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Question on mu(x) function

  1. Sep 6, 2004 #1
    let be the function given by f(x)=w(x)t^-3mu(x) where mu(x) is the Mobius function and w(x)=Sum(1<n<infinite)d(x-n) then my question is...does the Laplace transform of this function exist and is equal to

    L[f(x)]=Sum(1<n<Infinite)mu(n)/n^3
     
  2. jcsd
  3. Sep 6, 2004 #2
    sorry i made a mistake it should be L[f(x)]=Sum(1<n<Infinite)mu(n)exp(-sn)/n^3
     
  4. Sep 6, 2004 #3

    matt grime

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    It isn't a function (from R to R) so asking if its Laplace transform exists as a *function* seems a little moot.
     
  5. Sep 7, 2004 #4

    matt grime

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    the answer is, mutatis mutandis, yes, by the way, since that infinite sum obviously converges, for some range of s, to a real number. whether or not that is meaningful is a different question
     
    Last edited: Sep 7, 2004
  6. Sep 7, 2004 #5
    The question is interesting when related the generating function of MOebius function

    Sum(n)mu(n)/n^(4-s)=R(4-s) where the sum is from 1 to infinite then according to our formula:

    R(4-s)=M[w(x))/x^3] or M^-1[R(4-s)]=w(x)mu(x)/x^3 now integrating from k-1/2 to k+1/2 we have that Int(k-1/2,k+1/2)M^-1[R(4-s)]=mu(k)/k^3
     
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