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Insights Some Misconceptions about Indefinite Integrals - Comments

  1. Jul 3, 2016 #1
  2. jcsd
  3. Jul 6, 2016 #2
    Great Insight MM!
     
  4. Jul 11, 2016 #3

    Ssnow

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    Gold Member

    very interesting!
     
  5. Jul 20, 2016 #4
  6. Aug 25, 2016 #5
    The first paradox was very cool but where did the equation come from that the second paradox starts with?
     
  7. Aug 25, 2016 #6
    When we did integrals, we didn't look "backward" and consider the integral to be "interpreted as all the functions...". Instead, it was interpreted as :the area under the curve of the specified function; in this case, the function is 1/x - in which case, the RHS is correct.
    Going back the other way doesn't work, b/c the integral doesn't actually represent "all the functions" - just the specified one. And you can't take the derivative of a discontinuous function (or a "not smooth" function) as represented by your RHS?
     
  8. Aug 25, 2016 #7
    That is indeed true for a definite integral, but this insight is considering indefinite integrals, which deals with the problem of finding antiderivatives.
     
  9. Sep 10, 2016 #8
    Can someone explain where the equation comes from that the second paradox starts with? The one where the integrals on both sides cancel leaving 0 = 1?
     
  10. Sep 13, 2016 #9

    Dale

    Staff: Mentor

    Moderator's note: a fairly long series of posts based on a (now resolved) confusion have been removed.
     
  11. Sep 17, 2016 #10
    Good information, this.
     
  12. Sep 24, 2016 #11
    If you pick two separate constants for the 1/x integral, the formula won't work in the complex plane. log(x)+C continues to work in the complex plane as long as it is the same (single) value of C over all branches.

    For the second case, the paradox arises from the habit of leaving out the indefinite constants until the very end. Ordinarily, we work ahead without including the arbitrary constants which arise from each new integral, gumming them all up in the final +C we add at the very end. When the equation here finally reads 0=1, we have to remember that it should really read 0=1+C, where the single C is actually a combination of several arbitrary constants.

    If you want to be super pedantic about the C, you must remember that it arises from up to three indefinite integrals from the derivation of integration by parts, which STRICTLY reads

    int u dv = int d(uv) - int v du

    tl;dr there's always a +C floating around at the end of your equations when you integrate by parts. But for the same function, it had better be the same +C.
     
  13. Sep 24, 2016 #12

    thierrykauf

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    I like to think that this has always been known and clear to everybody.
     
  14. Sep 24, 2016 #13
    It really depends on people's backgrounds and how recently they've studied all this.

    This turns out to be because micromass can do integration by parts in his sleep and I had to think about it a bit :)

    In case it helps any other newbs out there, we want to solve:
    $$\int \frac{1}{x} dx$$
    Using integration by parts, whose formula is:
    $$\int u dv=uv-\int v du$$
    We substitute ##u=\frac {1}{x}##, ##du=-\frac{1}{x^2}dx## and ##dv=dx##, ##v=x## into the formula and find:
    $$= \frac {1}{x}x-\int x \left( -\frac {1}{x^2} dx\right)$$
    And we've now arrived at the second paradox that micromass explains in his Insight:
    $$\int \frac{1}{x} dx = 1 + \int \frac{1}{x} dx$$
     
    Last edited: Sep 24, 2016
  15. Sep 24, 2016 #14

    Mark44

    Staff: Mentor

    I hope you aren't trying to conclude from this that 0 = 1.
     
  16. Sep 24, 2016 #15
    Ha - very funny! o0)
     
  17. Sep 24, 2016 #16

    thierrykauf

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    Just like absolute value of x stands for two (not one) expressions, so does the shortcut C. When I learned this eons ago my teacher made it clear that C stood for two constants, one for each domain x < 0 and x>0.
     
  18. Sep 24, 2016 #17
    Thanks for pointing this out - I had to look this up again because I'd long forgotten why. For the benefit of others in my position...

    There are two constants when integrating ##\frac {1}{x}## because applying the fundamental theorem of calculus requires that the function be continuous over the interval we are interested in. For the indefinite integral, we are working with +/-##\infty## BUT ##\frac {1}{x}## is undefined when x=0, so its not continuous over our entire interval. To account for this, the integral must be evaluated on either side of the discontinuity, once for ##(-\infty,0)## and ##(0,+\infty)##. Each time this is done, we must use a different constant of integration.

    Had the function been something like ##\frac {1}{(x-1)}## we would have had one constant for x < 1 and another for x > 1
     
  19. Sep 24, 2016 #18
  20. Sep 24, 2016 #19

    jbriggs444

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    Start with the idea of an "equivalence relation". This is a relation that is reflexive, symmetric and transitive. Equality is one example of an equivalence relation. Modulus arithmetic is another. Two numbers are "equivalent modulo n" if they each have the same remainder upon division by n. In the case at hand, the equivalence relation is that two functions are equivalent if their first derivatives are identical as functions.

    Given a set (a set of functions in this case) and an equivalence relation, one can divide the set up into "equivalence classes" where every member of any particular equivalence class is equivalent to every other member.

    One can pick out a particular equivalence class by selecting a member, an "exemplar" of that class. The equivalence class selected in this manner is the set of all elements that are equivalent to the exemplar.

    The notation "[alpha]' in the article denotes the equivalence class whose exemplar is the constant function f(x) = ##\alpha##
    The notation "[0]" denotes the equivalence class whose exemplar is the constant function f(x) = 0.

    And indeed, those two equivalence classes are identical.
     
  21. Sep 24, 2016 #20
    thanks for explanation.

    further - so given;

    "In the case at hand, the equivalence relation is that two functions are equivalent if their first derivatives are identical as functions."

    then;

    y1 = 5x + 1

    and

    y2 = 5x + 7

    y1 and y2 are equivalent but y3 = 5x + 1 and y4 = 7x + 1 are not equivalent?

    are you free to make up any equivalence relationship?

    ie can I say that;

    given two functions are equivalent if the value of their first derivative is zero.

    eg;

    y1 = 2

    and

    y2 = Pi

    are equivalent? or are there pre-defined laws on equivalence?
     
  22. Sep 24, 2016 #21

    jbriggs444

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    Yes, you are free to make up any equivalence relationship. It just has to satisfy the properties. It has to be a relation and it has to be reflexive, symmetric and transitive.

    For instance, you could decide to use the relation where everything is equivalent to everything else. That is not likely to lead anywhere very interesting, but there is no rule saying that you cannot use that relation. That is how [much of] mathematics works. You set up rules and definitions, stick to them carefully and see where they lead.

    Edit to add:

    Note that nothing stops someone else from defining things differently. If you define equivalence one way and declare that sin x is equivalent to cos x for the purposes of one argument and if I define equivalence a different way so that sin x is not equivalent to cos x, there is no contradiction. Any statement either of us makes about the equivalence or lack thereof between the two functions needs to be understood in context. Both of us would have a duty to make it clear what definition of "equivalent" we are using.
     
    Last edited: Sep 24, 2016
  23. Sep 24, 2016 #22
    in physics this would just be called sampling right?
     
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