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.

Answer 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))

Answer 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!

Answer 3

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