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Questions about definite integrals

  1. Sep 14, 2012 #1
    While reading my calc book, I have developed a few questions about the situations in which definite integrals can exist. I've thought about these questions, and I feel that if I am able to answer some of them, I can make some other problems much easier, such as testing for convergence of a definite integral.

    1) If a definite integral diverges, can we conclude that it does not exist?
    2) If a definite integral does not exist, can we conclude that it diverges?
    3) If [itex] f(a) [/itex] is not defined, does [itex]\int^{b}_{a}f(x)dx [/itex] necessarily not exist?
    3) If [itex] f(a) [/itex] is not defined, but but [itex]\lim_{x→a+}f(x) [/itex] is, does [itex]\int^{b}_{a}f(x)dx [/itex] necessarily not exist?
    5) If [itex] f(a) [/itex] is defined, but [itex]\lim_{x→a+}f(x) [/itex] does not, does [itex]\int^{b}_{a}f(x)dx [/itex] necessarily not exist?

    I assume all these questions can be answered using the definition of the definite integral, along with the definition of continuity, but how exactly?

    Thanks!

    Note: Assume that b>a for the above integrals

    BiP
     
  2. jcsd
  3. Sep 14, 2012 #2

    Simon Bridge

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    An integral is just a fancy way of doing a sum.
    So.

    1. yes and no - depends what you mean by "exist".
    2. no.
    Can you think of where diverging integrals may come from - say - in Nature?

    3. no.
    f(x) only needs only to exist in (a,b) not [a,b] to be integrable.

    4. Also answered by 3.

    5. Can you come up with an example to illustrate this? I think you are describing a situation where f is not a function or is not peicewise continuous within (a,b).

    Generally, in order to be integrable, f has to be a piecewise-continuous function in the range of the integration. Look these up.
     
  4. Sep 14, 2012 #3

    AlephZero

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    Those statements depend on how you define "integration", and you can define it so nether of them is required. But I guess the OP is thinking about the http://en.wikipedia.org/wiki/Riemann_integral (even if the calc textbook doesn't mention the name).
     
  5. Sep 14, 2012 #4
    I see, thank you for all your help. If time permits, could you please offer an example of functions for the questions 3, 4, and 5?

    Also, I have heard of something called Lipschitz continuity, does that have any bearing on this problem?

    Also, for a Riemann integral, does the function have to be continuous in (a,b) or in [a,b] ? wikipedia proves the FTC assuming continuity in [a,b] but I assume that is superfluous condition and that continuity in (a,b) suffices to determine the existence of a riemann integral?
    Thanks!

    BiP
     
    Last edited: Sep 14, 2012
  6. Sep 14, 2012 #5

    Simon Bridge

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    The simplest way to create an undefined value is to divide by zero.

    If I define a function u(x): u=0 for x≤0 and u=1 for x>0 then f(x)=1/u(x-a) would be undefined for x≤a ... but I can still do the integral from a to b if b>a. (It's b-a).

    I'm pretty sure you can do the rest.
     
  7. Sep 14, 2012 #6

    Simon Bridge

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    Yah. I kinda figured a definition something like:
    $$\int_a^bf(x)dx = \lim_{N\rightarrow \infty} \frac{b-a}{N} \sum_{n=0}^N f(\frac{b-a}{N}n + a)$$... but if I just bring the sum in each end by one unit ... in the limit N → ∞ it's the same.

    $$\int_a^bf(x)dx = \lim_{N\rightarrow \infty} \frac{b-a}{N} \sum_{n=1}^{N-1} f(\frac{b-a}{N}n + a)$$

    ... though there is danger in that kind of argument of shaving a bit of each end until there's nothing left.

    I can also take the sum to within a small distance ##\epsilon << (b-a)/N## of the limits
    ... so I do the sum in range ##[a+\epsilon, b-\epsilon]##, instead of ##[a,b]##, and see what happens in the limit ##\epsilon \rightarrow 0##.
    $$\int_a^bf(x)dx = \lim_{N\rightarrow \infty}_{\epsilon \rightarrow 0}\frac{b-a-2\epsilon}{N} \sum_{n=0}^N f(\frac{b-a-2\epsilon}{N}n + a + \epsilon)$$
    I'm sure AlephZero can refine this argument.

    Few beginning courses are even this explicit anyway.
     
    Last edited: Sep 14, 2012
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