Posted by chuck on Thursday, December 3, 2009 at 2:38pm.
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(tu)) du from 0 to t.

math  MathMate, Thursday, December 3, 2009 at 4:20pm
If f(t)=g(t)=sin(t)
The convolution would be
∫ sin(u)sin(tu) du from 0 to t
Use the identity:
sin(x)sin(y)=(1/2)(cos(xy)cos(x+y))
where x=u, y=tu
∫ sin(u)sin(tu) du from 0 to t
=∫ (1/2)(cos(ut+u)cos(u+tu))du
=∫ (1/2)(cos(2ut)cos(t))du
=(1/2)[(1/2)sin(2ut)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))

math  chuck, Thursday, December 3, 2009 at 5:10pm
I figured it out before I got your answer, and I used the trig identity sin(tu)=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!

math  MathMate, Thursday, December 3, 2009 at 5:23pm
You're most welcome!
Glad that it helped, and thank you for your feedback.
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