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Is f integrable?

  1. Nov 23, 2008 #1
    The function f is defined on [0,1] so that f(1/n)=n^(-1/2) for n=1,2,3,... and f(x)=0 if x is not a reciprocal of a positive integer. Is f integrable on [0,1]? If so, prove it and compute the integral. If not then give an argument for why not.

    See, I read this question over a hundred times, and the thing is... f(x) is always going to be a reciprocal of a positive integer... so the second statement is saying that any number inbetween the reciprocal of a positive integer, meaning all the irrational numbers inbetween 0 and 1 are equal to zero, no? Therefore this function is not continuous? Therefore this function cannot be integrable right? On top of that, if the function is 1 over a squareroot, then there will be two values for every reciprocal of a positive integer. (keeping in mind that a squareroot gives a positive and a negative answer right?

    Am I wrong?

    Jonnah Song
  2. jcsd
  3. Nov 23, 2008 #2
    Yes, you are wrong, m-1/2 is defined to be the positive square root of n.
    A discontinuous function may well be integrable. To discuss this further, you would have to communicate to us, if you are talking about Riemann or Lebesgue integrals; in both cases you should at least state a definition of when a function is said to be integrable.
  4. Nov 23, 2008 #3


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    Integrable in what sense? f is obviously Lesbeque integrable: since the set of all numbers of the form 1/n for n a positive integer is countable, f is 0 "almost everywhere" so it is integrable and the integral is 0.

    There is a rather deep theorem that says if a function is bounded and its set of discontinuities is countable, as is the case here, then the function is integrable.

    You seem to be completely misunderstanding this problem if not integration itself.

    No, that's not said any where and is not true. f(x) is 0 except when x is the reciprocal of a positive integer.

    No! No irrational numbers are equal to zero! I know that you meant that f(x)= 0 for x an irrational number, but say what you mean! Being precise is very important in mathematics.

    Yes, this function is not continuous at 1/2, 1/3, 1/4, etc. No, it does NOT follow that this function is not integrable. No one has ever told you that a function must be continuous in order to be integrable!

    NOT right! The square root, like any function on the real numbers, gives only a single positive value. The equation x2= 3 has two solutions, [itex]\sqrt{3}[/itex] and [itex]-\sqrt{3}[/itex] but the reason we have to write "[itex]\pm\sqrt{3}[/itex]" is because just [itex]\sqrt{3}[/itex] means only the positive solution.
  5. Nov 23, 2008 #4
    But, let's say you have a sqaureroot of 9. It can as well be negative 3 and positive 3 right?
    So, there are positive and negative numbers for a squareroot? Is there even such a thing called a positive squareroot or a negative squareroote?
  6. Nov 23, 2008 #5
    Can you explain what the question states then? err.. the actual given statement because I am rather confused on what they are telling me
  7. Nov 23, 2008 #6
    The square root of a positive integer n is defined to be the positive solution of x=n2.

    As such a square root is always positive.
  8. Nov 23, 2008 #7
    My TA said " this function is integrable if and only if the upper and lower limits are the same. Function is continuous, so consider the intermediate value theorem." Do you guys have any idea what he's talking about?

    I'm very confused.
  9. Nov 24, 2008 #8
    The function is clearly NOT continuous. To compute the upper and lower Riemann sums, you will first have to define a sequence of partitions with mesh size tending to zero, then write down the sum and use the specific form of your function.
    So. how does, in general, a Riemann sum look like?
  10. Nov 24, 2008 #9


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    I can't speak for your TA but that is non-sense. What "upper and lower limits" are you talking about? If you mean the limits of integration, if they are the same, the the integral is trivially zero! That has nothing to do with a function being continuous. What function are you talking about? The function you give here is not continuous as you said.

    By my count you have now asserted that you "know" 7 different statements that are simply not true.

    The square root of 9 is 3. The roots of the equation are 3 and -3 but only one of them is "the square root of 3". The other is "-1 times the square root of 3".
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