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The Indefinite Integral of Zero

  1. Nov 3, 2014 #1
    Argument A

    ##∫ 0 dx = 0x + C = C##

    Argument B

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

    Discuss.
     
  2. jcsd
  3. Nov 3, 2014 #2

    Mark44

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    The "Discuss" part makes me think this is a homework problem.
     
  4. Nov 3, 2014 #3
    It's not, I just wanted to know which argument is right and why.
     
  5. Nov 3, 2014 #4

    Mark44

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    Argument B is silly, as it factors 0 into 0 times 1, and then moves the 0 outside the integral.
     
  6. Nov 3, 2014 #5
    What about the rule ##∫a f(x) dx = a ∫ f(x) dx##?
     
  7. Nov 3, 2014 #6

    pwsnafu

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    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.
     
  8. Nov 3, 2014 #7

    Mark44

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    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.
     
  9. Nov 3, 2014 #8

    Mark44

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    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?
     
  10. Nov 3, 2014 #9

    PeroK

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    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.
     
  11. Nov 3, 2014 #10

    pwsnafu

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    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
  12. Nov 3, 2014 #11
    So the indefinite integral is basically a set of functions?
    A set of antiderivatives, more specifically.
     
    Last edited: Nov 3, 2014
  13. Nov 3, 2014 #12
    My gut tells me it's A.
     
  14. Nov 3, 2014 #13

    PeroK

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    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.
     
  15. Nov 3, 2014 #14
    And how exactly do I do things more formally?
    Does ##∫a f(x) dx = a ∫ f(x) dx## only apply for nonzero a?
     
  16. Nov 3, 2014 #15

    PeroK

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    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##
     
  17. Nov 3, 2014 #16
    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?
     
  18. Nov 3, 2014 #17

    Mark44

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    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.
     
  19. Nov 3, 2014 #18

    Stephen Tashi

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    As PeroK mentioned, such rules are not about symbols representing unique functions. The symbol [itex] \int f(x) [/itex] 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 [itex] \int x dx [/itex] represents unique function [itex] x^2/2 + 1 [/itex].
    and the rule you quoted says [itex] \int 2x [/itex] is the unique function [itex] 2( x^2/2 + 1) = x^2 + 2[/itex]. This would contradict any calculation that said [itex] \int 2 x = x^2 [/itex] or [itex] \int 2 x = x^2 + 5 [/itex] etc.
     
  20. Nov 3, 2014 #19

    mathman

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    For indefinite integrals, the arbitrary added constant is just there. All you are doing in B is omitting it.
     
  21. Nov 4, 2014 #20
    I have always tackled such problems by using two different constants of integration for the two antiderivatives, then relating them according to the two results.
     
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