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Math Help - integrability question

  1. #1
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    integrability question

    3b) show an example for a function wich is bounded and exists in [a,b]
    but its not integrabile.
    prove that its not integrabile by definition
    3c)calculate  \int\frac{ln(x^{2}+1}{x^{4}}dx
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  2. #2
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    Re: integrability question

    What have you done on this?
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  3. #3
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    Re: integrability question

    i dont know how to solve th first part
    but i know some laws for proving integrability
    (*) for some dividing P
    E is the set of uppers sums, T is the set of lower sums
    if infE=supT then its defferentiable.

    (*) S(p) is the sum of each supremum of the function in subsection mutiplied by the length of the subsection.
    s(p) is the sum of each supremum of the function in subsection mutiplied by the length of the subsection.
    in each subsection if inf{S(p)-s(p)}=0 then its defferentiable
    or if S(p)-s(p)<\epsilon

    ok i tried this one:
    f(x)=1 on x=0
    f(x)=0 on other wise

    for every diving of the section s(p)=0 so is supT=0 too
    but S(p) is obviosly not sero so here is my example.
    so inf{S(p)-s(p)}\leq 0 but i dont know how to express S(p)
    ?

    regarding the second one i thought of using by parts
    but its not solving it
    Last edited by transgalactic; September 2nd 2011 at 12:15 PM.
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  4. #4
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    e^(i*pi)'s Avatar
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    Re: integrability question

    For the second part integration by parts is a good start (you'll then want to try partial fractions)

    u = \ln(x^2+1) \rightarrow du = \dfrac{2x}{x^2+1} dx

    \frac{dv}{dx} = x^{-4} \rightarrow v = -\frac{1}{3x^3}


    \int \dfrac{\ln(x^2+1)}{x^4} dx = -\dfrac{\ln(x^2+1)}{3x^3} - \int \dfrac{2x}{3x^3(x^2+1)}dx

    = -\dfrac{\ln(x^2+1)}{3x^3} - \dfrac{2}{3} \int \dfrac{1}{x^2(x^2+1)} dx


    Now use partial fractions.

    Hint: \dfrac{1}{x^2(x^2+1)} = \dfrac{(x^2 +1) - x^2}{x^2(x^2+1)}
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  5. #5
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    Re: integrability question

    Quote Originally Posted by e^(i*pi) View Post
    For the second part integration by parts is a good start (you'll then want to try partial fractions)

    u = \ln(x^2+1) \rightarrow du = \dfrac{2x}{x^2+1} dx

    \frac{dv}{dx} = x^{-4} \rightarrow v = -\frac{1}{3x^3}


    \int \dfrac{\ln(x^2+1)}{x^4} dx = -\dfrac{\ln(x^2+1)}{3x^3} - \int \dfrac{2x}{3x^3(x^2+1)}dx

    = -\dfrac{\ln(x^2+1)}{3x^3} - \dfrac{2}{3} \int \dfrac{1}{x^2(x^2+1)} dx


    Now use partial fractions.

    Hint: \dfrac{1}{x^2(x^2+1)} = \dfrac{(x^2 +1) - x^2}{x^2(x^2+1)}
    \frac{1}{x^2(x^2+1)}=\frac{A}{x}+\frac{B}{x^2}+C/(1+x^2)
    did i splitted it correctly by theory
    ?
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: integrability question

    Use the hint of e^(ipi) which is very useful:
    \frac{1}{x^2(x^2+1)}=\frac{(x^2+1)-x^2}{x^2(x^2+1)}=\frac{x^2+1}{x^2(x^2+1)}-\frac{x^2}{x^2(x^2+1)}=\frac{1}{x^2}-\frac{1}{x^2+1}

    EDIT:
    If you would use the traditional method then it has to be:
    \frac{1}{x^2(x^2+1)}=\frac{Ax+B}{x^2}+\frac{Cx+D}{  x^2+1}=\frac{(Ax+B)(x^2+1)+(Cx+D)(x^2)}{x^2(x^2+1)  }=\frac{Ax^3+Bx^2+Ax+B+Cx^3+Dx^2}{x^2(x^2+1)}=
    \frac{(A+C)x^3+(B+D)x^2+Ax+B}{x^2(x^2+1)}

    That means:
    B=1 therefore D=-B \Leftrightarrow D=-1 and A=C=0
    And so:
    \frac{1}{x^2(x^2+1)}=\frac{1}{x^2}-\frac{1}{x^2+1}
    Last edited by Siron; September 2nd 2011 at 12:21 PM.
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  7. #7
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    Re: integrability question

    ok i could take care of this integral from here
    regarding the first part:
    i know some laws for proving integrability

    (*) for some dividing P

    E is the set of uppers sums, T is the set of lower sums

    if infE=supT then its defferentiable.



