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In this post, Qiaochu Yuan remarks that 'it is convenient but misleading to write $$ \int f(x) \, dx=g(x) $$ [where the derivative of $g$ is $f$]'. This sentiment seems to be shared by many contributors here, and I don't understand why. To me, both definite and indefinite integration are both valid operations you can perform on a function, and there is nothing suspect about indefinite integration.

I know about the fundamental theorem of calculus, which (as far as I understand) explains the link between indefinite and definite integration. If by integration we mean computing the area under the graph, the fundamental theorem of calculus shows us that integration is the opposite of differentiation, since $$ \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) $$ This shows that every continuous function has an antiderivative. Since a clear link between integration and antidifferentiation has been established, we give the antiderivative the convenient label 'indefinite integral'. (This also explains why the definite and indefinite integration notations are so similar.) This label is fine, so long as we remember that integration is defined as finding the area under the graph, while antidifferentiation is defined as finding the inverse of the derivative.

Another result of the fundamental theorem of calculus is that $$ \int_{a}^{x}f(t) \, dt=\int f(x) \, dx $$ So obviously every indefinite integral can be rewritten in terms of definite integrals, but I don't understand the motivation behind this. If $F$ is an antiderivative of $f$, then why is it more correct to write $$ \int_{a}^{x} f(t) \, dt = F(x) \, , $$ compared to $$ \int f(x) \, dx = F(x) \, ? $$

Joe
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    I don't imagine that anyone is questioning the validity of integration. But writing the indefinite integral that way $A$. obscures the nature of $x$ as a dummy variable for integration and $B.$ incorrectly suggests that there is only one function $f$ that might be used there. Would you, say, write both $\int x,dx = \frac {x^2}2$ and $\int x,dx = \frac {x^2}2+1$ ? I assume you see the problem with that. Much better to write, e.g., $\int_a^x g(x),dx = f(x)$ or, better, $\int_a^x g(t),dt = f(x)$ – lulu Oct 23 '20 at 18:06
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    Saying $\int f(x)dx = g(x)$ is equivalent to saying that $\int f(t)dt = g(x).$ Then, you are left wondering: how can this be, since there is no relationship between $x$ and the dummy variable $t$. Then, you avoid the formality of $\int_a^x f(t)dt = g(x)$ by the very narrow interpretation of $\int f(t)dt = g(x)$ as only signifying that $g'(x) = f(x).$ – user2661923 Oct 23 '20 at 18:48
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    Suppose $f(x)=\sin(x) \cos(x)$. One person might find an anti-derivative of $\frac12 \sin^2(x)$. Another might find $-\frac12 \cos^2(x)$. A third might get $-\frac14 \cos(2x)$. These are obviously different (the first is non-negative and the second non-positive) but they are equally correct (or incorrect) and they differ by constants – Henry Oct 26 '20 at 01:44
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    I like to think that you can use that notation, but then you have to be careful to realize that "$=$" means something different. You might find this answer to a previous similar question useful. – JonathanZ Oct 26 '20 at 13:09

7 Answers7

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Basically, there's a type error: "$\int f(x)\,dx$" is a perfectly meaningful thing, but that thing is not a single function - rather, it's a set of functions.

The point is that a function doesn't have a unique antiderivative. For example, ${x^2\over 2}$ is an antiderivative of $x$ (with respect to $x$ of course), but so is ${x^2\over 2}-4217$. It's not the indefinite integral which is suspect, but rather the notation we use around it - specifically, the way we use "$=$." Properly speaking, $\int f(x)dx$ refers to a set of functions.

This is generally addressed by including a constant of integration, so that we write $$\int x\,dx={x^2\over 2}+C$$ to mean "The set of antiderivatives of $x$ is the set of functions of the form ${x^2\over 2} + C$ for $C\in\mathbb{R}$."

