Laplace Transform Help

  • #1

Main Question or Discussion Point

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]
 

Answers and Replies

  • #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.
 
  • #3
hey, haven't thought about Euler's formula, thanks alot......
 
  • #4
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Partial integration would also have worked, but maybe takes a little longer.
 
  • #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.
 
  • #6
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dextercioby said:
Partial integration does not work.U try it and convince yourself i'm right.You expressed an opinion without checking the computations...

Daniel.
Check again, I did, got the same answer...
 
Last edited:
  • #7
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[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]
 
  • #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:

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