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According to Mathematica,

$$ \begin{aligned} \sum_{t=0}^{\infty}t\cos x t &= -\frac{1}{4} \csc ^2\left(\frac{x }{2}\right) \\ \sum_{t=0}^{\infty}t^2\cos x t &= 0 \\ \sum_{t=0}^{\infty}t^3\cos x t &= \frac{1}{8} (\cos x +2) \csc ^4\left(\frac{x}{2}\right) \end{aligned} $$

I'm not sure how the results are obtained. I tried using the fact that $\frac{\mathrm{d}}{\mathrm{d}x}\sin xt = t \cos xt$ and

$$\sum_{t=0}^{N} \sin xt = \csc \left(\frac{x}{2}\right) \sin \left(\frac{N x}{2}\right) \sin \left(\frac{1}{2} (N+1) x\right)$$

But I don't know how to find the limit of the derivative at $N\to\infty$

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1 Answer 1

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Consider $$\sum_{t=0}^\infty t\cos tx +i t\sin tx =\sum_{t=0}^\infty t e^{itx}.$$

Since \begin{align} \sum_{t=0}^\infty t e^{itx} &=\frac{e^{ix}}{\left(1-e^{ix}\right)^2}\\&= \frac{e^{ix}}{1-2e^{ix}+e^{i2x}}\\&=\frac{1}{e^{ix}+e^{-ix}-2}\\&=\frac{1}{2\cos x-2}\\&=-\frac{1}{2}\frac{1}{1-\cos x}\\&=-\frac{1}{4}\csc^2 \frac{x}{2}. \end{align}

We obtain what we want. Similar to $\alpha\in\mathbb{N}$.

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