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Why can't the indefinite integral be written with sigma notation?

  1. Oct 19, 2012 #1
    Hi, I've been wondering this since I started learning integration. I get that ∫ is basically an elongated S for "sum", because that is what it is basically doing. But then Ʃ does the same thing as well. If I'm understanding the difference, it is that Ʃ increments by finite measures, whereas ∫ increments by infinitesimal measures. But even if that is the difference, can't you 'simulate' infinitesimal incrementation with sigma notation? I gave it a shot below:

    [itex]∫f(x)dx=\sum^{∞}_{i=1}f(idx)dx+\sum^{∞}_{i=1}f(-idx)dx=\sum^{∞}_{i=1}f(idx)dx+f(-idx)dx=\sum^{∞}_{i=-∞}f(idx)dx[/itex]

    Which is basically saying that the indefinite integral is equal to a sum. The way to get the sum is to divide the real number line from 0 to ∞ into ∞ parts. I get that ∞ isn't any particular number, but shouldn't we at least be able to say that each of these ∞ parts has a length of dx?

    If that's the case, then you would take two sums, one from 0 to ∞, and the other from 0 to -∞ and increment by dx as you sum each area given by f(i*dx)*dx from the positive and negative ends of the number line. What do you think?

    [Edit: I see that the first two sums is actually not equal to the third, since the first two actually skip over 0. So technically, the indefinitely integral is equal to the last sum, not the first two?]
     
    Last edited: Oct 19, 2012
  2. jcsd
  3. Oct 19, 2012 #2

    jbriggs444

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    The serious problem with the formula you give is that the result of an indefinite integral is supposed to be a function. The formula you give evaluates to a number.

    The less serious problem is the one you already pointed out: ∞ is not a number. That caveat aside, the approach you give is the essense of the Riemann integral.

    http://en.wikipedia.org/wiki/Riemann_integral
     
  4. Oct 20, 2012 #3
    Yeah, as soon as I tried to solve out using that formula, I realized for myself that this is basically trying to sum up the whole are under the curve to infinity. Basically it is equal to the definite integral from -∞ to ∞, as below?

    [itex]∫^{∞}_{-∞}f(x)dx=\sum^{∞}_{i=-∞}f(idx)dx[/itex]
     
  5. Oct 20, 2012 #4
    And actually, I had another question that was related enough that I figured I would ask in the same thread. I sort of understand, I think, that the motivation for calculating the area under the curve and determining of the anti-derivative/indefinite integral were not necessarily related. And that the indefinite integral said something about the area under the curve wasn't known a priori?

    Well, anyway, what I'm wondering is when I see for example the following:

    [itex]f(x)=\frac{d}{dx}F(x)=\frac{F(x+dx)-F(x)}{dx}[/itex]

    I can apply it algebraically using infinitesimals to, for example, f(x)=x^2, as per:

    [itex]\frac{d}{dx}x^2=\frac{(x+dx)^2-x^2}{dx}[/itex]
    [itex]=\frac{x^2+2xdx+dx^2-x^2}{dx}[/itex]
    [itex]=\frac{2xdx+dx^2}{dx}[/itex]
    [itex]=\frac{dx(2x+dx)}{dx}[/itex]
    [itex]=2x+dx[/itex]
    [itex]=2x+0[/itex]
    [itex]=2x[/itex]

    But is there an equivalent algebraic process for indefinite integration that applies at least to polynomials? I get that integration is basically like taking an infinitesimal sum, which from the scale of real numbers would be akin to taking an infinite sum of infinitesimals, so the only method that I can think of is the algebraic equivalence of a series. But these just seem so clunky for me. Am I missing something, or is the only practical way to determine the indefinite integral to just follow the derivation process in reverse?

    And apologies if my terminology is a bit non-rigorous.
     
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