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Integration Work Check

  1. Dec 23, 2008 #1
    Can someone confirm that the following is correct:

    S [e^(-nx) ][x^(s-oo-2)] dx , limits from x=1 to x=infinity, n is some real integer>=1, and 0<Re(s)<1

    What I want to know is, because the equation vanishes for all "x" in the interval x=1 to x=infinity, excluding x=1, can I just bind (e^(-nx)) to e^(-n(1)), leaving:

    S [e^(-nx) ][x^(s-oo-2)] dx=(e^(-n))*S [x^(s-oo-2)] dx

    Any help would be much appreciated.
     
    Last edited: Dec 23, 2008
  2. jcsd
  3. Dec 23, 2008 #2

    HallsofIvy

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    1. There are symbols in that my webreader cannot read.
    2. I don't know what "s-∞-2" could possibly mean.
    3. I don't know what "bind (e^(-nx)) to e^(-n(1))" means.
     
  4. Dec 23, 2008 #3
    Sorry about the confusion:

    First, the s-oo-2 is read as "s" (the variable s) minus infinity minus 2.

    Next, when I say bind it to a constant, I mean:

    Say you have:

    S f(x)*g(x) dx, limits from x=x1 to x=x2

    Now if you know that the equation f(x) on the interval x1 to x2 is at most f(k1) and at least f(k2), then the integral:

    S f(x)*g(x) dx

    Will have a maximum absolute value of:

    abs( S f(k1)*g(x) dx )

    And a minimum of:

    abs( S f(k2)*g(x) dx)

    Now what I'm asking is, because within my original equation, the only value in the interval that doesn't go to zero is when x=1, as a result, can I repeat the above process for e^(-nx) and "bind" the value to the interval:

    e^-n<=e^(-nx)<=e^-n

    Which implies:

    e^(-nx)=e^(-n)


    I'm not trying to solve the integral; I'm merely attempting to establish whether or not it converges, and if so, how small an interval can I confine the possible solution of the actual integral to.
     
    Last edited: Dec 23, 2008
  5. Dec 23, 2008 #4

    HallsofIvy

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    Yes, even I could read[/i] it that way but what does it MEAN? Since "infinity" is not a number, that makes no sense to me. Is it intended as a limit? And if so what could "infinity minus 2" mean?

     
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