Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integral calculation help

  1. Feb 24, 2006 #1
    hi,
    I'm working on thiis:
    [tex].25 = \int_{-\infty}^m\frac{e^{x}}{(1+e^{-x})^2}dx[/tex]
    I let [tex]u=1+e^{-x} , du=-e{-x}dx[/tex]
    which gives:
    [tex]\frac{1}{u}=\frac{1}{1+e^{-x}}|_{-\infty}^m[/tex]
    then I solve it out and I get
    [tex]\frac{1}{1+e^{-m}}-1=.25[/tex]
    I have tried to solve this and I keep getting a negative ln. I moved the 1 to the RHS and then took ln of both sides. I can't get it to work.
    help
    CC
     
  2. jcsd
  3. Feb 24, 2006 #2

    StatusX

    User Avatar
    Homework Helper

    But you don't have an e-x in the numerator. Try multiplying the top and bottom by e2x, then using the substitution u=1+ex, which will leave you with something like (u-1) to some power divided by u2.

    By the way, there is no solution to the equation you mention at the end since 1/(1+ex)<1 for any real x.
     
  4. Feb 24, 2006 #3
    ok
    Statusx, you are absolutely correct. I have forgotten my - sign. It should read

    [tex].25 = \int_{-\infty}^m\frac{e^{-x}}{(1+e^{-x})^2}dx[/tex]
    sorry...
    then let

    [tex]u=1+e^{-x} , du=-e^{-x}dx[/tex]
    which gives
    [tex]\frac{1}{1+e^{-m}}-1=.25[/tex]
    so then you get
    [tex]\frac{1}{1+e^{-m}}=1.25[/tex]
    from there I took to ln of both sides and tried to solve for m, but nothing is working out.
    CC
     
  5. Feb 24, 2006 #4
    then,
    [tex]1=1.25+1.25e^{-m}[/tex]
    then
    [tex]-.2=e^{-m}[/tex]
    ln makes no sense
    CC
     
  6. Feb 24, 2006 #5

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    It makes no sense because you've been sloppy! :grumpy:
    We have:
    [tex]\int_{-\infty}^{m}\frac{e^{-x}dx}{(1+e^{-x})^{2}}=\frac{1}{1+e^{-x}}\mid_{-\infty}^{m}=\frac{1}{1+e^{-m}}-\frac{1}{1+e^{\infty}}=\frac{1}{1+e^{-m}}[/tex]
     
  7. Feb 24, 2006 #6
    der duh der duh
    [tex]\frac{1}{e^{-{-\infty}}}=0[/tex]
    I am a sloppy, sloppy student.:redface:
    I'm all good now.
    Thanks
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook