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Question Number 161815 by HongKing last updated on 22/Dec/21

Answered by Lordose last updated on 23/Dec/21

$$\mathrm{\Omega }\left(\mathrm{x}\right)=\mathrm{\Gamma }\left(\frac{\mathrm{x}}{2}\right)\mathrm{\Gamma }(1-\frac{\mathrm{x}}{2})\mathrm{\Gamma }(\frac{\mathrm{x}}{2}+\frac{1}{2})\mathrm{\Gamma }(1-(\frac{\mathrm{x}}{2}+\frac{1}{2}))\sin \left(\pi \mathrm{x}\right)$$$$\mathrm{\Omega }\left(\mathrm{x}\right)=\mathit{\pi }\csc \left(\frac{\mathit{\pi }\mathrm{x}}{2}\right)\cdot \mathit{\pi }\csc \left(\frac{\mathit{\pi }(1+\mathrm{x})}{2}\right)\sin \left(\pi \mathrm{x}\right)$$$$\mathrm{\Omega }\left(\mathrm{x}\right)=2{\mathit{\pi }}^2$$$${\mathrm{x}}^2-\frac{4\mathrm{x}}{\mathrm{\Omega }\left(\mathrm{x}\right)}+\frac{1}{{\mathit{\pi }}^4}=0$$$${\mathrm{x}}^2-2\left(\frac{1}{{\mathit{\pi }}^2}\right)\mathrm{x}+{\left(\frac{1}{{\pi }^2}\right)}^2=0$$$${(\mathrm{x}-\frac{1}{{\mathit{\pi }}^2})}^2=0$$$$\mathrm{x}=\frac{1}{{\mathit{\pi }}^2}$$$$\mathit{\varnothing }\mathbf{sE}$$

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