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Integral equation

  1. Mar 21, 2005 #1
    what does an equation need so that the equation can be integrated?
    not all of equation can be integarted right?
    I was quite confuse if it can be integrated or not whether I found a difficult integration. Thanx
  2. jcsd
  3. Mar 21, 2005 #2


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    The function must be continuous on the interval you are integrating (x1 to x2) or continuous everywhere if you want an indefinite integral. The series must also converge
  4. Mar 21, 2005 #3


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    the answer is not correct, but the question is also not precise. what do you really want to know?
  5. Mar 21, 2005 #4


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    From the top of my head,the theory of intergral equations has been established about 100 yrs ago and we have enough tools to solve them.Give us an example & we'll see what to do.

  6. Mar 21, 2005 #5
    I think I know what he means, and I don't think he's talking about Integral Equations. One word: Differential Galois Theory :smile:
  7. Mar 22, 2005 #6
    That's not one word!
  8. Mar 25, 2005 #7
    I thought the same, it's what I assumed (I heard from my teacher) before but I'm not quite sure. By the way, what does "The series must also converge" mean? and I don't know about Differential Galois Theory. Thanx for your reply :smile:
  9. Mar 25, 2005 #8


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    Any function f(x) that maps all x to a single f(x) can be integrated. Defining the integral of f(x) other than simple the integral of f(x) and doing anything with it can provide very tricky in general though.
  10. Mar 25, 2005 #9
    That's not true.

    [tex] f(x) = \biggl\{ \begin{array}{ccc}0 & & x \ \mbox{rational} \\ 1 & & x \ \mbox{irrational}\end{array}[/tex]

    is not integrable.
    Last edited: Mar 25, 2005
  11. Mar 27, 2005 #10


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    I have no idea what it means for f(x) to map "all x to a single f(x)" means!

    Certainly, every continuous function is integrable but that is not necessary.

    If I remember correctly "every bounded function whose set of discontinuities is a set of measure 0" is Riemann integrable. Oddly enough, you need Lebesque measure theory to define "set of measure 0"!
  12. Mar 27, 2005 #11


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    To be honest, neither do I it was 4:30AM when I wrote it :rolleyes: Perhaps I should stop relying on theorems I come up with after 3:30AM and a bit of too much intoxicating drink :wink:

    Why can't that be integrated?
    Last edited: Mar 27, 2005
  13. Mar 27, 2005 #12
    Take a look at its upper and lower sums.
  14. Mar 27, 2005 #13


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    If you're talking to me that means nothing to me.
  15. Mar 27, 2005 #14
    Well, if we're talking about Riemann integrability, then it's not integrable because, letting [itex]P = \{ I_1, I_2, ..., I_n\}[/itex] be a partitioning of the interval [itex]I = [a, b][/itex], [itex]m_i = \min_{x \in I_i} f(x)[/itex], [itex]M_i = \max_{x \in I_i} f(x)[/itex] (Edit: really I should have used greatest lower bound and least upper bound here but it doesn't make a difference for this function), [itex]s[/itex] to be the mesh fineness of [itex]P[/itex], and [itex]L_i[/itex] to be the length of the interval [itex]I_i[/itex], we find

    [tex]\lim_{s \rightarrow 0} \sum_{i=1}^n M_iL_i \neq \lim_{s \rightarrow 0} \sum_{i=1}^n m_iL_i[/tex]

    Edit 2: I should probably also make the important note that [itex]f(x)[/itex] is Lebesgue-integrable.
    Last edited: Mar 27, 2005
  16. Apr 2, 2005 #15
    thanx for your replies
  17. Apr 2, 2005 #16


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    One-D it would help to know for which class you want to know this.

    Obviously if its for Real Analysis - my answer is not complete - and its more of a Galois theory you looking for

    If it's for Calculus 2 then thats the correct answer - a function must be continuous in order to be integrated

    If its for Multivariable Calculus then there are limits and continuity that you must check, for all variables

    If its for Discrete Math its something along the lines of what Data said
  18. Apr 8, 2005 #17
    now I m learning calculus 2, I just want to know, cause I got that questions on my mind. Do you know where I can find web so I can learn about calculus more, especially cal 2. thanx Cronxeh
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