Otto Dunkel Memomorial April 17, 2020April 17, 2020 Tolaso Prove that Solution We have successively: since for it holds that Read more They might interest you:A productThe line is the shortest path between two pointsInverse zeta(3) limitTrigonometric sumA non existent limitAn integral with rootsDifferential equationParametric improper integralLimit of a multiple integral
![\[\sum_{n=1}^{\infty}\frac{1}{n}\int_{2n\pi}^{\infty}\frac{\sin z}{z}\,\mathrm{d}z=\pi-\frac{\pi \ln 2 \pi}{2}\]](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-12fbc3a69f049f8f4cf6449c89c031ac_l3.png "Rendered by QuickLaTeX.com")
![\begin{align*} \sum_{n=1}^{\infty} \frac{1}{n} \int_{2\pi n}^{\infty} \frac{\sin z}{z} \,\mathrm{d}z &= \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\infty} \frac{\sin nt}{t+2\pi} \, \mahrm{d}t \\ &=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n} \int_{0}^{2\pi} \frac{\sin nt}{t + 2 m \pi} \, \mathrm{d}t \\ &= \sum_{m=1}^{\infty} \int_{0}^{2\pi} \sum_{n=1}^{\infty} \frac{1}{n} \frac{\sin nt}{t+ 2 m \pi} \, \mathrm{d}t \\ &= \sum_{m=1}^{\infty} \int_{0}^{2\pi} \frac{\pi -t}{2\left (t + 2 m\pi \right )} \, \mathrm{d}t\\ &= \pi \sum_{n=1}^{\infty} \left [ \left ( 1 + \frac{1}{2} \right ) \log \left ( 1 + \frac{1}{n} \right ) -1 \right ] \\ &=\pi \log \left [ \lim_{N \rightarrow +\infty} e^{-N} \prod_{n=1}^{N} \left ( \frac{n+1}{n} \right )^{n+1/2} \right ] \\ &= \pi \log \left ( \lim_{N \rightarrow +\infty} \frac{\sqrt{N+1}\left ( N+1 \right )^N e^{-N}}{N!} \right ) \\ &= \pi \log \left ( \frac{e}{\sqrt{2\pi}} \right ) \\ &= \pi - \frac{ \pi \log 2\pi}{2} \end{align*}](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-c1cd853f99c207f562fbbedd9d4b31e8_l3.png "Rendered by QuickLaTeX.com")
it holds that![\[\sum_{n=1}^{\infty} \frac{\sin nt}{n} = \frac{\pi-t}{2}\]](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-c462fbb56d1e9cda2f42013d28553d4a_l3.png "Rendered by QuickLaTeX.com")
it holds that