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Laplace Transform Help

  1. Jan 1, 2005 #1
    I can't seem to integrate this properly and can't find the proper algebraic substituition for it. There's a table of laplace transforms and sin ax is included but I'll really like to do it myself.

    [tex] {\cal L} (sin ax) [/tex]
     
  2. jcsd
  3. Jan 1, 2005 #2

    dextercioby

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    [tex] L (\sin ax)=:\int_{0}^{+\infty} e^{-sx}\sin ax dx [/tex]
    Use Euler's formula to get
    [tex] L (\sin ax)=Im[\int_{0}^{+\infty} e^{-x(s-ia)} dx] [/tex]
    Make the substitution
    [tex] -x(s-ia)\rightarrow -z [/tex],with [tex] |z|=x^{2}(a^{2}+s^{2}) [/tex]
    and solve the integral to get:
    [tex] L (\sin ax)=Im(\frac{1}{s-ia})=\frac{a}{a^{2}+s^{2}}[/tex]
    ,s>0.

    Daniel.
     
  4. Jan 2, 2005 #3
    hey, haven't thought about Euler's formula, thanks alot......
     
  5. Jan 2, 2005 #4
    Partial integration would also have worked, but maybe takes a little longer.
     
  6. Jan 2, 2005 #5

    dextercioby

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    Partial integration does not work.U try it and convince yourself i'm right.You expressed an opinion without checking the computations...

    Daniel.
     
  7. Jan 2, 2005 #6
    Check again, I did, got the same answer...
     
    Last edited: Jan 2, 2005
  8. Jan 2, 2005 #7
    [tex]\int_0^\infty e^{-sx}sin(ax)dx=-\frac{1}{a} cos(ax)e^{-sx}|_0^\infty - \frac{s}{a} \int_0^\infty cos(ax)e^{-sx}dx [/tex]
    [tex]= \frac{1}{a}-\frac{s}{a} [\frac{1}{a}sin(ax)e^{-sx}|_0^\infty + \frac{s}{a} \int_0^\infty sin(ax)e^{-sx}dx ] [/tex]
    [tex]= \frac{1}{a}-\frac{s^2}{a^2} \int_0^\infty sin(ax)e^{-sx}[/tex]

    so
    [tex](1+\frac{s^2}{a^2}) \int_0^\infty e^{-sx}sin(ax)dx = \frac{1}{a}[/tex]
    [tex]\int_0^\infty e^{-sx}sin(ax)dx = \frac{a}{a^2+s^2}[/tex]
     
  9. Jan 2, 2005 #8

    dextercioby

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    I'm sorry,u were right. :redface: There are 2 ways to part integrate.I chose the one that didn't go anywhere. :redface:

    [tex] \int_{0}^{+\infty} e^{-sx}\sin ax dx=-\frac{1}{a}(\cos ax) e^{-sx}|_{0}^{+\infty}-\frac{s}{a}\int_{0}^{+\infty} (\cos ax) e^{-sx} dx [/tex]
    [tex] =\frac{1}{a}-\frac{s}{a}[\frac{1}{a}(\sin ax e)^{-sx}|_{0}^{+\infty}+\frac{s}{a}\int_{0}^{+\infty} (\sin ax) e^{-sx} dx] [/tex]

    Denote the first integral by 'I' and u'll get:
    [tex] I=\frac{1}{a}-\frac{s^{2}}{a^{2}}I\Rightarrow I=\frac{a}{s^{2}+a^{2}} [/tex]

    I don't know why,but i like the complex integral version much more... :tongue2:

    Daniel.

    EDIT:At the same time...U missed a 'dx' in your antepenultimate integral... :tongue2:
     
    Last edited: Jan 2, 2005
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