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Given the following integral: $I(N)=\int_{0}^{t}\prod_{k=1}^{N}\sin(k\omega \tau)d\tau$ does someone know if is it possible to find the solution of $I(N)$ in a closed form? I'm able to find the formula for $N=2$ $I(2)=\frac{1}{2}\frac{\sin(\omega t)}{\omega}-\frac{1}{6}\frac{\sin(3\omega t)}{\omega}$ and for $N=3$ $I(3)=-\frac{1}{16}\frac{\cos(4\omega t)}{\omega}-\frac{1}{8}\frac{\cos(2\omega t)}{\omega}+\frac{1}{24}\frac{\cos(6\omega t)}{\omega}$ but I'm not able to find $I(N)$ Thank you in advance.

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    Integrating by parts one can get $I_N(t) = I_{N-1}(t) \sin(N\omega t) - \int_0^t N\omega \cos (N\omega \tau) I_{N-1}(\tau) \mathrm{d}\tau$ Don't know whether the recursive formula is helpful though.2012-06-01

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