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Let us consider this small problem: $$ \int0\;dx = 0\cdot\int1\;dx = 0\cdot(x+c) = 0 \tag1 $$ $$ \frac{dc}{dx} = 0 \qquad\iff\qquad \int 0\;dx = c, \qquad\forall c\in\mathbb{R} \tag2 $$

These are two conflicting results. Based on this other question, Sam Dehority's answer seems to indicate: $$ \int\alpha f(x)\;dx\neq\alpha\int f(x)\;dx,\qquad\forall\alpha\in\mathbb{R} \tag3 $$

However, this clearly implies that indefinite integration is nonlinear, since a linear operator $P$ must satisfy $P(\alpha f) = \alpha Pf, \forall\alpha\in\mathbb{R}$, including $\alpha=0$. After all, a linear combination of elements of a vector space $V$ may have zero valued scalars: $f = \alpha g + \beta h, \forall\alpha,\beta\in\mathbb{R}$ and $g, h\in V$. This all seems to corroborate that zero is not excluded when it comes to possible scalars of linear operators.

To take two examples, both matrix operators in linear algebra and derivative operators are linear, even when the scalar is zero. In a matrix case for instance, let the operator $A$ operate a vector: $A\vec{x} = \vec{y}$. Now: $A(\alpha\vec{x}) = \alpha A\vec{x} = \alpha\vec{y}$. This works even for $\alpha = 0$.

Why is $(3)$ true? Can someone prove it formally? If $(3)$ is false, how do we fix $(1)$ and $(2)$? When exactly does the following equality hold (formal proof)? $$ \int\alpha f(x)\;dx = \alpha\int f(x)\;dx,\qquad\forall\alpha\in\mathbb{R} $$

I would appreciate formal answers and proofs.

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3 Answers 3

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The issue here is that the indefinite integral operator, when applied to a function, does not give a function as its output. It gives an equivalence class of functions, equivalent up to constant translation.

We usually write $\int 2x ~dx = x^2 +C$, but what we really mean is that $\int 2x ~dx = [x^2]$, where $$[x^2]=\{x^2+C | C\in \mathbb{R}\}.$$

The space of such equivalent classes also forms a vector space with vector addition and scalar multiplication defined as follows:

$$[f]+[g]=[f+g]$$ $$c[f]=[cf]$$

(Note that while addition is equivalent to point-wise adding the functions in each equivalence class, multiplication is not when the scalar is $0$: $$0[f]=\{f+\overline{C}: C\in \mathbb{R}\} =\{0+C | C\in \mathbb{R}\} = [\overline{0}]$$ $$\ne \{0(f+\overline{C}) | C\in \mathbb{R}\} = \{\overline{0}\}.$$

That latter, besides not following from the definition, is a not an equivalence class in the right form.)

Now, the apparent problem disappears: $$\int \overline{ 0} ~dx = [\overline{0}]$$ and $$0\int \overline{1} ~dx = 0[x] = [0x]=[\overline{0}]$$

In the above, $\overline{-}$ maps a scalar to the constant function returning that scalar. $\overline{0}$ denotes the zero function, while $0$ denotes the zero scalar. (Similarly $\overline{1}$ denotes the identically-one function, and $\overline{c}$ denotes the everywhere-$c$ function.) This notation is just for clarity, to emphasize the difference between the two zero symbols.

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    $\begingroup$ This is especially nicely put! $\endgroup$ Dec 15, 2014 at 21:16
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    $\begingroup$ It's a very nice answer.+1 $\endgroup$
    – mfl
    Dec 15, 2014 at 21:18
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    $\begingroup$ I want to like this answer, but I don't see how $0[x]=[0]$. Zero times any function in $[0]$ is identically zero, so the result collapses to the singleton $\{0\}$, not $\{C:C\in\mathbb R\}$. $\endgroup$
    – user856
    Dec 15, 2014 at 21:27
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    $\begingroup$ @Rahul: The set of all functions $X\to F$ modulo constant offsets is a vector space with the operations $[f]+[g]=[f+g]$ and $c[f]=[cf]$. It's easy to check that this is well-defined. Alternatively it's the quotient of the vector space of functions, by the subspace of constant functions. $\endgroup$ Dec 15, 2014 at 21:35
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    $\begingroup$ @Rahul It's not even so much that it's defined that way, as that the vector space axioms force it to be so. $\endgroup$ Dec 16, 2014 at 2:04
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$\int f(x)dx$ is the usual notation for the set of all antiderivatives of $f(x).$ Thus, if $F(x)$ is a particular derivative of $f(x)$ then

$$\int f(x)dx=\{F(x)+c:c\in\mathbb{R}\},$$

if the domain of $f$ is connected. Thus,

$$\int \alpha f(x)dx=\{\alpha F(x)+c:c\in\mathbb{R}\}$$ if $\alpha\ne 0$ and

$$\int \alpha f(x)dx=\{c:c\in\mathbb{R}\}$$ if $\alpha=0.$

You are trying to translate linearity of definite integral to indefinite integral. But a definite integral gives you a number and an indefinite integral gives you a set of functions. So, linearity must be understood in this sense, and not with particular antiderivatives.

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My "answer" should be a comment under the two posted answers, but I don't have enough reputation to post comments. There is a (very popular) mistake in both answers. Indefinite integral operator does NOT give a class of functions equal up to constant translation. Let's look at an example. Someone might evaluate integral of 1/x like this: $$ \int\frac{1}{x}dx = \begin{cases} logx + C & \text{if x > 0}\\ log(-x) + C & \text{if x < 0}\\ \end{cases} = log|x| + C \\ \text{where C} \in \mathbb{R}. $$ However, this is not an exhaustive answer. The constants for negative and positive x can differ. The exhaustive answer would be $$ \int\frac{1}{x}dx = \begin{cases} logx + C & \text{if x > 0}\\ log(-x) + D & \text{if x < 0}\\ \end{cases},\\ \text{where C,D} \in \mathbb{R}. $$ Thus it is not true to say that $$ \int f(x) dx = \{ F(x) + C : C \in \mathbb{R} \}\\ \text{for a particular $F(x)$}. $$

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  • $\begingroup$ What goes wrong here is that the domain of the function you integrate here is not connected. As long as you assume that all domains are connected, everything is ok. $\endgroup$ Apr 22 at 6:23

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