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If we have an indefinite integral $g(x)+c$ of a function $f,$ can we treat $c$ (entirely independent of $x$) as being some changing value?

For example, if we want to use the parameter $c$ to classify the set of antiderivatives of $f,$ can we define $h_c(x)=g(x)+c=j(x,c)$ under the understanding that $h_c(x)$ represents the value of a different function for different $c$?

ryang
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user37577
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  • If $c$ is a second variable we have $h$ as a function of two variables, and $f$ as its partial derivative with respect to $x$. – Peter Sep 02 '22 at 13:15
  • If $c$ changed with $x$, it would be a function of $x$, i.e. $c(x)$, and if you think about it for a while, it would have $c'(x) =0$. But a function with zero derivative is constant on any interval it is defined on. – JonathanZ Sep 02 '22 at 13:19
  • @Peter I think it would, the partial derivative of $j$ at $(x,c)$ would be equal to the derivative of $h_c(x)$ which itself would be the value of the derivative function $h'_c$ at $x$ which would be the derivative of a different function $h_c$ for every $c$? the point is that $h_c$ is no particular function, or alternatively if we call it $h$, then $h$ denotes a different function for different $c$. – user37577 Sep 02 '22 at 13:38
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    "For example if we want to use it as a parameter to classify the set of antiderivatives of $f$?"... that's exactly what we use $c$ for. – 5xum Sep 02 '22 at 13:57
  • @5xum in which case, $c$ can change depending on which function of interest is? – user37577 Sep 02 '22 at 14:01
  • @Peter I don't believe you need a function with a variable as it's argument to allow that variable to change but obviously if $f(x)=c$ and $f(x)=4$ then $c=5$ and $f(x)=5$ then the definition of the constant function $f$ has changed. – user37577 Sep 02 '22 at 14:03
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    @user37577 No, $c$ doesn't "change", $c$ is a free parameter. In other words, when we say $\int f(x)dx = F(x)+c$, this is just shorthand for saying "If the the set of all antiderivatives of $f$ is the set ${x\mapsto F(x) + c| c\in\mathbb R}$", or in other words, "If $G$ is an antiderivative of $f$, then $G(x)=F(x)+c$ for some $c\in\mathbb R$. – 5xum Sep 02 '22 at 14:09
  • @5xum, I'm allowing for it's differing values by saying for 'some $c$' surely? Changing in this case just means I consider it for different assignments, $c=4$, $c=2$ etc, not that it is an argument of any function. Is your definition of 'changing' that it is like $x$, the argument of some function or something at the center of investigations for which we define limits etc? – user37577 Sep 02 '22 at 14:11
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    @user37577 Sure. For every selection of $c$, you get one possible antiderivative of $f$. And it goes the other way round, i.e. every antiderivative of $f$ can be obtained by some value of $c$. But $c$ does not "depend on $f$". I don't really know what you meant by "funciton of interest" in your comment. If you mean $f$, then no, $c$ doesn't "change depending on $f$". – 5xum Sep 02 '22 at 14:13
  • @5xum so $c$ is 'changing' (we are allowing or at least exploring for different numbers that it can take) What I meant was, if we define the set as the set of functions $f$ such that $f(x)=x+c$, then under the $c=3$ assignment we are at an element in the set where $f$ is a different function than under the $c=2$ assignment. So if I want to find the element in the set at $c=2$ I may refer to it as $f$ to allow me to substitute it, so the function denoted by $f$ may be different depending on which $c$ value is. I lazily replaced $h_c(x)$ with $f$ without noticing, sorry. – user37577 Sep 02 '22 at 14:18
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    @user37577 Oh, that's what you mean by "function of interest"! Yes, then you are correct. For example, $x+1, x+2, x+3$ are all antiderivatives of $x\mapsto 1$, however, if you say $\int 1dx = x+1+c$, then you need $c=2$ to get the function $x+3$, while if you say $\int 1dx = x+2+c$, then you need $c=1$ to get $x+3$. However, note that both solutions, $x+1$ and $x+2$ are correct, because $${x\mapsto (x+1)+c| c\in\mathbb R} = {x\mapsto (x+2)+c|c\in\mathbb R},$$ i.e., the two sets above are identical. – 5xum Sep 02 '22 at 14:25
  • @5xum Thank you. – user37577 Sep 05 '22 at 10:23

3 Answers3

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If we have an indefinite integral $g(x)+c$ of a function $f,$ can we treat $c$ as being some changing value?

can we define $h_c(x)=g(x)+c$ under the understanding that $h_c(x)$ represents the value of a different function for different $c$?

An indefinite integral is more a useful notational shorthand than a mathematically important object. It is conventionally called a family of antiderivatives; as such, if its integrand is of the form $g'$ and has an interval domain, then, literally, $$\int g'\,\mathrm dx=\{g(x)+C\mid C\in\mathbb R\},$$ where $C$ represents uncountably many values. How to understand the "indefinite integral" notation in calculus? contains several justifications for manipulating and adding indefinite integrals as actual sets by reading $\text‘=\text’$ as an equivalence relation such that writing \begin{align}x^2+C=x^2+3+C\quad&\implies0=3\\x^2+C=x^2+3+D\quad&\implies E=C-D=3\end{align} make sense.

More simply, we can frame an indefinite integral as the general representation of its integrand's antiderivatives, its specification containing one independent parameter (arbitrary constant) $C_i$ per maximal interval of its integrand's domain. Here, each instantiation of $$\int g'\,\mathrm dx=g(x)+C$$ has an unimportant value of $C.$ Even though this object is not a particular antiderivative, we manipulate it as if $C$ is merely undetermined. This seems to match your above description, and hopefully answers your question.

In the context of solving differential equations with given conditions, $C$ becomes an unknown whose value is to be determined.

ryang
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No, $c$ is not a changing value. It represents any real number.

"For example if we want to use it as a parameter to classify the set of antiderivatives of $f$?" Yes, that's exactly why we add the $+c$.

If you mean "change" as in "let's change our answer $\int f(x)dx = g(x) + c_1$ to $\int f(x)dx = g(x) + c_2$," then sure, that would work too because $c_1$ and $c_2$ are just constants. Those constants by themselves do not change, though.

Does that answer your question?

Accelerator
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  • 'Changing' is ambiguous, here when I mean 'changing' I imply that $c$ defines a different function, but is constant with respect to a particular function, if $h_c$ is a particular function the $c$ is constant, if we are allowed to consider different functions we call $h_c$ then $c$ changes, but I like to consider that as switching between 'frames' where $c$ is a constant and independent of $x$ every time. But does doing so affect the integration process? – user37577 Sep 05 '22 at 10:25
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We must keep $c$ constant during the integration. We may sometimes consider how the problem is different for different values of $c$.

An interesting example is $\int\frac{1}{x}dx$. For $x>0$ we have $$\int\frac{1}{x}dx=\log(x)+C$$ and for $x<0$ we have $$\int\frac{1}{x}dx=\log(-x)+C'$$ It is only when we allow $C$ to be complex that we can see the two forms are equivalent.

If $c$ is allowed to vary it must be independent of $x$, to ensure it does not change during the integration.

Peter
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