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Question Number 145911 by Mrsof last updated on 09/Jul/21

Commented by Mrsof last updated on 09/Jul/21

$$helpmesirpleasebycomplexnumber$$

Answered by Dwaipayan Shikari last updated on 09/Jul/21

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sin \left(x\right)}{{(x+1)}^2+1}dx$$$$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sin (u-1)}{u^2+1}du$$$$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sin \left(u\right)\cos \left(1\right)-\sin \left(1\right)\cos \left(u\right)}{u^2+1}du$$$$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}oddfuctiondu-{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\cos \left(u\right)}{u^2+1}du$$$$=0-2{\int }_0^{\mathrm{\infty }}\frac{\cos \left(u\right)}{u^2+1}du$$$$=-2\frac{\pi }{2}{e}^{-1}=-\frac{\pi }{e}$$

Commented by Mrsof last updated on 09/Jul/21

$$sircanyouhelpmebyresideo$$

Answered by Mathspace last updated on 09/Jul/21

$$areyoustudentorteachermsof..$$

Answered by mathmax by abdo last updated on 09/Jul/21

$$\mathrm{{\rm Y}}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{sinx}}{{\mathrm{x}}^2+2\mathrm{x}+2}\mathrm{dx}\Rightarrow \mathrm{{\rm Y}}=\mathrm{Im}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{ix}}}{{\mathrm{x}}^2+2\mathrm{x}+2}\mathrm{dx}\right)$$$$\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iz}}}{{\mathrm{z}}^2+2\mathrm{z}+2}\mathrm{poles}\mathrm{of}\phi ?$$$${\mathrm{\Delta }}^{{}^{\prime }}=1-2=-1\Rightarrow {\mathrm{z}}_1=-1+\mathrm{i}\mathrm{and}{\mathrm{z}}_2=-1-\mathrm{i}$$$$\mathrm{residus}\mathrm{theorem}\mathrm{give}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,{\mathrm{z}}_1)$$$$\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iz}}}{(\mathrm{z}-{\mathrm{z}}_1)(\mathrm{z}-{\mathrm{z}}_2)}\Rightarrow \mathrm{Res}(\phi ,{\mathrm{z}}_1)={\lim }_{\mathrm{z}\to {\mathrm{z}}_1}(\mathrm{z}-{\mathrm{z}}_1)\phi \left(\mathrm{z}\right)$$$$={\lim }_{\mathrm{z}\to {\mathrm{z}}_1}\frac{{\mathrm{e}}^{{\mathrm{iz}}_{}}}{{\mathrm{z}}_1-{\mathrm{z}}_2}=\frac{{\mathrm{e}}^{{\mathrm{iz}}_1}}{2\mathrm{i}}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi .\frac{{\mathrm{e}}^{{\mathrm{iz}}_1}}{2\mathrm{i}}$$$$=\pi {\mathrm{e}}^{\mathrm{i}(-1+\mathrm{i})}=\pi {\mathrm{e}}^{-1-\mathrm{i}}=\pi {\mathrm{e}}^{-1}\{\cos \left(1\right)-\mathrm{isin}\left(1\right)\}\Rightarrow$$$$=\pi {\mathrm{e}}^{-1}\cos \left(1\right)-\mathrm{i}\pi {\mathrm{e}}^{-1}\sin \left(1\right)\Rightarrow \mathrm{{\rm Y}}=-\frac{\pi }{\mathrm{e}}\sin \left(1\right)$$$$$$

Answered by mathmax by abdo last updated on 09/Jul/21

$$\mathrm{\Psi }={\int }_0^{\mathrm{\infty }}\frac{\mathrm{cosx}}{{\mathrm{x}}^4+1}\mathrm{dx}\Rightarrow 2\mathrm{\Psi }={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{cosx}}{{\mathrm{x}}^4+1}\mathrm{dx}=\mathrm{Re}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{ix}}}{{\mathrm{x}}^4+1}\mathrm{dx}\right)$$$$\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iz}}}{{\mathrm{z}}^4+1}\Rightarrow \phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iz}}}{({\mathrm{z}}^2-\mathrm{i})({\mathrm{z}}^2+\mathrm{i})}=\frac{{\mathrm{e}}^{\mathrm{iz}}}{(\mathrm{z}-{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})(\mathrm{z}+{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})(\mathrm{z}-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})(\mathrm{z}+{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})}$$$$\mathrm{the}\mathrm{poles}\mathrm{of}\phi \mathrm{are}\stackrel{-}{+}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}\mathrm{and}\stackrel{-}{+}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\phi ,{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})+\mathrm{Res}(\phi ,-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})\}$$$$\mathrm{Res}(\phi ,{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}})=\frac{{\mathrm{e}}^{{\mathrm{ie}}^{\frac{\mathrm{i}\pi }{4}}}}{{2\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}\left(2\mathrm{i}\right)}=\frac{1}{4\mathrm{i}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}{\mathrm{e}}^{\mathrm{i}(\frac{1}{\sqrt{2}}+\frac{\mathrm{i}}{\sqrt{2}})}$$$$=\frac{1}{4\mathrm{i}}{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}{\mathrm{e}}^{\frac{\mathrm{i}}{\sqrt{2}}}{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}=\frac{{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}}{4\mathrm{i}}{\mathrm{e}}^{\mathrm{i}(\frac{1}{\sqrt{2}}-\frac{\pi }{4})}$$$$\mathrm{Res}(\phi ,-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})=\frac{{\mathrm{e}}^{-{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}}}{-{2\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}}(-2\mathrm{i})}=\frac{1}{4\mathrm{i}}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}{\mathrm{e}}^{-(\frac{1}{\sqrt{2}}-\frac{\mathrm{i}}{\sqrt{2}})}$$$$=\frac{{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}}{4\mathrm{i}}{\mathrm{e}}^{\mathrm{i}(\frac{1}{\sqrt{2}}+\frac{\pi }{4})}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\frac{{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}}{4\mathrm{i}}{\mathrm{e}}^{\mathrm{i}(\frac{1}{\sqrt{2}}-\frac{\pi }{4})}+\frac{{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}}{4\mathrm{i}}{\mathrm{e}}^{\mathrm{i}(\frac{1}{\sqrt{2}}+\frac{\pi }{4})}\}$$$$=\frac{\pi }{2}{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}\left\{{\mathrm{e}}^{\frac{\mathrm{i}}{\sqrt{2}}}({\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}+{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{4}})\right\}$$$$=\frac{\pi }{2}{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}\left(2\frac{1}{\sqrt{2}}\right){\mathrm{e}}^{\frac{\mathrm{i}}{\sqrt{2}}}=\frac{\pi }{2\sqrt{2}}{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}(\cos \left(\frac{1}{\sqrt{2}}\right)+\mathrm{isin}\left(\frac{1}{\sqrt{2}}\right))\Rightarrow$$$$2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{cosx}}{1+{\mathrm{x}}^4}\mathrm{dx}=\frac{\pi }{2\sqrt{2}}{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}\cos \left(\frac{1}{\sqrt{2}}\right)\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{cosx}}{1+{\mathrm{x}}^4}\mathrm{dx}=\frac{\pi }{4\sqrt{2}}{\mathrm{e}}^{-\frac{1}{\sqrt{2}}}\cos \left(\frac{1}{\sqrt{2}}\right)$$

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