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Question Number 146898 by mathmax by abdo last updated on 16/Jul/21

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{cosx}}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+2)({\mathrm{x}}^2+3)}\mathrm{dx}$$

Answered by qaz last updated on 16/Jul/21

$$\mathcal{L}\left({\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{tx}\right)}{{\mathrm{x}}^2+{\mathrm{a}}^2}\mathrm{dx}\right)\left(\mathrm{s}\right).........(\mathrm{a}>0)$$$$={\int }_0^{\mathrm{\infty }}\frac{\mathrm{s}}{({\mathrm{x}}^2+{\mathrm{a}}^2)({\mathrm{s}}^2+{\mathrm{x}}^2)}\mathrm{dx}$$$$=\frac{\mathrm{s}}{{\mathrm{s}}^2-{\mathrm{a}}^2}{\int }_0^{\mathrm{\infty }}(\frac{1}{{\mathrm{x}}^2+{\mathrm{a}}^2}-\frac{1}{{\mathrm{s}}^2+{\mathrm{x}}^2})\mathrm{dx}$$$$=\frac{\mathrm{s}}{{\mathrm{s}}^2-{\mathrm{a}}^2}(\frac{1}{\mathrm{a}}{\tan }^{-1}\frac{\mathrm{x}}{\mathrm{a}}-\frac{1}{\mathrm{s}}{\tan }^{-1}\frac{\mathrm{x}}{\mathrm{s}}){\mid }_0^{\mathrm{\infty }}$$$$=\frac{\pi }{2\mathrm{a}(\mathrm{s}+\mathrm{a})}$$$${\int }_0^{\mathrm{\infty }}\frac{\cos \mathrm{x}}{{\mathrm{x}}^2+{\mathrm{a}}^2}\mathrm{dx}={\mathcal{L}}^{-1}\left(\frac{\pi }{2\mathrm{a}(\mathrm{s}+\mathrm{a})}\right)(\mathrm{t}=1)=\frac{\pi }{2\mathrm{a}}{\mathrm{e}}^{-\mathrm{at}}{\mid }_{\mathrm{t}=1}=\frac{\pi }{2\mathrm{a}}{\mathrm{e}}^{-\mathrm{a}}$$$${\int }_0^{\mathrm{\infty }}\frac{\cos \mathrm{x}}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+2)({\mathrm{x}}^2+3)}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}(\frac{\cos \mathrm{x}}{2({\mathrm{x}}^2+1)}-\frac{\cos \mathrm{x}}{{\mathrm{x}}^2+2}+\frac{\cos \mathrm{x}}{2({\mathrm{x}}^2+3)})\mathrm{dx}$$$$=\frac{\pi }{4}{\mathrm{e}}^{-1}-\frac{\pi }{2\sqrt{2}}{\mathrm{e}}^{-\sqrt{2}}+\frac{\pi }{4\sqrt{3}}{\mathrm{e}}^{-\sqrt{3}}$$

Commented by mathmax by abdo last updated on 16/Jul/21

$$\mathrm{thank}\mathrm{sir}\mathrm{answer}\mathrm{correct}..$$

Answered by mathmax by abdo last updated on 16/Jul/21

$$\mathrm{{\rm Y}}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{cosx}}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+2)({\mathrm{x}}^2+3)}\mathrm{dx}\Rightarrow 2\mathrm{{\rm Y}}=\mathrm{Re}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{ix}}}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+2)({\mathrm{x}}^2+3)}\right)$$$$\mathrm{let}\mathrm{\Lambda }\left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iz}}}{({\mathrm{z}}^2+1)({\mathrm{z}}^2+2)({\mathrm{z}}^2+3)}\Rightarrow \mathrm{\Lambda }\left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{iz}}}{(\mathrm{z}-\mathrm{i})(\mathrm{z}+\mathrm{i})(\mathrm{z}-\sqrt{2}\mathrm{i})(\mathrm{z}+\sqrt{2}\mathrm{i})(\mathrm{z}-\mathrm{i}\sqrt{3})(\mathrm{z}+\mathrm{i}\sqrt{3})}$$$${\int }_{\mathrm{R}}\mathrm{\Lambda }\left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\mathrm{\Lambda },\mathrm{i})+\mathrm{Res}(\mathrm{\Lambda },\mathrm{i}\sqrt{2})+\mathrm{Res}(\mathrm{\Lambda },\mathrm{i}\sqrt{3})\}$$$$\mathrm{Res}(\mathrm{\Lambda },\mathrm{i})=\frac{{\mathrm{e}}^{-1}}{\left(2\mathrm{i}\right)(-1+2)(-1+3)}=\frac{{\mathrm{e}}^{-1}}{4\mathrm{i}}$$$$\mathrm{Res}(\mathrm{\Lambda },\mathrm{i}\sqrt{2})=\frac{{\mathrm{e}}^{-\sqrt{2}}}{\left(2\sqrt{2}\right)\mathrm{i}(-2+1)(-2+3)}=-\frac{{\mathrm{e}}^{-\sqrt{2}}}{2\sqrt{2}\mathrm{i}}$$$$\mathrm{Res}(\mathrm{\Lambda },\mathrm{i}\sqrt{3})=\frac{{\mathrm{e}}^{-\sqrt{3}}}{\left(2\mathrm{i}\sqrt{3}\right)(-3+1)(-3+2)}=\frac{{\mathrm{e}}^{-\sqrt{3}}}{4\mathrm{i}\sqrt{3}}\Rightarrow$$$${\int }_{\mathrm{R}}\mathrm{\Lambda }\left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\frac{{\mathrm{e}}^{-1}}{4\mathrm{i}}-\frac{{\mathrm{e}}^{-\sqrt{2}}}{2\sqrt{2}\mathrm{i}}+\frac{{\mathrm{e}}^{-\sqrt{3}}}{4\mathrm{i}\sqrt{3}}\}$$$$=\frac{\pi }{2}{\mathrm{e}}^{-1}-\frac{\pi }{\sqrt{2}}{\mathrm{e}}^{-\sqrt{2}}+\frac{\pi }{2\sqrt{3}}{\mathrm{e}}^{-\sqrt{3}}\Rightarrow$$$$\mathrm{{\rm Y}}=\frac{\pi }{4}{\mathrm{e}}^{-1}-\frac{\pi }{2\sqrt{2}}{\mathrm{e}}^{-\sqrt{2}}+\frac{\pi }{4\sqrt{3}}{\mathrm{e}}^{-\sqrt{3}}$$

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