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A brief solution with the help of complex analysis techniques would be by defining the function \(\displaystyle f(z)=\frac{\pi \csc(\pi z)\sin(z)}{n}\), then \[\sum_{n=-\infty,\,n\neq 0}^{\infty}\frac{(-1)^{n}\sin(n)}{n}=-\textrm{Residue}\left(f(z);z=0\right)=-1\Rightarrow\sum_{n=-\infty,\,n\neq 0}^...

We will use the identity \(\displaystyle \Gamma(z+1)=z\cdot \Gamma(z)\), so \[ \int_{0}^{1}\log \left(\Gamma(x+1)\right)\;\mathbb{d}x = \int_{0}^{1}\ln\left(x\cdot \Gamma(x)\right)\;\mathbb{d}x=\int_{0}^{1}\log(x)\;\mathbb{d}x+\int_{0}^{1}\log\left(\Gamma(x)\right)\;\mathbb{d}x=-1+\underbrace{\int_{...

Consider the sawtooth wave \(f(x)=x,\;\;x\in(-\pi,\pi)\) and \(f(x+2n\pi)=f(x)\) for \(x\in \mathbb{R}\) and \(n\in\mathbb{Z}\). sawtoothwave.gif [/centre] Recall the fourier series formula \(\displaystyle f(x)=\frac{a_{0}}{2}+\sum_{k=1}^{N}\left[a_{k}\cos(kx)+b_{k}\sin(kx)\right]\), with \(\display...