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Question Number 122791 by mohammad17 last updated on 19/Nov/20

Answered by mathmax by abdo last updated on 19/Nov/20

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{mx}\right)}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}\Rightarrow 2\mathrm{I}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\cos \left(\mathrm{mx}\right)}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}=\mathrm{Res}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{imx}}}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}\right)$$$$\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{imz}}}{{({\mathrm{z}}^2+1)}^2}\mathrm{we}\mathrm{have}{\lim }_{\mathrm{z}\to \mathrm{\infty }}\mid \mathrm{z}\phi \left(\mathrm{z}\right)\mid =0\mathrm{and}$$$$\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{imz}}}{{(\mathrm{z}-\mathrm{i})}^2{(\mathrm{z}+\mathrm{i})}^2}[\mathrm{rssidus}\mathrm{theorem}\mathrm{give}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,\mathrm{i})\mathrm{and}\mathrm{Res}(\phi ,\mathrm{i})={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{1}{(2-1)!}{\left\{{(\mathrm{z}-\mathrm{i})}^2\phi \left(\mathrm{z}\right)\right\}}^{\left(1\right)}$$$$={\lim }_{\mathrm{z}\to \mathrm{i}}{\left\{\frac{{\mathrm{e}}^{\mathrm{imz}}}{{(\mathrm{z}+\mathrm{i})}^2}\right\}}^{\left(1\right)}={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{{\mathrm{ime}}^{\mathrm{imz}}{(\mathrm{z}+\mathrm{i})}^2-2(\mathrm{z}+\mathrm{i}){\mathrm{e}}^{\mathrm{imz}}}{{(\mathrm{z}+\mathrm{i})}^4}$$$$={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{{\mathrm{ime}}^{\mathrm{imz}}(\mathrm{z}+\mathrm{i})-{2\mathrm{e}}^{\mathrm{imz}}}{{(\mathrm{z}+\mathrm{i})}^3}={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{(\mathrm{im}(\mathrm{z}+\mathrm{i})-2){\mathrm{e}}^{\mathrm{imz}}}{{(\mathrm{z}+\mathrm{i})}^3}$$$$=\frac{2\mathrm{i}\left(\mathrm{im}\right)-2){\mathrm{e}}^{-\mathrm{m}}}{{\left(2\mathrm{i}\right)}^3}=\frac{(-2\mathrm{m}-2){\mathrm{e}}^{-\mathrm{m}}}{-8\mathrm{i}}=\frac{(\mathrm{m}+1){\mathrm{e}}^{-\mathrm{m}}}{4\mathrm{i}}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \times \frac{(\mathrm{m}+1){\mathrm{e}}^{-\mathrm{m}}}{4\mathrm{i}}=\frac{\pi }{2}(\mathrm{m}+1){\mathrm{e}}^{-\mathrm{m}}\Rightarrow$$$$2\mathrm{I}=\frac{\pi }{2}(\mathrm{m}+1){\mathrm{e}}^{-\mathrm{m}}\Rightarrow ★\mathrm{I}=\frac{\pi }{4}(\mathrm{m}+1){\mathrm{e}}^{-\mathrm{m}}★$$

Answered by TANMAY PANACEA last updated on 19/Nov/20

$$\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{(4x^2+4)-(x^2+4)}{(x^2+1)(x^2+4)}dx$$$$\frac{4}{3}{\int }_0^{\mathrm{\infty }}\frac{dx}{x^2+4}-\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{dx}{x^2+1}$$$$\mid \frac{4}{3}\times \frac{1}{2}{tan}^{-1}\left(\frac{x}{2}\right)-\frac{1}{3}{tan}^{-1}x{\mid }_{}^{\mathrm{\infty }}$$$$\frac{2}{3}\times \frac{\pi }{2}-\frac{1}{3}\times \frac{\pi }{2}=\frac{\pi }{6}$$