    (*) S(p) is the sum of each supremum of the function in subsection mutiplied by the length of the subsection.

    s(p) is the sum of each supremum of the function in subsection mutiplied by the length of the subsection.

    in each subsection if inf{S(p)-s(p)}=0 then its defferentiable

    or if S(p)-s(p)<\epsilon



    ok i tried this one:

    f(x)=1 on x=0

    f(x)=0 on other wise



    for every diving of the section s(p)=0 so is supT=0 too

    but S(p) is obviosly not sero so here is my example.

    so inf{S(p)-s(p)}\leq 0 but i dont know how to express S(p)

    ?
    Last edited by transgalactic; September 2nd 2011 at 10:04 PM.
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  8. #8
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    Re: integrability question

    Quote Originally Posted by transgalactic View Post
    \frac{1}{x^2(x^2+1)}=\frac{A}{x}+\frac{B}{x^2}+C/(1+x^2)
    did i splitted it correctly by theory
    ?
    At the level you appear to be studying at, you are most surely expected to know how to perform a partial fraction decomposition. You can at least work backwards from your answer and see if it works.

    And please show all of your working in your first posting of future questions. We should not have to be asking for it.
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  9. #9
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    Re: integrability question

    my last post states that i have solved the integral and moved to the next one.

    regarding the first part:

    i know some laws for proving integrability



    (*) for some dividing P



    E is the set of uppers sums, T is the set of lower sums



    if infE=supT then its integrible.







    (*) S(p) is the sum of each supremum of the function in subsection mutiplied by the length of the subsection.



    s(p) is the sum of each supremum of the function in subsection mutiplied by the length of the subsection.



    in each subsection if inf{S(p)-s(p)}=0 then its defferentiable



    or if S(p)-s(p)<\epsilon







    ok i tried this one:



    f(x)=1 on x=0



    f(x)=0 on other wise







    for every diving of the section s(p)=0 so is supT=0 too



    but S(p) is obviosly not sero so here is my example.



    so inf{S(p)-s(p)}\leq 0 but i dont know how to express S(p)



    ?
    Last edited by transgalactic; September 3rd 2011 at 01:19 PM.
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  10. #10
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    Re: integrability question

    Quote Originally Posted by transgalactic View Post
    3b) show an example for a function wich is bounded and exists in [a,b]
    but its not integrabile.
    prove that its not integrabile by definition
    It is known that a function is Riemann integrable if and only if it is bounded and continuous almost everywhere, i.e., the set of its points of discontinuity has measure zero. A set has measure zero if it can be covered by intervals whose combined length can be made arbitrarily small; however, it is not necessary to understand this precisely for this problem. Every finite set and even every countable infinite set has measure zero, so the function you tried: f(x)=1 on x=0 f(x)=0 otherwise, is integrable (it has a single point of discontinuity). You need a lot more points of discontinuity. Consider Dirichlet function, which is nowhere continuous. It is easy to show that the upper sum, say, on [0, 1] is always 1 and the lower sum is 0.

    (*) for some dividing P
    E is the set of uppers sums, T is the set of lower sums
    if infE=supT then its defferentiable.
    Why are you saying "differentiable" instead of "integrable"?
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  11. #11
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    Re: integrability question

    dirichlet functio is defined
    D(x) =1 for rational x
    D(x) =0 for irrational x

    S(p)=\sum_{i=1}^{n}m_{i}(x_{i}-x_{i-1})=0 so supT=0
    S(p)=\sum_{i=1}^{n}M_{i}(x_{i}-x_{i-1})=1 so infE=1
    supT is not equal infE so its not integrible
    correct?
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  12. #12
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    Re: integrability question

    To say whether this is correct, I need to know what are m_i, M_i, T and E, as well as why all m_i=0 and all M_i=1.
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