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Dirichlet regularization

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hbghlyj Posted at 2024-4-24 19:21:12 |Read mode
$$\sum_{n=1}^k \sin (n x)=\csc \left(\frac{x}{2}\right) \sin \left(\frac{k x}{2}\right) \sin \left(\frac{1}{2}(k+1) x\right)$$
从而$\sum_{n=1}^\infty \sin (n x)$发散。
但是可以Dirichlet regularization:
$$\lim _{s \rightarrow 0}\left(\sum_{n=1}^{\infty} n^{-s} \sin (n x)\right)=\frac{1}{2} \cot \left(\frac{x}{2}\right)$$为什么


类似:regulator method $1+2+3+\cdots=-\frac1{12}$

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 Author| hbghlyj Posted at 2024-6-2 21:52:04
Last edited by hbghlyj at 2024-6-3 19:01:00$ \int_2^{\infty} \frac{1}{k}\sin(kr) \mathrm{d}r $发散。但如果乘以$e^{-\varepsilon r}$:
mathematica.stackexchange.com/questions/22986 … -to-compute-integral
$Q(k) = \lim \limits_{\varepsilon \to 0} \int_2^{\infty} \frac{e^{-\varepsilon r}}{k}\sin(kr) \mathrm{d}r = \frac{\cos(2k)}{k^2}$
$$\int_2^{\infty} \frac{e^{-\varepsilon r}}{k}\sin(kr) \mathrm{d}r =[-\frac{e^{-\varepsilon r}(\varepsilon \sin (k r)+k \cos (k r))}{k\left(\varepsilon^2+k^2\right)}]^\infty_{r=2}=\frac{e^{-2\varepsilon}(\varepsilon \sin (2k)+k \cos (2k))}{k\left(\varepsilon^2+k^2\right)}\xrightarrow[\varepsilon\to0]{}\frac{\cos(2k)}{k^2}$$

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2025-4-20 21:55 GMT+8

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