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\begin{aligned} \cos \dfrac{z}{2}=\displaystyle \prod_{k\geq 0}\left(1-\frac{z^2}{\pi^2(2k+1)^2}\right)\Rightarrow \tan \frac{z}{2}=-2\frac{d}{dz}\ln \cos \frac{z}{2}=\sum_{k\geq 0}\frac{4z}{\pi^2(2k+1)^2-z^2}
\begin{aligned}\displaystyle \int_0^{\infty}\frac{2-2\cos x-x\sin x}{x^4}\,dx=\frac{1}{3} \int_0^{\infty}\frac{\sin x-x\cos x}{x^3}\,dx=\frac{1}{6}\int_0^{ \infty} \frac{\sin x}{x}\,dx =\frac{\pi}{12}
\begin{aligned}\displaystyle \int_0^{\infty}\frac{2-2\cos x-x\sin x}{x^4}\,dx=\frac{1}{3} \int_0^{\infty}\frac{x\cos x-\sin x}{x^3}\,dx
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