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Prove that f is not integrable on [0,1]

  1. Mar 30, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##f(0) = 0## and ##f(x) = 1/x ## if ##0 < x \leqslant 1##. Show that ##f## is not integrable on ##[0,1]##.
    Hint: Show that the first term in the Riemann sum, ##f(x_1^*) ~\Delta x##, can be made arbitrarily large

    2. Relevant equations
    Definition of integral using Riemann sum

    3. The attempt at a solution
    Using the definition, we have $$\int_0^1 f(x) dx = \lim_{n \to\infty} \sum_{i=1}^n f(x_i^*) \Delta x $$ Now I am not sure how the hint could be used here. Should I try to go for a proof by contradiction ?
     
  2. jcsd
  3. Mar 30, 2017 #2

    haruspex

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    You need to consider specific choices for x1 and pick corresponding values for x1*.
     
  4. Mar 30, 2017 #3
    If we divide ##[0,1]## into ##n## intervals, then ##\Delta x = \frac{1}{n}## and the first interval would be ##[x_0, x_1] = [0, \frac{1}{n}]##. If we take ##x_1^*## to be right endpoint of the interval, then ##f(x_1^*) = 1/(1/n) = n## and hence ##f(x_1^*)\Delta x = 1## How can this be made arbitrarily large ?
     
  5. Mar 30, 2017 #4
    Nvm, didn't notice your last post :/
    The definition of the Riemann integral eventually hinges on a limit. Specifically when all of your segments' width goes to zero. In this case, though, the tighter you pick your partition, the larger the sum will become, unbounded. Resulting series diverges.
     
    Last edited: Mar 30, 2017
  6. Mar 30, 2017 #5

    PeroK

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    The hint looks like a hindrance to me. Try looking at the Riemann sum for ##n##, not just one term.
     
  7. Mar 30, 2017 #6

    haruspex

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    True, but...
    So pick a different x1*.
     
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