  • That said, blindly adding a constant of integration still doesn't always fix the problem: let $f(x)=-{1\over x^2}+1$ if $x>0$ and $-{1\over x^2}-1$ if $x<0$; what's the derivative of $f$, and does $f$ have the form $-{1\over x^2}+C$ for some fixed real number $C$?
Michael Hardy
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Noah Schweber
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  • Hi Noah. Thank you for posting this answer. Could you please clarify that I understand it correctly? I'll do my best to summarise your points. If we interpret the statement $$\int f(x) , dx = F(x)$$ literally, then it is false: the LHS represents a set of functions, whereas the RHS refers to one particular function. Even if we write $$\int f(x) , dx = F(x)+C$$ we still have to interpret the RHS as referring to all possible values of $C$, rather than just one value. – Joe Oct 23 '20 at 18:49
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    @Joe Yes. And even then a single constant of integration might not be enough; see the end of my answer, or see Randall's answer. – Noah Schweber Oct 23 '20 at 18:53
  • Oh yes. I forgot about that. Also, it occurred to me that I might be abusing notation in another way. Technically, $f$ is the function, whereas $f(x)$ refers to $f$ evaluated at a particular point $x$. So, if we were being really pedantic, $$\int x^2 dx = \frac{x^3}{3}+C$$ means that the set of functions, where each function is defined on connected interval(s) by $\frac{x^3}{3}$ plus some constant, have a slope of $x^2$ at the point $x$ for all $x$. Is that right? – Joe Oct 23 '20 at 19:30
  • @Joe No - in my experience when we write "$\int f$" with no further decoration, we're looking at antiderivatives which are defined on as large a domain as possible. The set you're describing is the union of the sets of the form $\int_If(x)dx$ for $I$ a connected interval. – Noah Schweber Oct 23 '20 at 20:10
  • So "$\int x^2dx={x^3\over 3}+C$" means that $(i)$ every function of the form ${x^3\over 3}+C$ for $C\in\mathbb{R}$ is an antiderivative of $x^2$ w/r/t $x$ and is not the proper restriction of any antiderivative of $x^2$ (that last bit is vacuous here since each such function happens to be defined on all of $\mathbb{R}$), and $(ii)$ every such "maximally defined antiderivative" is of the form ${x^3\over 3}+C$ for some $C\in\mathbb{R}$. – Noah Schweber Oct 23 '20 at 20:13
  • Thanks, that clears up my confusion. So the function we are looking for has to be defined on the same domain as $f$ (or on a larger domain). Is that correct? – Joe Oct 23 '20 at 20:21
  • Isn't this the same kind of "suspect" as $\sin(x) = x + O(x^3)$? – Federico Poloni Oct 24 '20 at 16:48
  • @FedericoPoloni Yup. I hate hate hate hate HATE that notation. (hem) – Noah Schweber Oct 24 '20 at 17:10
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    I like this example calculation: $$ \int \frac1x , dx = \int 1 \cdot \frac1x , dx = x \cdot \frac1x - \int x \cdot \left( -\frac1{x^2}\right) , dx = 1 + \int \frac1x , dx $$ It looks like this gives $0=1$ after subtraction of $\int \frac1x , dx,$ but this calculation rather exemplifies that the equal signs should rather be equivalence signs; $u \equiv v$ if $u'=v'$. – md2perpe Oct 25 '20 at 01:02
  • Out of curiosity, why the downvote? – Noah Schweber Jan 12 '21 at 00:36
  • I realise that this is quite an old post now, but I'm still not exactly sure about the status of the variable $x$ in the integral $\int f(x) , dx$. I'm sorry if what I'm asking doesn't make much sense. After Qiaochu posted his answer, he said that one of the main problems with writing $\int f(x) , dx = F(x) + C$ is that $x$ is being used in two different ways—as a dummy variable, and as a free variable representing the input of a set of functions. So writing $\int x^2 , dx = \frac{x^3}{3} + C$ is misleading because... – Joe Mar 02 '21 at 18:45
  • ...the $x$ on the LHS is doing two jobs: it is a dummy variable that serves to tell us that we are trying to find the antiderivatives of the function $f: x \mapsto x^2$; but it is also the point at which we are evaluating the antiderivatives. Hence, it might be more logical to write something such as $\left(\int t^2 , dt\right)(x)=\frac{x^3}{3}$, meaning that $x$ is not being used in two different ways. I tried asking Qiaochu about this but he didn't reply. Do I understand things correctly? – Joe Mar 02 '21 at 18:45
  • @NoahSchweber: And if we wanted to define the indefinite integral formally, we could define it as the set ${F\mid F' = f}$. Then, it can be shown as a theorem that if $S$ is one antiderivative of $f$, then every antiderivative of $f$ must be of the form $F:x \mapsto S(x)+C$, where $C$ is a function that is constant over a connected interval. – Joe Mar 07 '21 at 14:36
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The other answers have made good points about constants of integration but this is not actually what I meant, although it is related. What I meant is what lulu says in the comments: writing antiderivatives this way misleads you about the relationship between the $x$ on the LHS (which is a dummy variable) and the $x$ on the RHS (which is not). The "real" $x$ on the LHS is one of the bounds of integration, which is being suppressed in the notation.

