math

I'm trying to find the convolution f*g where f(t)=g(t)=sin(t). I set up the integral and proceed to do integration by parts twice, but it keeps working out to 0=0 or sin(t)=sin(t). How am I supposed to approach it? integral (sin(u)sin(t-u)) du from 0 to t.

asked by chuck
  1. If f(t)=g(t)=sin(t)
    The convolution would be
    ∫ sin(u)sin(t-u) du from 0 to t

    Use the identity:
    sin(x)sin(y)=(1/2)(cos(x-y)-cos(x+y))
    where x=u, y=t-u

    ∫ sin(u)sin(t-u) du from 0 to t
    =∫ (1/2)(cos(u-t+u)-cos(u+t-u))du
    =∫ (1/2)(cos(2u-t)-cos(t))du
    =(1/2)[(1/2)sin(2u-t)-ucos(t)] (from 0 to t)
    =(1/2)[(1/2)sin(t-(1/2)sin(-t)-tcos(t)]
    =(1/2)(sin(t)- t*cos(t))

    posted by MathMate
  2. I figured it out before I got your answer, and I used the trig identity sin(t-u)=sin(t)cos(u)-cos(t)sin(u). I ended up having to use four or five more trig substitutions before finally getting to the answer you have there. Your substitution is much easier to compute. Too bad I didn't check back here before going through all of that! Thank you!

    posted by chuck
  3. You're most welcome!
    Glad that it helped, and thank you for your feedback.

    posted by MathMate

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