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In the process to compute $\int_0^{+\infty} \frac{t^{m-1}}{1+t^{2n}}{\rm d}t$ (see this thread), I got stuck on this sum : $$S_{m,n} = \sum_{k=0}^{n-1} \sin(m\theta_k)(\pi-\theta_k)$$ where $m$ and $n$ are integers, $1\le m<2n$, and $\theta_k=\dfrac{(2k+1)\pi}{2n}$.

Knowing the result, I tried to multiply this sum by $\sin\frac{m\pi}{2n}$ to get a telescopic form, but I got stuck on something horrible, like this : $$\frac{1}{2n}\left[\pi-\theta_0 -\frac\pi n \sum_{k=1}^{n-1} \cos\frac{mk\pi}{n} + \cos(m\pi)(\pi-\theta_{n-1})\right]$$ and I don't know what to do, because the sum in the middle does only "simplifies" in $$\frac12\frac{(-1)^{m+1}\sin\frac{\pi}{2m}-1}{\sin\frac{m\pi}{2n}}$$ if I'm not mistaken (I should be !).

Do you see anything simpler than what I did ?

Thanks for reading.

Nicolas FRANCOIS
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2 Answers2

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$$S_{m,n}=\sum_{k=0}^{n-1} \sin \left(m\frac{\pi (2 k+1) }{2 n}\right)\left(\pi -\frac{\pi (2 k+1)}{2 n}\right)$$ Before any simplification (given by a CAS) $$S_{m,n}=\frac{2 \pi n \csc \left(\frac{\pi m}{2 n}\right)-\pi \sin \left(\frac{\pi m (2 n+1)}{2 n}\right) \csc ^2\left(\frac{\pi m}{2 n}\right)-\pi \sin \left(\frac{1}{2} (2 \pi m-\pi )\right) \csc \left(\frac{\pi m}{2 n}\right)}{4 n}$$ After simplification (have a look here) $$S_{m,n}=\frac{\pi \csc \left(\frac{\pi m}{2 n}\right) \left(2 n-\sin (\pi m) \cot \left(\frac{\pi m}{2 n}\right)\right)}{4 n}$$ Since $m$ is an integer $$S_{m,n}=\frac{1}{2} \pi \csc \left(\frac{\pi m}{2 n}\right)$$

Claude Leibovici
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$S_{m, n}$ is just the imaginary part of $\sum e^{im\theta_k}(\pi -\theta_k)$.

$\sum e^{im\theta_k}$ is just a geometric series, so a closed form is available.

As for $\sum e^{im \theta_k} \theta_k$, note that it's closely related to the derivative of $\sum e^{im \theta_k}$ with respect to $m$.

Just a user
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