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|sin(t)| = sin(t) on [0, pi], and |sin(t)| = -sin(t) on [pi, 2pi] or on [-pi, 0]

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|sin(t)| = sin(t) on [0, pi], and |sin(t)| = -sin(t) on [pi, 2pi] or on [-pi, 0]

Thanks - I thought of this as well, but this would mean I have to integrate on each interval, and I get sum(n=0, n=inf) ((1+exp(pi*s)/exp(n*pi*s)*(s^2+1)). Is there a way to simplify this? I'm supposed to be using Laplace transforms to solve a differential equation with |sint| as the inhomogeneous part.

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Think geometric series where r=exp(-pi*s).

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Ah of course, thanks :)Think geometric series where r=exp(-pi*s).

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