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Homework Help: Integration problem

  1. Oct 14, 2006 #1
    Can someone give me a hint on how to solve

    [tex]I = \int_0^\infty e^{-ax} \sin{bx} dx[/tex]

    ?
     
  2. jcsd
  3. Oct 14, 2006 #2
    Use integration by parts, and set the upper limit as t. Then let t--> infinity (i.e. improper integral)
     
    Last edited: Oct 14, 2006
  4. Oct 14, 2006 #3
    But integration by parts will only give me a similar integral to solve...?
     
  5. Oct 14, 2006 #4
    you have to use it multiple times, so that you can solve the integral
     
  6. Oct 14, 2006 #5
    You can write the trig function as a complex exponential:

    [tex]
    \sin(bx) = \frac{1}{2i}\left[e^{ibx}-e^{-ibx}\right]\,.
    [/tex]

    Then you will have two similar integrals to solve, which are straight forward.
     
  7. Oct 16, 2006 #6
    jprO: Of course. That will do it.

    courtrigad: Partial integration multiple times will give me a loop between [tex]e^{-ax} \sin{bx}[/tex] and [tex]e^{-ax} \cos{bx}[/tex].
     
  8. Oct 16, 2006 #7

    HallsofIvy

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    Science Advisor

    Yes, it does:
    [tex]I = \int_0^\infty e^{-ax} \sin{bx} dx[/tex]
    Let [itex]u= e^{-ax}[/itex], [itex]dv= sin(bx)dx[/itex]
    Then [itex]du= -ae^{-ax}dx[/itex], [itex]v= -\frac{1}{b}cos(bx)[/itex]
    so the integral becomes
    [tex]I= -\frac{1}{b}e^{-ax}cos(bx)\left|_0^\inftjy- \frac{a}{b}\int_0^\inty}e^{-ax}cos(bx)dx[/tex]
    Since e-ax goes to 0 as x goes to infinity, while cos(bx) is bounded, that is
    [tex]I= \frac{1}{b}- \frac{a}{b}\int_0^\inty}e^{-ax}cos(bx)dx[/tex]
    Now let [itex]u= e^{-ax}[/itex], [itex]dv= cos(bx)[/itex]. Again we have [itex]du= -a e^{-ax}dx[/itex], [itex]v= \frac{1}{b}sin(bx)[/tex]
    The integral is now:
    [tex]I= \frac{1}{b}- \frac{a^2}{b^2}\int_0^\infty e^{-ax}sin(bx)dx[/tex]
    But remember that [itex]I= \int_0^\infty e^{-ax}sin(bx)dx[/itex]
    so that equation says
    [itex]\int_0^\infty e^{-ax}sin(bx)dx= \frac{1}{b}- \frac{a^2}{b^2}\int_0^\infty e^{-ax}sin(bx)dx[/tex]
    so
    [tex]\left(1+frac{a^2}{b^2}\right)\int_0^\infty e^{-ax}sin(bx)dx= \frac{1}{b}[/tex]
    [tex]\int_0^\infty e^{-ax}sin(bx)dx= \frac{b}{a^2+b^2}[/tex]
    (Modulo any silly litte errors from working too fast!)
     
  9. Oct 16, 2006 #8
    I don't understand why you do the substitutions. Can't you just solve the integral by parts without doing the substitutions?
     
  10. Oct 16, 2006 #9
    do you agree that integration by parts if of the form:

    [tex] \int udv = uv-\int vdu [/tex]? You need substitutions to do integration by parts.
     
    Last edited: Oct 16, 2006
  11. Oct 17, 2006 #10
    I see what you mean. I'm just not used to do it your way. Actually, I'm doing it your way. I just don't think about what I'm really doing. ;)
     
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