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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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My first thought is to find formulas for a couple more cases ($N=4$, $N=5$) and see if you can find a pattern. – Ailurus Jun 1 '12 at 9:55
If you look to $\prod_{k=1}^N \sin(k \omega \tau)$ up to $N=6$, the following pattern emerges: you can rewrite the product as a summation of sines (when $N$ is odd) or as a summation of cosines (when $N$ is even). The constant in front of the summation is $\frac{1}{2^{N-1}}$. Not sure about the pattern for the arguments of the sines (or cosines). – Ailurus Jun 1 '12 at 10:24
Hmm one other idea, perhaps you can write a recursive expression for the mentioned summation. I mean, if you know for instance the summation for $N=3$, the one for $N=4$ can of course be found by multiplying this summation by $\sin(4\omega \tau)$. Using the equalities $\sin(a)\sin(b) = \frac{1}{2}(\cos(a-b)-\cos(a+b))$ and $\cos(a)\sin(b) = \frac{1}{2}(\sin(a+b)-\sin(a-b))$, you can then write this expression again as a summation of purely sines or cosines. – Ailurus Jun 1 '12 at 10:45
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. – Willie Wong Jun 1 '12 at 11:06

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