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Question Number 126494 by mathmax by abdo last updated on 21/Dec/20

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{sinx}}{\mathrm{x}({\mathrm{x}}^4+1)}\mathrm{dx}$$

Answered by mathmax by abdo last updated on 22/Dec/20

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{sinx}}{\mathrm{x}({\mathrm{x}}^4+1)}\mathrm{dx}\Rightarrow 2\mathrm{I}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{sinx}}{\mathrm{x}({\mathrm{x}}^4+1)}\mathrm{dx}=\mathrm{Im}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{ix}}}{\mathrm{x}({\mathrm{x}}^4+1)}\mathrm{dx}\right)$$$$\mathrm{let}\mathrm{consider}\mathrm{the}\mathrm{complex}\mathrm{function}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iz}}}{\mathrm{z}({\mathrm{z}}^4+1)}\mathrm{poles}\mathrm{of}\phi ?$$$$\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iz}}}{\mathrm{z}({\mathrm{z}}^2-\mathrm{i})({\mathrm{z}}^2+\mathrm{i})}=\frac{{\mathrm{e}}^{\mathrm{iz}}}{\mathrm{z}(\mathrm{z}-{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})(\mathrm{z}+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})(\mathrm{z}-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})(\mathrm{z}+{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})}$$$$\mathrm{so}\mathrm{the}\mathrm{poles}\mathrm{are}0,\stackrel{-}{+}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}\mathrm{and}\stackrel{-}{+}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}\mathrm{residus}\mathrm{theorem}\mathrm{give}$$$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\phi ,0)+\mathrm{Res}(\phi ,{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})+\mathrm{Res}(\phi ,-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})\}$$$$\mathrm{Res}(\phi ,0)=1$$$$\mathrm{Res}(\phi ,{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})=\frac{{\mathrm{e}}^{{\mathrm{ie}}^{\frac{\mathrm{i}\pi }{4}}}}{{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}.{2\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}\left(2\mathrm{i}\right)}=\frac{{\mathrm{e}}^{\mathrm{i}(\frac{1}{\sqrt{2}}+\frac{\mathrm{i}}{\sqrt{2}})}}{\left(2\mathrm{i}\right)\left(2\mathrm{i}\right)}=-\frac{1}{4}{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}(\cos \left(\frac{1}{\sqrt{2}}\right)+\mathrm{isin}\left(\frac{1}{\sqrt{2}}\right))$$$$\mathrm{Res}(\phi ,-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})=\frac{{\mathrm{e}}^{-{\mathrm{ie}}^{-\frac{\mathrm{i}\pi }{4}}}}{(-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})(-{2\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})(-2\mathrm{i})}=\frac{{\mathrm{e}}^{-{\mathrm{ie}}^{-\frac{\mathrm{i}\pi }{4}}}}{-4\mathrm{i}(-\mathrm{i})}$$$$=-\frac{1}{4}{\mathrm{e}}^{-\mathrm{i}(\frac{1}{\sqrt{2}}-\frac{\mathrm{i}}{\sqrt{2}})}=-\frac{1}{4}{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}{\mathrm{e}}^{-\frac{\mathrm{i}}{\sqrt{2}}}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{1-\frac{1}{4}{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}{\mathrm{e}}^{\frac{\mathrm{i}}{\sqrt{2}}}-\frac{1}{4}{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}{\mathrm{e}}^{-\frac{\mathrm{i}}{\sqrt{2}}}\}$$$$=2\mathrm{i}\pi \{1-\frac{1}{4}{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}({\mathrm{e}}^{\frac{\mathrm{i}}{\sqrt{2}}}+{\mathrm{e}}^{-\frac{\mathrm{i}}{\sqrt{2}}})\}$$$$=2\mathrm{i}\pi \{1-\frac{1}{2}{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}\cos \left(\frac{1}{\sqrt{2}}\right)\}\Rightarrow$$$$2\mathrm{I}=2\pi \{1-\frac{1}{2}{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}\cos \left(\frac{1}{\sqrt{2}}\right)\}\Rightarrow \mathrm{I}=\pi -\frac{\pi }{2}{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}\cos \left(\frac{1}{\sqrt{2}}\right)$$

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