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Question Number 145745 by mathmax by abdo last updated on 07/Jul/21

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}\frac{\cos \left(2\mathrm{x}\right)}{{({\mathrm{x}}^2+1)}^2({\mathrm{x}}^2+4)}\mathrm{dx}$$

Answered by mathmax by abdo last updated on 11/Jul/21

$$\mathrm{\Psi }={\int }_0^{\mathrm{\infty }}\frac{\cos \left(2\mathrm{x}\right)}{{({\mathrm{x}}^2+1)}^2({\mathrm{x}}^2/+4)}\mathrm{dx}\Rightarrow 2\mathrm{\Psi }=\mathrm{Re}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{2\mathrm{ix}}}{{({\mathrm{x}}^2+1)}^2({\mathrm{x}}^2+4)}\mathrm{dx}\right)$$$$\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{2\mathrm{iz}}}{{({\mathrm{z}}^2+1)}^2({\mathrm{z}}^2+4)}\Rightarrow \phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{2\mathrm{iz}}}{{(\mathrm{z}-\mathrm{i})}^2{(\mathrm{z}+\mathrm{i})}^2(\mathrm{z}-2\mathrm{i})(\mathrm{z}+2\mathrm{i})}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\phi ,\mathrm{i})+\mathrm{Res}(\phi ,2\mathrm{i})\}$$$$\mathrm{Res}(\phi ,\mathrm{i})={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{1}{(2-1)!}{\left\{{(\mathrm{z}-\mathrm{i})}^2\phi \left(\mathrm{z}\right)\right\}}^{\left(1\right)}$$$$={\lim }_{\mathrm{z}\to \mathrm{i}}{\left\{\frac{{\mathrm{e}}^{2\mathrm{iz}}}{{(\mathrm{z}+\mathrm{i})}^2({\mathrm{z}}^2+4)}\right\}}^{\left(1\right)}$$$$={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{{2\mathrm{ie}}^{2\mathrm{iz}}{(\mathrm{z}+\mathrm{i})}^2({\mathrm{z}}^2+4)-(2(\mathrm{z}+\mathrm{i})({\mathrm{z}}^2+4)+2\mathrm{z}{(\mathrm{z}+\mathrm{i})}^2){\mathrm{e}}^{2\mathrm{iz}}}{{(\mathrm{z}+\mathrm{i})}^4{({\mathrm{z}}^2+4)}^2}$$$$=\frac{{2\mathrm{ie}}^{-2}(-4)\left(3\right)-(4\mathrm{i}\left(3\right)+2\mathrm{i}(-4)){\mathrm{e}}^{-2}}{{(-4)}^2.3^2}$$$$=\frac{-{24\mathrm{e}}^{-2}-{4\mathrm{ie}}^{-2}}{16.9}$$$$\mathrm{Res}(\phi ,2\mathrm{i})={\lim }_{\mathrm{z}\to 2\mathrm{i}}(\mathrm{z}-2\mathrm{i})\phi \left(\mathrm{z}\right)={\lim }_{\mathrm{z}\to 2\mathrm{i}}\frac{{\mathrm{e}}^{-4}}{4\mathrm{i}{(-3)}^2}=\frac{{\mathrm{e}}^{-4}}{36\mathrm{i}}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\frac{-{24\mathrm{e}}^{-2}-{4\mathrm{ie}}^{-2}}{144}+\frac{{\mathrm{e}}^{-4}}{36\mathrm{i}}\}$$$$=2\mathrm{i}\pi (-\frac{24}{144}{\mathrm{e}}^{-2})+\frac{8\pi }{144}{\mathrm{e}}^{-2}+\frac{\pi }{18}{\mathrm{e}}^{-4}$$$$2\mathrm{\Psi }=\mathrm{Re}(....)\Rightarrow 2\mathrm{\Psi }=\frac{4\pi }{72}{\mathrm{e}}^{-2}+\frac{\pi }{18}{\mathrm{e}}^{-4}\Rightarrow$$$$\mathrm{\Psi }=\frac{\pi }{36}{\mathrm{e}}^{-2}+\frac{\pi }{36}{\mathrm{e}}^{-4}$$

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