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Question Number 147309 by puissant last updated on 19/Jul/21

Answered by mathmax by abdo last updated on 19/Jul/21

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{cosx}}{{({\mathrm{x}}^2+1)}^2}\Rightarrow 2\mathrm{I}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{cosx}}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}=\mathrm{Re}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{ix}}}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}\right)$$$$\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iz}}}{{({\mathrm{z}}^2+1)}^2}\Rightarrow \phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iz}}}{{(\mathrm{z}-\mathrm{i})}^2{(\mathrm{z}+\mathrm{i})}^2}$$$$\mathrm{residus}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,\mathrm{i})$$$$\mathrm{Res}(\phi ,\mathrm{i})={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{1}{(2-1)!}{\left\{{(\mathrm{z}-\mathrm{i})}^2\phi \left(\mathrm{z}\right)\right\}}^{\left(1\right)}$$$$={\lim }_{\mathrm{z}\to \mathrm{i}}{\left\{\frac{{\mathrm{e}}^{\mathrm{iz}}}{{(\mathrm{z}+\mathrm{i})}^2}\right\}}^{\left(1\right)}={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{{\mathrm{ie}}^{\mathrm{iz}}{(\mathrm{z}+\mathrm{i})}^2-2(\mathrm{z}+\mathrm{i}){\mathrm{e}}^{\mathrm{iz}}}{{(\mathrm{z}+\mathrm{i})}^4}$$$$={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{{\mathrm{ie}}^{\mathrm{iz}}(\mathrm{z}+\mathrm{i})-{2\mathrm{e}}^{\mathrm{iz}}}{{(\mathrm{z}+\mathrm{i})}^3}=\frac{\left(2\mathrm{i}\right){\mathrm{ie}}^{-1}-{2\mathrm{e}}^{-1}}{{\left(2\mathrm{i}\right)}^3}=\frac{-{4\mathrm{e}}^{-1}}{-8\mathrm{i}}=\frac{{\mathrm{e}}^{-1}}{2\mathrm{i}}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi .\frac{{\mathrm{e}}^{-1}}{2\mathrm{i}}=\frac{\pi }{\mathrm{e}}\Rightarrow 2\mathrm{I}=\frac{\pi }{\mathrm{e}}\Rightarrow \mathrm{I}=\frac{\pi }{2\mathrm{e}}$$

Commented by puissant last updated on 19/Jul/21

$$thanks$$

Answered by mathmax by abdo last updated on 19/Jul/21

$$\mathrm{parametric}\mathrm{method}\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{cosx}}{{\mathrm{x}}^2+{\mathrm{a}}^2}\mathrm{dx}(\mathrm{a}>0)\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-2\mathrm{a}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{cosx}}{{({\mathrm{x}}^2+{\mathrm{a}}^2)}^2}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(1\right)=-2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{cosx}}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{cosx}}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}=-\frac{1}{2}{\mathrm{f}}^{{}^{\prime }}\left(1\right)$$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{1}{2}\mathrm{Re}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{iz}}}{{\mathrm{z}}^2+{\mathrm{a}}^2}\mathrm{dx}\right)\mathrm{and}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{iz}}}{{\mathrm{z}}^2+{\mathrm{a}}^2}\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\mathrm{h},\mathrm{ia})=2\mathrm{i}\pi .\frac{{\mathrm{e}}^{-\mathrm{a}}}{2\mathrm{ia}}=\frac{\pi }{\mathrm{a}}{\mathrm{e}}^{-\mathrm{a}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{2\mathrm{a}}{\mathrm{e}}^{-\mathrm{a}}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=\frac{\pi }{2}(-\frac{1}{{\mathrm{a}}^2}{\mathrm{e}}^{-\mathrm{a}}-\frac{1}{\mathrm{a}}{\mathrm{e}}^{-\mathrm{a}})\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(1\right)=\frac{\pi }{2}(-{2\mathrm{e}}^{-1})=-\pi {\mathrm{e}}^{-1}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{cosx}}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}=-\frac{1}{2}{\mathrm{f}}^{{}^{\prime }}\left(1\right)=-\frac{1}{2}(-\pi {\mathrm{e}}^{-1})=\frac{\pi }{2\mathrm{e}}$$

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