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Question Number 139036 by mnjuly1970 last updated on 21/Apr/21

$$$$$$prove::$$$$\mathit{\varphi }={\int }_0^{\mathrm{\infty }}\frac{\sqrt{x}}{x^2+2x+5}dx=\frac{\pi }{2\sqrt{\phi }}$$$$\phi :=goldenratio...$$

Answered by Dwaipayan Shikari last updated on 21/Apr/21

$$\frac{1}{4i}{\int }_0^{\mathrm{\infty }}\frac{\sqrt{x}}{(x+1-2i)}-\frac{\sqrt{x}}{(x+1+2i)}dx$$$$=\frac{1}{2i}{\int }_0^{\mathrm{\infty }}\frac{t^2}{t^2+1-2i}-\frac{t^2}{t^2+1+2i}dt$$$$=\frac{1}{2i}{\int }_0^{\mathrm{\infty }}\frac{2i-1}{t^2+1-2i}+\frac{2i+1}{t^2+1+2i}dt=\frac{\pi }{4i}(\frac{2i-1}{\sqrt{1-2i}}+\frac{2i+1}{\sqrt{1+2i}})$$$$=\frac{\pi }{4i}(-\sqrt{1-2i}+\sqrt{1+2i})=\frac{\pi }{4i}\left(2i\sqrt{\frac{\sqrt{5}-1}{2}}\right)=\frac{\pi }{2\sqrt{\phi }}$$$$==$$$$\sqrt{x+\sqrt{-y}}=\sqrt{\frac{x+\sqrt{x^2+y}}{2}}+\sqrt{\frac{x-\sqrt{x^2+y}}{2}}$$$$\sqrt{x-\sqrt{-y}}=\sqrt{\frac{x+\sqrt{x^2+y}}{2}}-\sqrt{\frac{x-\sqrt{x^2+y}}{2}}$$$$\sqrt{1+2i}=\sqrt{\frac{1+\sqrt{5}}{2}}+i\sqrt{\frac{\sqrt{5}-1}{2}}$$$$\sqrt{1-2i}=\sqrt{\frac{1+\sqrt{5}}{2}}-i\sqrt{\frac{\sqrt{5}-1}{2}}$$

Commented by mnjuly1970 last updated on 21/Apr/21

$$thanksalot...$$

Answered by mnjuly1970 last updated on 21/Apr/21

Answered by mathmax by abdo last updated on 21/Apr/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{\sqrt{\mathrm{x}}}{{\mathrm{x}}^2+2\mathrm{x}+5}\mathrm{dx}\Rightarrow \mathrm{\Phi }{=}_{\sqrt{\mathrm{x}}=\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{t}\left(2\mathrm{t}\right)}{{\mathrm{t}}^4+{2\mathrm{t}}^2+5}\mathrm{dt}$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{t}}^2}{{\mathrm{t}}^4+{2\mathrm{t}}^2+5}\mathrm{dt}\mathrm{let}\mathrm{w}\left(\mathrm{z}\right)=\frac{{\mathrm{z}}^2}{{\mathrm{z}}^4+{2\mathrm{z}}^2+5}\mathrm{poles}\mathrm{of}\mathrm{w}!$$$${\mathrm{z}}^4+{2\mathrm{z}}^2+5=0\to {\mathrm{x}}^2+2\mathrm{x}+5=0({\mathrm{t}}^2=\mathrm{x})$$$$\mathrm{\Delta }=4-20=-16\Rightarrow {\mathrm{x}}_1=\frac{-2+\mathrm{i}\sqrt{20}}{2}=\frac{-2+2\mathrm{i}\sqrt{5}}{2}=-1+\mathrm{i}\sqrt{5}$$$${\mathrm{x}}_2=-1-\mathrm{i}\sqrt{5}\Rightarrow \mathrm{w}\left(\mathrm{z}\right)=\frac{{\mathrm{z}}^2}{({\mathrm{z}}^2+1-\mathrm{i}\sqrt{5})({\mathrm{z}}^2+1+\mathrm{i}\sqrt{5})}$$$$=\frac{{\mathrm{z}}^2}{(\mathrm{z}-\sqrt{{\mathrm{x}}_1})(\mathrm{z}+\sqrt{{\mathrm{x}}_1})(\mathrm{z}-\sqrt{{\mathrm{x}}_2})(\mathrm{z}+\sqrt{{\mathrm{x}}_2})}$$$${\mathrm{x}}_1=\sqrt{6}{\mathrm{e}}^{-\mathrm{iarctan}\left(\sqrt{5}\right)}\mathrm{and}{\mathrm{x}}_2=\sqrt{6}{\mathrm{e}}^{\mathrm{iarctan}\left(\sqrt{5}\right)}$$$$\Rightarrow \mathrm{w}\left(\mathrm{z}\right)=\frac{{\mathrm{z}}^2}{(\mathrm{z}-\sqrt{\sqrt{6}}{\mathrm{e}}^{-\frac{\mathrm{i}}{2}\arctan \left(\sqrt{5}\right)})(\mathrm{z}+\sqrt{\sqrt{6}}{\mathrm{e}}^{-\frac{\mathrm{i}}{2}\arctan \left(\sqrt{5}\right)})(\mathrm{z}-\sqrt{\sqrt{6}}{\mathrm{e}}^{\frac{\mathrm{i}}{2}\arctan \left(\sqrt{5}\right)})(\mathrm{z}+\sqrt{\sqrt{6}}{\mathrm{e}}^{\frac{\mathrm{i}}{2}\arctan \left(\sqrt{5}\right)})}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{w}\left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\mathrm{w},-\sqrt{\sqrt{6}}{\mathrm{e}}^{-\frac{\mathrm{i}}{2}\arctan \left(\sqrt{5}\right)}+\mathrm{Res}(\mathrm{w},\sqrt{\sqrt{6}}{\mathrm{e}}^{\frac{\mathrm{i}}{2}\arctan \left(\sqrt{5}\right)}\}$$$$\mathrm{Res}(\mathrm{w},\sqrt{\sqrt{6}}{\mathrm{e}}^{\frac{\mathrm{i}}{2}\arctan \left(\sqrt{5}\right)})=\frac{\sqrt{6}{\mathrm{e}}^{\mathrm{iarctan}\left(\sqrt{5}\right)}}{2\sqrt{\sqrt{6}}{\mathrm{e}}^{\frac{\mathrm{i}}{2}\arctan \left(\sqrt{5}\right)}\sqrt{6}(2\mathrm{isin}(\arctan \left(\sqrt{5}\right)}$$$$=\frac{1}{4\mathrm{i}}\frac{{\mathrm{e}}^{\frac{\mathrm{i}}{2}\arctan \left(\sqrt{5}\right)}}{\sqrt{\sqrt{6}}\sin \left(\arctan \left(\sqrt{5}\right)\right)}....\mathrm{be}\mathrm{continued}...$$

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