The sense in which this is misleading becomes clearer once you start considering double integrals, which is the context of the question you link to. If it makes sense to write $\int f(x) \, dx = g(x)$, then surely it also makes sense to write $\int g(x) \, dx = h(x)$, right? Then does it make sense to write

$$\iint f(x) \, dx \, dx = h(x)$$

or not? What do you think?

Qiaochu Yuan
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    Agree – the problem with the antiderivative-notion of indefinite integral wouldn't be so bad if it only were for integration on the real line, but it just doesn't scale to higher dimensions. In more advanced context, one very often considers something of the form $\int_\Omega!\mathrm{d}x:f(x)$, where $\Omega$ may be some subset of $\mathbb{R}^n$ or some manifold / Lie group or whatever – no problem for integration, but antiderivatives? No way. – leftaroundabout Oct 24 '20 at 12:32
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    Another way of looking at this might be: when we write $f(x)=g(x)$ there's an implicit quantifier that this is true for all $x$. So we have $f(1)=g(1)$, $f(7)=g(7)$, $f(\pi)=g(\pi)$, and so on. But when we write $\int x,dx = x^2/2+C$ we wouldn't then conclude that $\int 3,d3=9/2+C$... – Carmeister Oct 24 '20 at 14:59
  • Hi Qiaochu. Thank you for clarifying your comments. I have one question, though: why is the $x$ in $\int f(x) , dx$ considered a dummy variable? To me, the statement $$\int x , dx=\frac{x^2}{2}+C$$ means that the function $f$ defined by $f(x)=\frac{x^2}{2}+C$ has a slope of $x$ at the point $x$ (for all $x$). When you say 'dummy variable', do you mean that I could just as well write that the function $f$ defined by $f(y)=\frac{y^2}{2}+C$ has a slope of $y$ at the point $y$? – Joe Oct 24 '20 at 19:39
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    @Joe: the $x$ in $\int f(x) , dx$ is a dummy variable because it's been integrated over; the "real" $x$ is the upper bound of integration, which is why it's more accurate to write this integral as $\int_a^x f(t) , dt$ (and $t$ can be replaced with any other dummy variable). This issue already arises when you consider definite and indefinite sums; the partial sum $\sum_{k=0}^n f(k)$ is a function of $n$ because $n$ is the upper bound of the sum, and $k$ is a dummy variable; after it's been summed over it no longer appears. – Qiaochu Yuan Oct 24 '20 at 19:46
  • @QiaochuYuan I understand what you mean in the summation example, but I'm still a little unsure about integrals. I'm sorry to bother you, but could you go into a bit more detail about how the $x$ in $\int f(x) , dx$ is a dummy variable? I don't see how that immediately follows from the fact that the $x$ is being integrated over. – Joe Oct 24 '20 at 20:04
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    @Joe: it's exactly the same as the summation example. Maybe it's clearer if we change the "real" variable: $\int_0^y f(x) , dx$ is a function of $y$, not $x$, because $y$ is what appears in the bounds of the integral. The $x$ has completely disappeared. Yes? – Qiaochu Yuan Oct 24 '20 at 20:06
  • @QiaochuYuan Yes, that makes sense. But I don't understand how this makes the other notation inaccurate. If we write $$\int x , dx = \frac{x^2}{2}+C , \text{,}$$ then the LHS does not appear to contain a dummy variable, and I don't understand why it should contain one. Doesn't this notation correctly suggest that the slope of $f$, defined by $f(x)=\frac{x^2}{2}+C$ has a slope of $x$ at the point $x$? Here, $x$ seems to mean a genuine point on the graph of $y=\frac{x^2}{2}+C$—it's just that it's arbitrary which point we choose. And so to me it doesn't seem like $x$ is a dummy variable. – Joe Oct 24 '20 at 20:18
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    @Joe: please pay attention to the words I'm using; you've claimed I said indefinite integration is "suspect" or "inaccurate" and that's not what I said, I said "misleading" and I meant "misleading." It's misleading precisely because the LHS does not appear to contain a dummy variable, but in fact it does. The LHS is shorthand for an integral of the form $\int_a^x t , dt$ where $a$ is a constant and $t$ can be replaced by any other dummy variable. People implicitly understand this and it's mostly fine for single integrals but gets confusing for double integrals. – Qiaochu Yuan Oct 24 '20 at 20:28
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    And double integrals were the context for the question you linked to; you haven't addressed that part of my response at all. – Qiaochu Yuan Oct 24 '20 at 20:29
