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Question Number 152464 by mathdanisur last updated on 28/Aug/21

Answered by mindispower last updated on 29/Aug/21

$$\frac{sin\left(z\right)}{z}=\prod _{n⩾1}(1-\frac{z^2}{n^2{\pi }^2})$$$$\frac{sh\left(t\right)}{t}=\prod _{n⩾1}(1+\frac{t^2}{n^2{\pi }^2})$$$$ln\left(\frac{sh\left(t\right)}{t}\right)=\sum _{n⩾1}ln(1+\frac{t^2}{n^2{\pi }^2})$$$$\sum _{n⩾1}{\int }_0^1\frac{{ln}^{2m}\left(t\right)\sum _{k⩾0}{(-1)}^k{\left(\frac{t^2}{n^2{\pi }^2}\right)}^{k+1}\frac{1}{k+1}}{t}$$$$=\sum _{n⩾1}\sum _{k⩾1}\frac{1}{k}.\frac{1}{{\left(n\pi \right)}^{2k}}{\int }_0^1{(-1)}^{k-1}{t}^{2k-1}{ln}^{2m}\left(t\right)dt$$$$=\sum _{k⩾1}\frac{{(-1)}^{k-1}}{k}\sum _{n⩾1}\frac{1}{{\left(n\pi \right)}^{2k}}{\int }_0^{\mathrm{\infty }}{e}^{-\left(2k\right)x}{x}^{2m}dx$$$$=\sum _{k⩾1}\frac{{(-1)}^{k-1}}{k+1}\sum _{n⩾1}{\left(\frac{1}{n\pi }\right)}^{2k}\frac{\mathrm{\Gamma }(2m+1)}{{\left(2k\right)}^{2m+1}}$$$$=\sum _{k⩾1}\frac{{(-1)}^{k-1}}{k+1}.\frac{\left(2m\right)!}{{\left(2k\right)}^{2m+1}.{\pi }^{2k}}\zeta \left(2k\right)...A$$$$\zeta \left(2k\right)=B_{2k}.\frac{2^{2k-1}}{\left(2k\right)!}{\pi }^{2k}$$$$A\iff \sum _{k⩾1}\frac{{(-1)}^{k-1}}{k}.\frac{\left(2m\right)!}{{\left(2k\right)}^{2m+1}{\pi }^{2k}}.2^{2k-1}{\pi }^{2k}.\frac{1}{\left(2k\right)!}{B}_{2k}$$$$=\sum _{k⩾1}\frac{{(-1)}^{k-1}}{k}.\frac{2^{2(k-m-1)}\left(2m\right)!}{k^{2m+1}\left(2k\right)!}{B}_{2k}$$$$=\sum _{k⩾1}\frac{{(-1)}^{k-1}{2}^{2(k-\mathit{m}-1)}\left(2\mathit{m}\right)!}{k^{2(m+1)}\left(2k\right)!}{B}_{2k}$$$$$$$$$$

Commented by mathdanisur last updated on 29/Aug/21

$$\mathrm{Thank}\mathrm{you}\mathrm{Ser}$$

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