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.


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 $$[x^2]=\{x^2+C: C\in \mathbb{R}\}.$$

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, $\overline{0}$ denotes the zero function, while $0$ denotes the zero scalar. (similarly $\overline{1}$ denotes the identically-one function). This notation is just for clarity, to emphasize the difference between the two zero symbols.

| cite | improve this answer | |
  • 1
    $\begingroup$ This is especially nicely put! $\endgroup$ – Travis Willse Dec 15 '14 at 21:16
  • 2
    $\begingroup$ It's a very nice answer.+1 $\endgroup$ – mfl Dec 15 '14 at 21:18
  • 2
    $\begingroup$ 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 '14 at 21:27
  • 5
    $\begingroup$ @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. $\endgroup$ – hmakholm left over Monica Dec 15 '14 at 21:35
  • 3
    $\begingroup$ @Rahul It's not even so much that it's defined that way, as that the vector space axioms force it to be so. $\endgroup$ – Steven Stadnicki Dec 16 '14 at 2:04

$\int f(x)dx$ 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.

| cite | improve this answer | |

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 $$ \int f(x) dx = \{ F(x) + C : C \in \mathbb{R} \}\\ \text{for a particular $F(x)$}. $$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.