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    @Joe: A helpful way of seeing whether a variable is a “dummy” is asking whether it would make sense to plug in a value for it. Does it make any sense to say “take $\int x,dx$, then plug in $x=3$, to get $\int 3, d3$”? No! If you’re thinking of $\int x, dx$ as a function (or, better, a set of functions), and you want to see what the function does at 3, you don’t plug 3 in for $x$, you put 3 in as the upper limit of integration. So $x$ is not the variable representing the input of the function. – Peter LeFanu Lumsdaine Oct 25 '20 at 09:43
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    The last point you make in the post is one of the key objections to this notation IMO. I see new student quite often do this mistake: they substitute $f(x) = \int g(x)dx$ into $\int f(x) dx$ and end up with something that makes absolutely no sense and don't see where things went wrong. So even if the expression is fine to use when communicating with people that really understands the meaning of it, it is a source of error and leads to confusion for people new to integration so "misleading" is a very appropriate characterization. – Winther Oct 25 '20 at 15:24
  • @QiaochuYuan I'm sorry, I didn't mean to misrepresent what you were saying. The reason I haven't mentioned double integrals is because I haven't studied multivariable calculus at all. So are you suggesting that we define $\int f(x) , dx$ as $\int_{a}^{x} f(t) , dt$? If so, what are the advantages of this, and is it possible to define the indefinite integral as a set of antiderivatives instead? – Joe Dec 11 '20 at 21:38
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    @Joe: it's not a matter of definitions, it's a matter of notation. The advantages are as I described: the $x$ is in the correct place, and the dummy variable is correctly treated as a dummy variable, which is less confusing for various reasons. It's the same reason we write $f(n) = \sum_{k=1}^n g(k)$ and not $f(n) = \sum_{n=1}^n g(n)$; here $k$ is the dummy variable. – Qiaochu Yuan Dec 11 '20 at 21:41
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    @QiaochuYuan OK, I think I understand now. So we want $\int f(x) , dx$ to represent a function (or, to be more precise, a function evaluated at $x$—I think). But there is no way to plug in $x=3$, for instance, into $\int f(x) , dx$. By contrast, we can write $\int_{a}^{3} f(t) , dt$ if we use the notation you suggested. Unfortunately, as someone who has never studied multivariable calculus before, I don't see much of a problem with $\iint f(x) , dx , dx = h(x)$, other than the problems in the single integral case. Are you saying that things get much more confusing with double integrals? – Joe Dec 11 '20 at 21:56
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    @QiaochuYuan And thank you, this has really helped me get a better grip on the notation surrounding integrals. – Joe Dec 11 '20 at 21:57
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    @Joe: yes, that's a good example, you can't plug a number into a dummy variable because it's already been "used up." It isn't much of a problem in the single-variable case; as I have already said, the original context of the quote you asked this question about was about a multivariable integral. If that isn't relevant to you then don't worry about it. – Qiaochu Yuan Dec 11 '20 at 22:58
  • @QiaochuYuan: Sorry to bother you again, but are the following hypothetical notations also less misleading? If $f$ is an integrable function, then we could denote its set of antiderivatives as $\left(\int f\right)(x)$ (where $\int f$ is not a function but a multifunction). Or, perhaps, $\left(\int f(t) dt \right)(x)$, where $t$ is a dummy variable, meaning that the variable $x$ is not being used in two different ways. For instance, we could write $\left(\int t^2 \right)(x) = \frac{x^3}{3}+C$ (where the $C$ notation implicitly denotes a set of antiderivatives). – Joe Feb 24 '21 at 14:55
  • If you find the time to respond to my question, then I'd be very grateful. But don't worry if not. – Joe Mar 02 '21 at 19:24
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For one example, the familiar old "formula" $$ \int \frac{1}{x} \ dx = \ln|x| + C $$ is false (unless you define the indefinite integral VERY carefully). This purports to say that any antiderivative of $f(x) = \frac{1}{x}$ must take the form $F(x) = \ln|x|+C$ for some fixed constant $C$. But this is only true over a connected interval. For example, the function $$ G(x) = \begin{cases} \ln|x| +1, & x < 0\\ \ln|x|-1, & x > 0\end{cases} $$ satisfies $G'=f$, even though it is not expressible in the form $\ln|x|+C$. Done right, we should only define indefinite integrals over intervals (this is due to the Mean Value Theorem).

