三角関数の部分分数展開

三角関数は以下のように部分分数展開できる。

(1)

\begin{align*} \pi\tan\pi x & =-\sum_{k=0}^{\infty}\frac{2x}{x^{2}-\left(k+\frac{1}{2}\right)^{2}}\\ & =-\sum_{k=-\infty}^{\infty}\frac{1}{x+\frac{1}{2}+k} \end{align*}

(2)

\begin{align*} \pi\cot\pi x & =\frac{1}{x}+\sum_{k=1}^{\infty}\frac{2x}{x^{2}-k^{2}}\\ & =\sum_{k=-\infty}^{\infty}\frac{1}{x+k} \end{align*}

(3)

\begin{align*} \frac{\pi}{\sin\pi x} & =\sum_{k=-\infty}^{\infty}\frac{(-1)^{k}}{x+k}\\ & =\frac{1}{x}+\sum_{k=1}^{\infty}\frac{(-1)^{k}2x}{x^{2}+k^{2}} \end{align*}

(4)

\begin{align*} \frac{\pi}{\cos\pi x} & =\sum_{k=-\infty}^{\infty}\frac{(-1)^{k}}{x+\frac{1}{2}+k}\\ & =\sum_{k=0}^{\infty}\frac{(-1)^{k}(2k+1)}{x^{2}-\left(\frac{1}{2}+k\right)^{2}} \end{align*}

(1)

\begin{align*} \pi\tan\pi x & =-\frac{d}{dx}\log\left(\cos\pi x\right)\\ & =-\frac{d}{dx}\log\prod_{k=1}^{\infty}\left(1-\frac{x^{2}}{\left(k-\frac{1}{2}\right)^{2}}\right)\\ & =-\sum_{k=1}^{\infty}\frac{d}{dx}\log\left(1-\frac{x^{2}}{\left(k-\frac{1}{2}\right)^{2}}\right)\\ & =-\sum_{k=1}^{\infty}\frac{2x}{x^{2}-\left(k-\frac{1}{2}\right)^{2}}\\ & =-\sum_{k=0}^{\infty}\frac{2x}{x^{2}-\left(k+\frac{1}{2}\right)^{2}}\tag{*}\\ & =-\sum_{k=0}^{\infty}\left\{ \frac{1}{x-\left(k+\frac{1}{2}\right)}+\frac{1}{x+\left(k+\frac{1}{2}\right)}\right\} \\ & =-\sum_{k=-\infty}^{\infty}\frac{1}{x+\frac{1}{2}+k} \end{align*}

(2)

\begin{align*} \pi\cot\pi x & =\frac{d}{dx}\log\left(\sin\pi x\right)\\ & =\frac{d}{dx}\log\left\{ \pi x\prod_{k=1}^{\infty}\left(1-\frac{x^{2}}{k^{2}}\right)\right\} \\ & =\frac{d}{dx}\log\pi x+\sum_{k=1}^{\infty}\frac{d}{dx}\log\left(1-\frac{x^{2}}{k^{2}}\right)\\ & =\frac{1}{x}+\sum_{k=1}^{\infty}\frac{2x}{x^{2}-k^{2}}\tag{*}\\ & =\frac{1}{x}+\sum_{k=1}^{\infty}\frac{1}{x-k}+\frac{1}{x+k}\\ & =\sum_{k=-\infty}^{\infty}\frac{1}{x+k} \end{align*}

(3)

\begin{align*} \frac{\pi}{\sin\pi x} & =\frac{\pi\left(\sin^{2}\frac{\pi x}{2}+\cos^{2}\frac{\pi x}{2}\right)}{2\sin\frac{\pi x}{2}\cos\frac{\pi x}{2}}\\ & =\frac{\pi}{2}\cot\frac{\pi x}{2}+\frac{\pi}{2}\tan\frac{\pi x}{2}\\ & =\frac{1}{2}\sum_{k=-\infty}^{\infty}\frac{2}{x+2k}-\frac{1}{2}\sum_{k=-\infty}^{\infty}\frac{2}{x+1+2i}\\ & =\sum_{k=-\infty}^{\infty}\frac{(-1)^{k}}{x+i}\tag{*}\\ & =\frac{1}{x}+\sum_{k=1}^{\infty}\left\{ \frac{(-1)^{k}}{x+i}+\frac{(-1)^{-k}}{x-i}\right\} \\ & =\frac{1}{x}+\sum_{k=1}^{\infty}\frac{(-1)^{k}2x}{x^{2}+i^{2}} \end{align*}

