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Integral OH YES!

  1. May 9, 2005 #1
    Ok, here is the integral i seem to be having some issues with. I know there's a very simple step im missing.

    [tex]\int_{}^{} e^x \sin(\pi x) dx [/tex]

    i attempted to do this using by parts integration.

    I tried u = [tex] \sin(\pi x) [/tex] so du= [tex] \pi \cos(\pi x) dx [/tex]
    so then dv= [tex] e^x dx [/tex] and v= [tex] e^x [/tex]

    after using [tex] uv- \int v du [/tex] I seem to be going in circles...can someone help?
     
  2. jcsd
  3. May 9, 2005 #2

    quasar987

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    Do integration by parts on [itex]\int v du [/itex]. You'll end up with

    [tex]\int_{}^{} e^x \sin(\pi x) dx = something - \int_{}^{} e^x \sin(\pi x) dx [/tex]

    Aaah.. add [itex]\int_{}^{} e^x \sin(\pi x) dx [/itex] on both sides. Divide by two. ta-dam.
     
  4. May 9, 2005 #3

    shmoe

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    Small warning, it won't look quite like you've described, the constant won't be two.
     
  5. May 10, 2005 #4

    quasar987

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    Oh right.. because of the pies!
     
  6. May 10, 2005 #5

    Galileo

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    There's a quick alternative way using complex exponentials.

    Since
    [tex]e^x \sin (\pi x)= \Im (e^{x+i\pi x})=\Im (e^{x(1+i\pi)})[/tex]

    [tex]\int e^{x(1+i\pi)} dx=\frac{e^{x(1+i\pi)}}{1+i\pi}=\frac{(1-i\pi)e^{x(1+i\pi)}}{1+\pi^2}[/tex]

    Now take the imaginary part.
     
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