Randall
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    It can be repaired by allowing $C$ to be a locally constant function and not just a constant. This is a very simple instance of de Rham cohomology, in this case the zeroth de Rham cohomology $H^0_{dR}(\mathbb{R} \setminus { 0 }, \mathbb{R}) \cong \mathbb{R}^2$. – Qiaochu Yuan Oct 24 '20 at 08:17
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    And, of course, things get even worse if you try to extend this antiderivative to the complex plane. – Michael Seifert Oct 24 '20 at 13:41
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    @QiaochuYuan Discussed in detail here. – J.G. Oct 24 '20 at 20:58
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I think one clear way to see this is that the most pedantic notation would be something like this:

$$ f(x) \in \int f'(x)\,dx $$

You just have to think of $f$ as some element in a vector space, e.g. $f \in C^1(a,b)$, the set of continuous functions with continuous first derivative. So the integral operator is a linear transformation, a map of the kind $\int_a^x: C(a, b) \rightarrow C^1(a, b)$. This makes evident that the writing $$F(x) = \int f(x)\,dx,$$ where $F'(x) = f(x)$, could easily be notation abuse, though most of the time it's not the case, as the others have pointed out. The writer knows what they're writing: they assumed $F$ as a stand-in for a whole set of functions, but this is not always clear to the reader. Another problem this brings to surface is that the integral operator $\int_a^x$ as it has been written above is ill-defined, as it should point one element $f \in C(a,b)$ to one element $g \in C^1(a,b)$, not to a whole set of them. How to define this integral operator soundly is a question above my paygrade.

Michael Hardy
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Coral Bleaching
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If we define a function $F(x)$ as the antiderivate of the function $f(x)$ then by FTC we see that: $$\int_a^xf(t)\,dt=F(x)-F(a)\neq F(x)\forall f(x)$$ The problem lies in that for a given known derivative, $F'(x)=f(x)$, there are many different equations which satisfy $F(x)$ which differ by a constant, often notated by${}+C$. This can be seen in many cases where people neglect this "constant of integration": $$\int\frac{1}{ax}\,dx=\frac{1}{a}\int\frac 1x\,dx=\frac1a\ln|x|+C_1$$ $$\int\frac1{ax}\,dx=\frac1a\ln|ax|+C_2$$ which at first seem completely different but when the correct values of $C_1,C_2$ are chosen these are the same, see?

I think the point that the post is trying to make is that while in many scenarios it is easy to define a new function as the antiderivative of another, some may find this as misleading by thinking there is only one function $g(x)$ which satisfies these conditions, when in reality it is what we call a "family" of similar functions

Michael Hardy
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Henry Lee
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Setting the math of it aside, there is a real pedagogical/linguistic problem.

New students (reasonably) think the "Definite" and "Indefinite" integral are two variations on the same phenemonenon, whereas in fact the former is the phenemonon and the latter is notation that makes the former easier to compute.

In particular, the Fundamental Theorem of Calculus looks tautological the first time students see it, since they already "know" that integrals are anti-derivatives.

hunter
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  • Do you mean that the only reason antiderivatives are useful in the first place is that they help us compute areas? If so, that would make sense: both derivatives and integrals tells us about slopes and areas, but the antiderivative, in of itself, doesn't seem to be very useful. – Joe Oct 25 '20 at 18:05
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Mark. Basically, the information provided by that book has been summarized in former answers to the original question of this conversation. If you can consult Chapter 5 of that book, you may find the main ideas:

  1. there is a distinction between "antiderivatives" and "primitives": an antiderivative is ONE solution to the equation $\int f(x) dx = g(x)$ where g(x) is the unknown solution, while the primitive is the set of all solutions to the same equation. Moreover, it is convenient to regard the equation $\int f(x) dx =g(x)$ as a differential equation f(x) = g'(x), which, formally, has nothing to do with (definite) integration, although definite integration is a very important tool in developping theoretic antidifferentiation results (such as the existence of a solution to f(x) =g'(x) when f(x) is continuous on an interval).

  2. the set of all solutions to f(x) = g'(x) is connected with the constant of integrations, and constant of integrations are realted to the shape of the domain of f(x). A basic example has been already provided: if f(x) = 1/x, then all solutions are g(x)= log|x| +C(x), where C(x) is a function constant on (-\infty,0) and constant on (0,\infty) (this type of functions are called locally constant in that book); for instance, g(x) = log|x| if x<0 and g(x) = log|x| - 5 if x>0 is one of the solutions.

  3. chapters 9, 10, 11 and 12 deal with standard functions (trigonometric, binomial differentials, etc) and there, dealing with integrations constants is the main issue. Summarizing a lot, is something else than a mechanical implementation of formulas as most of books do.

Parcly Taxel
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Antonio
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