(4)

\begin{align*} \frac{\pi}{\cos\pi x} & =\frac{\pi}{\sin\pi\left(x+\frac{1}{2}\right)}\\ & =\sum_{i=-\infty}^{\infty}\frac{(-1)^{k}}{x+\frac{1}{2}+k}\tag{*}\\ & =\frac{1}{x+\frac{1}{2}}+\sum_{k=1}^{\infty}\left\{ \frac{(-1)^{k}}{x+\frac{1}{2}+k}+\frac{(-1)^{-k}}{x+\frac{1}{2}-k}\right\} \\ & =\sum_{k=1}^{\infty}\left\{ \frac{(-1)^{k-1}}{x-\frac{1}{2}+k}+\frac{(-1)^{-k}}{x+\frac{1}{2}-k}\right\} \\ & =\sum_{k=1}^{\infty}\frac{(-1)^{k-1}(2k-1)}{x^{2}-\left(\frac{1}{2}-k\right)^{2}}\\ & =\sum_{k=0}^{\infty}\frac{(-1)^{k}(2k+1)}{x^{2}-\left(\frac{1}{2}+k\right)^{2}} \end{align*}

1つ証明出来ればもう一つは以下を使って簡単に証明できる。
\begin{align*} \text{・} & \tan x=-\cot\left(x+\frac{1}{2}\right)\\ \text{・} & \cot x=-\tan\left(x-\frac{1}{2}\right)\\ \text{・} & \sin x=\cos\left(x-\frac{1}{2}\right)\\ \text{・} & \cos x=\sin\left(x+\frac{1}{2}\right) \end{align*}

双曲線関数は以下のように部分分数展開できる。

(1)

\begin{align*} \pi\tanh\pi x & =\sum_{k=0}^{\infty}\frac{2x}{x^{2}+\left(k+\frac{1}{2}\right)^{2}}\\ & =\sum_{k=-\infty}^{\infty}\frac{1}{x-\left(\frac{1}{2}+k\right)i} \end{align*}

(2)

\begin{align*} \pi\coth\pi x & =\frac{1}{x}+\sum_{k=1}^{\infty}\frac{2x}{x^{2}+k^{2}}\\ & =\sum_{k=-\infty}^{\infty}\frac{1}{x-ki} \end{align*}

(3)

\begin{align*} \frac{\pi}{\sinh\pi x} & =-\sum_{k=-\infty}^{\infty}\frac{(-1)^{k}}{x-ki}\\ & =-\frac{1}{x}-\sum_{k=1}^{\infty}\frac{(-1)^{k}2x}{x^{2}-k^{2}} \end{align*}

(4)

\begin{align*} \frac{\pi}{\cosh\pi x} & =-i\sum_{k=-\infty}^{\infty}\frac{(-1)^{k}}{x-\left(\frac{1}{2}+k\right)i}\\ & =-\sum_{k=0}^{\infty}\frac{(-1)^{k}(2k+1)}{x^{2}+\left(\frac{1}{2}+k\right)^{2}} \end{align*}

(1)

\begin{align*} \pi\tanh\pi x & =-i\pi\tan(ix\pi)\\ & =\sum_{k=0}^{\infty}\frac{2x}{x^{2}+\left(k+\frac{1}{2}\right)^{2}}\\ & =\sum_{k=-\infty}^{\infty}\frac{1}{x-\left(\frac{1}{2}+k\right)i} \end{align*}

(2)

\begin{align*} \pi\coth\pi x & =i\pi\coth(ix\pi)\\ & =\frac{1}{x}+\sum_{k=1}^{\infty}\frac{2x}{x^{2}+k^{2}}\\ & =\sum_{k=-\infty}^{\infty}\frac{1}{x-ki} \end{align*}

(3)

\begin{align*} \frac{\pi}{\sinh\pi x} & =\frac{\pi}{-i\sin(i\pi x)}\\ & =-\sum_{k=-\infty}^{\infty}\frac{(-1)^{k}}{x-ki}\\ & =-\frac{1}{x}-\sum_{k=1}^{\infty}\frac{(-1)^{k}2x}{x^{2}-k^{2}} \end{align*}

(4)

\begin{align*} \frac{\pi}{\cosh\pi x} & =\frac{\pi}{\cos(i\pi x)}\\ & =-i\sum_{k=-\infty}^{\infty}\frac{(-1)^{k}}{x-\left(\frac{1}{2}+k\right)i}\\ & =-\sum_{k=0}^{\infty}\frac{(-1)^{k}(2k+1)}{x^{2}+\left(\frac{1}{2}+k\right)^{2}} \end{align*}

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三角関数の部分分数展開

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