Commented by mohammad17 last updated on 19/Nov/20

$$thankyoumis$$

Answered by mathmax by abdo last updated on 19/Nov/20

$$\mathrm{A}={\int }_0^{\mathrm{\infty }}\frac{{\sin }^2\mathrm{x}}{{\mathrm{x}}^2}\mathrm{dx}\mathrm{by}\mathrm{parts}\mathrm{A}={[-\frac{{\sin }^2\mathrm{x}}{\mathrm{x}}]}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}\frac{2\mathrm{sinx}\mathrm{cosx}}{\mathrm{x}}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{\sin \left(2\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}{=}_{2\mathrm{x}=\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{\sin \left(\mathrm{t}\right)}{\frac{\mathrm{t}}{2}}\times \frac{\mathrm{dt}}{2}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{sint}}{\mathrm{t}}\mathrm{dt}=\frac{\pi }{2}$$

Answered by mathmax by abdo last updated on 19/Nov/20

$$4)\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+4)}\mathrm{dx}\Rightarrow \mathrm{I}=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{x}}^2}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+4)}\mathrm{dx}$$$$\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{z}}^2}{({\mathrm{z}}^2+1)({\mathrm{z}}^2+4)}\Rightarrow \phi \left(\mathrm{z}\right)=\frac{{\mathrm{z}}^2}{(\mathrm{z}-\mathrm{i})(\mathrm{z}+\mathrm{i})(\mathrm{z}-2\mathrm{i})(\mathrm{z}+2\mathrm{i})}$$$$\mathrm{residus}\mathrm{theorem}\to {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\phi ,\mathrm{i})+\mathrm{Res}(\phi ,2\mathrm{i})\}$$$$\mathrm{Res}(\phi ,\mathrm{i})={\lim }_{\mathrm{z}\to \mathrm{i}}(\mathrm{z}-\mathrm{i})\phi \left(\mathrm{z}\right)=\frac{-1}{2\mathrm{i}\left(3\right)}=-\frac{1}{6\mathrm{i}}$$$$\mathrm{Res}(\phi ,2\mathrm{i})={\lim }_{\mathrm{z}\to 2\mathrm{i}}(\mathrm{z}-2\mathrm{i})\phi \left(\mathrm{z}\right)=\frac{-4}{4\mathrm{i}(-3)}=\frac{1}{3\mathrm{i}}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{-\frac{1}{6\mathrm{i}}+\frac{1}{3\mathrm{i}}\}=\frac{\pi }{3}+\frac{2\pi }{3}=\pi \Rightarrow \mathrm{I}=\frac{\pi }{2}$$$$\mathrm{error}\mathrm{at}\mathrm{the}\mathrm{question}...!$$

Answered by mathmax by abdo last updated on 19/Nov/20

$$\mathrm{A}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{xsin}\left(\mathrm{ax}\right)}{{\mathrm{x}}^2+4}\mathrm{dx}\Rightarrow \mathrm{A}=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{xsin}\left(\mathrm{ax}\right)}{{\mathrm{x}}^2+4}\mathrm{dx}\mathrm{let}$$$$=\frac{1}{2}\mathrm{Im}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{x}}{{\mathrm{x}}^2+4}{\mathrm{e}}^{\mathrm{iax}}\mathrm{dx}\right)\mathrm{let}\mathrm{w}\left(\mathrm{z}\right)=\frac{{\mathrm{ze}}^{\mathrm{iaz}}}{{\mathrm{z}}^2+4}$$$${\lim }_{\mathrm{z}\to \mathrm{\infty }}\mid \mathrm{zw}\left(\mathrm{z}\right)\mid =0\mathrm{and}\mathrm{w}\left(\mathrm{z}\right)=\frac{{\mathrm{ze}}^{\mathrm{iaz}}}{(\mathrm{z}-2\mathrm{i})(\mathrm{z}+2\mathrm{i})}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{W}\left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\mathrm{w},2\mathrm{i})=2\mathrm{i}\pi .\frac{2\mathrm{i}{\mathrm{e}}^{\mathrm{ia}\left(2\mathrm{i}\right)}}{4\mathrm{i}}=\mathrm{i}\pi {\mathrm{e}}^{-2\mathrm{a}}\Rightarrow$$$$\mathrm{A}=\frac{\pi }{2}{\mathrm{e}}^{-2\mathrm{a}}\mathrm{error}\mathrm{at}\mathrm{the}\mathrm{question}{}^{\prime }.!$$

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