Results 1 to 3 of 3

Math Help - Integration Help

  1. #1
    Junior Member
    Joined
    Jun 2008
    Posts
    61

    Integration Help

    Can someone confirm that the following is correct:

    ∫〖[e^(-nx) ][x^(s-∞-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:

    ∫〖[e^(-nx) ][x^(s-∞-2)] 〗 dx=(e^(-n))*∫〖[x^(s-∞-2)] 〗 dx

    Any help would be much appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by rman144 View Post
    Can someone confirm that the following is correct:

    ∫〖[e^(-nx) ][x^(s-∞-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:

    ∫〖[e^(-nx) ][x^(s-∞-2)] 〗 dx=(e^(-n))*∫〖[x^(s-∞-2)] 〗 dx

    Any help would be much appreciated.
    You can't take  e^{-n} out as a constant. It's a function of x  e^{-nx} .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jun 2008
    Posts
    61

    Reply

    I know the value is a function, but on that particular interval, isn't the substitution allowed? Basically, if on that interval, the only meaningful solution occurs when x=1, why can't I just substitute the one in and let the value act as a constant e^-n?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 3rd 2010, 01:54 AM
  2. Replies: 2
    Last Post: November 2nd 2010, 05:57 AM
  3. Replies: 8
    Last Post: September 2nd 2010, 01:27 PM
  4. Replies: 2
    Last Post: February 19th 2010, 11:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 07:58 AM

Search Tags


/mathhelpforum @mathhelpforum