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Notation
$T_n(x)$:Chebyshev polynomials of the first kind
$\zeta(s)$:Riemann zeta function
$$\ds\sum_{0< n}\frac{\cos nx}{n}=-\ln\left|2\sin\frac{x}{2}\right|\qquad(x\in\mathbb{R})$$
\begin{align} \sum_{0< n}\frac{\cos nx}{n}&=\sum_{0< n}\frac1n\Re(e^{inx}) \\ &=\Re(-\ln(1-e^{ix})) \\ &=-\Re\left(\ln\left(\frac{e^{\frac{ix}{2}}-e^{-\frac{ix}{2}}}{2i}\cdot\frac{2e^{\frac{ix}{2}}}{i}\right)\right) \\ &=-\Re\left(\ln\left(2\sin\frac{x}{2}\cdot\frac{e^{\frac{ix}{2}}}{i}\right)\right) \\ &=-\ln\left(\left|2\sin\frac{x}{2}\right|\cdot\left|\frac{e^{\frac{ix}{2}}}{i}\right|\right) \\ &=-\ln\left|2\sin\frac{x}{2}\right| \end{align}
$$\sum_{0< n}\frac{T_n(t)}{n}=-\frac12\ln(1-t)-\frac{\ln2}{2}\qquad(-1\leq t\leq1)$$
let$t$as$\cos x$in theorem1, and use equality$T_n(\cos x)=\cos(nx)$and$\ds\sin^2\frac{x}{2}=\frac{1-\cos x}{2}$
$$\int_0^{\pi/2}\tan^x\theta d\theta=\frac\pi2\sec\frac{\pi x}{2}\qquad(0\leq x<1)$$
\begin{align} \int_0^{\pi/2}\tan^x\theta d\theta&=\int_0^{\pi/2}\sin^x\theta\cos^{-x}\theta d\theta \\ &=\int_0^{\pi/2}\sin^{2\cdot\frac{1+x}{2}-1}\theta\cos^{2\cdot\frac{1-x}{2}-1}\theta d\theta \\ &=\frac12{\rm B}\left(\frac{1+x}{2},\frac{1-x}{2}\right) \\ &=\frac{\Gamma(\frac{1+x}{2})\Gamma(\frac{1-x}{2})}{2\Gamma(1)} \\ &=\frac{\pi}{2\sin\frac{\pi(1+x)}{2}} \\ &=\frac\pi2\csc\left(\frac\pi2+\frac{\pi x}{2}\right) \\ &=\frac\pi2\sec\frac{\pi x}{2} \end{align}
$$\sum_{0\leq n< m< l}\frac{(-1)^{m+l}}{(n+1/2)ml}$$