# The Indefinite Integral of Zero

1. Nov 3, 2014

Argument A

$∫ 0 dx = 0x + C = C$

Argument B

$∫ 0 dx = ∫ (0)(1) dx = 0 ∫ 1 dx = 0(x+C) = 0$

Discuss.

2. Nov 3, 2014

### Staff: Mentor

The "Discuss" part makes me think this is a homework problem.

3. Nov 3, 2014

It's not, I just wanted to know which argument is right and why.

4. Nov 3, 2014

### Staff: Mentor

Argument B is silly, as it factors 0 into 0 times 1, and then moves the 0 outside the integral.

5. Nov 3, 2014

What about the rule $∫a f(x) dx = a ∫ f(x) dx$?

6. Nov 3, 2014

### pwsnafu

The indefinite integral does not calculate a function. It calculates an equivalence class of function. All constant functions are in the same equivalence class.

Further, $\int 0 \, dx$ being constant is wrong. If for example the domain is $\mathbb{R}\setminus\{0\}$ then you would have two constants, not necessarily equal.

7. Nov 3, 2014

### Staff: Mentor

There's no need to write 0x in the above.
$\int 0 dx = C$
$\int 0 dx = 0\int dx = 0x + C = C$
You don't get rid of the arbitrary constant as you showed above.

Since d/dx(C) = 0, then any antiderivative of 0 is C, for some arbitrary constant.

8. Nov 3, 2014

### Staff: Mentor

Can you elaborate on this? I'm not following you. How does the domain enter into this problem, given that we're working with indefinite integrals?

9. Nov 3, 2014

### PeroK

For what it's worth, I think this is an interesting question. What do you think? A or B?

Hint: one answer is clearly wrong, but it's not so easy to explain exactly why.

10. Nov 3, 2014

### pwsnafu

This is not easy to see when you have $f(x) = 0$, so the example I give my students is
$f : \mathbb{R}\setminus\{0\} \rightarrow \mathbb{R}$ where $f(x) = -x^{-2}$ and
$g: (0, \infty) \rightarrow \mathbb{R}$ where $g(x) = -x^{-2}$.
Then the function
$h(x) = x^{-1} + 1$ for $x > 0$ and $h(x) = x^{-1} - 1$ for $x < 0$ is an antiderivative of f. But it isn't expressible as $F(x) + C$ where C is an "arbitrary constant". The "constant" changes as you pass over the "hole". On the other hand, h isn't an antiderivative of g (because the domains don't match), but the restriction of h to $(0,\infty)$ is an antiderivative of g and expressible in the form $G(x) + C$.

The difference lies in the number of connected components of the domain. The former has two so there are two constants of integration, one for x<0 and another for x>0. The latter only has one connected component, and there is one constant of integration. In general $\int f(x) \, dx = F(x) + H(x)$ where $F' = f$ and $H$ is constant on individual connected components of the domain of f.*

Writing $f(x) = 0$ is ambiguous. It most likely has domain of all of reals. But it could be any subset of R. OP did not specifiy the domain, hence he cannot make the conclusion that the constant of integration is indeed a constant value always.

*Edit: I'm ignoring functions like Cantor's functions here.

Last edited: Nov 3, 2014
11. Nov 3, 2014

So the indefinite integral is basically a set of functions?
A set of antiderivatives, more specifically.

Last edited: Nov 3, 2014
12. Nov 3, 2014

My gut tells me it's A.

13. Nov 3, 2014

### PeroK

Your head should tell you that as well. In general, you have to be careful with the constant of integration. It's an informal way of specifying an equivalence class of functions (as pointed out above). So, that's where the multiplication by 0 breaks down. If you do things more formally, you still have an equivalence class after multiplication by 0 - so you don't get rid of the constant of integration.

And, yes, an indefinite integral is actually a set of functions.

14. Nov 3, 2014

And how exactly do I do things more formally?
Does $∫a f(x) dx = a ∫ f(x) dx$ only apply for nonzero a?

15. Nov 3, 2014

### PeroK

No. You work with equivalence classes of functions, not with a constant of integration.

An analogy would be $0 \cdot x(mod \ n) = 0 (mod \ n) ≠ 0$

16. Nov 3, 2014

So what you're saying is I should treat the indefinite integral as a set of functions ${F(x) + C | C ∈ ℝ}$, but not as a single function?

17. Nov 3, 2014

### Staff: Mentor

Yes.

If a function f has two distinct antiderivatives F1 and F2 (IOW F1] = f and F2] = f) then F1(x) - F2(x) ≡ C.

An example in the same vein is this:
$$\int sin(x)cos(x)dx$$
One student uses substitution to evaluate this integral, using u = sin(x), so du = cos(x)dx.
Then integral becomes $\int udu = (1/2)u^2 = (1/2)sin^2(x)$.

Another student also uses substitution, but with u = cos(x), du = -sin(x)dx
With this substitution, the integral becomes $-\int udu = -(1/2)u^2 = -(1/2)cos^2(x)$.

(Notice that both students omitted the constant of integration.)

Here we have two distinct antiderivatives for the same integrand. As it turns out, the two antiderivatives differ by a constant: (1/2)sin2(x) = -(1/2)cos2(x) + 1/2, independent of the value of x.

18. Nov 3, 2014

### Stephen Tashi

As PeroK mentioned, such rules are not about symbols representing unique functions. The symbol $\int f(x)$ does not represent a unique function. Such rules don't work if you interpret the anti-derivative symbol to be a unique function.

For example If suppose we say that $\int x dx$ represents unique function $x^2/2 + 1$.
and the rule you quoted says $\int 2x$ is the unique function $2( x^2/2 + 1) = x^2 + 2$. This would contradict any calculation that said $\int 2 x = x^2$ or $\int 2 x = x^2 + 5$ etc.

19. Nov 3, 2014

### mathman

For indefinite integrals, the arbitrary added constant is just there. All you are doing in B is omitting it.

20. Nov 4, 2014