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| Nov 21, 2015 at 1:08 | review | Reopen votes | |||
| Nov 21, 2015 at 3:07 | |||||
| Oct 13, 2015 at 15:00 | review | Reopen votes | |||
| Oct 13, 2015 at 15:56 | |||||
| Aug 23, 2014 at 18:27 | comment | added | Anixx | @Vectornaut the other formula is here: mathoverflow.net/questions/179202/… 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. | |
| Aug 23, 2014 at 18:23 | comment | added | Vectornaut | 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. | |
| Aug 23, 2014 at 18:10 | comment | added | Anixx | @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. | |
| Aug 23, 2014 at 18:09 | review | Reopen votes | |||
| Aug 23, 2014 at 20:43 | |||||
| Aug 23, 2014 at 18:07 | comment | added | Vectornaut | @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. | |
| Aug 23, 2014 at 18:00 | comment | added | Anixx | @Vectornaut yes, I think it is very useful, at least, very interesting. | |
| Aug 23, 2014 at 17:59 | comment | added | Vectornaut | @Anixx, is my guess about the motivation for your question correct? Do you think the answer I suggested is useful? | |
| Aug 23, 2014 at 17:51 | comment | added | Anixx | @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. | |
| Aug 23, 2014 at 17:45 | comment | added | Vectornaut | @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. | |
| Aug 23, 2014 at 17:44 | comment | added | Vectornaut | @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?" | |
| Aug 23, 2014 at 17:34 | comment | added | KConrad | @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. | |
| Aug 23, 2014 at 17:30 | comment | added | Vectornaut | @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. | |
| Aug 23, 2014 at 17:13 | comment | added | Vectornaut | 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. | |
| Aug 23, 2014 at 14:46 | comment | added | Anixx | @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. | |
| Aug 23, 2014 at 14:44 | comment | added | KConrad | @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. | |
| Aug 23, 2014 at 14:37 | comment | added | Anixx | @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. | |
| Aug 23, 2014 at 14:32 | history | closed |
Eric Wofsey abx Andy Putman Stefan Kohl♦ Lucia |
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| Aug 23, 2014 at 14:19 | comment | added | Andy Putman | 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. | |
| Aug 23, 2014 at 13:26 | answer | added | KConrad | timeline score: 12 | |
| Aug 23, 2014 at 13:12 | review | Close votes | |||
| Aug 23, 2014 at 14:32 | |||||
| Aug 23, 2014 at 9:41 | history | edited | Anixx | CC BY-SA 3.0 |
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| Aug 23, 2014 at 9:34 | history | edited | Anixx | CC BY-SA 3.0 |
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| Aug 23, 2014 at 9:24 | comment | added | Joonas Ilmavirta | @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. | |
| Aug 23, 2014 at 9:16 | comment | added | Anixx | @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 $x$. | |
| Aug 23, 2014 at 9:13 | comment | added | Anixx | @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. | |
| Aug 23, 2014 at 9:02 | comment | added | Lutz Mattner | In what sense is the proposed "definition" a definition? For example, if $f(x)=1$? And what properties can then be proved for it? | |
| Aug 23, 2014 at 8:53 | comment | added | Joonas Ilmavirta | 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. | |
| Aug 23, 2014 at 8:47 | comment | added | Anixx | @Joonas Ilmavirta yes | |
| Aug 23, 2014 at 8:46 | comment | added | PVAL | 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). | |
| Aug 23, 2014 at 8:46 | comment | added | Joonas Ilmavirta | Does that constant prescription method give constants consistently for $\int f'g=fg-\int fg'$? If $g(x)=f(x+a)$, do we have $g^{(-1)}(x)=f^{(-1)}(x+a)$? | |
| Aug 23, 2014 at 8:34 | history | asked | Anixx | CC BY-SA 3.0 |