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Question Number 119752 by Bird last updated on 26/Oct/20

$$find{I}_{\lambda }={\int }_0^{\mathrm{\infty }}\frac{ch(1+\lambda cosx)}{{(x^2+1)}^2}dx$$ $$(\lambda real>0)$$

Answered by mathmax by abdo last updated on 27/Oct/20

$${\mathrm{I}}_{\lambda }={\int }_0^{\mathrm{\infty }}\frac{\mathrm{ch}(1+\lambda \mathrm{cosx})}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}\Rightarrow {\mathrm{I}}_{\lambda }={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{1+\lambda \mathrm{cosx}}+{\mathrm{e}}^{-(1+\lambda \mathrm{cosx})}}{2{({\mathrm{x}}^2+1)}^2}\mathrm{dx}$$ $$=\frac{\mathrm{e}}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\lambda \mathrm{cosx}}}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}+\frac{1}{2\mathrm{e}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\lambda \mathrm{cosx}}}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}=\frac{\mathrm{e}}{2}\mathrm{A}+\frac{1}{2\mathrm{e}}\mathrm{B}$$ $$\mathrm{A}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\lambda \mathrm{cosx}}}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}\Rightarrow \mathrm{A}=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\lambda \mathrm{cosx}}}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}$$ $$\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\lambda \mathrm{cosz}}}{{({\mathrm{z}}^2+1)}^2}\Rightarrow \phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\lambda \mathrm{cosz}}}{{(\mathrm{z}-\mathrm{i})}^2{(\mathrm{z}+\mathrm{i})}^2}$$ $${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,\mathrm{i})$$ $$\mathrm{Res}(\phi ,\mathrm{i})={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{1}{(2-1)!}{\left\{{(\mathrm{z}-\mathrm{i})}^2\phi \left(\mathrm{z}\right)\right\}}^{\left(1\right)}$$ $$={\lim }_{\mathrm{z}\to \mathrm{i}}{\left\{\frac{{\mathrm{e}}^{\lambda \mathrm{cosz}}}{{(\mathrm{z}+\mathrm{i})}^2}\right\}}^{\left(1\right)}={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{-\lambda {\mathrm{sinze}}^{\lambda \mathrm{cosz}}{(\mathrm{z}+\mathrm{i})}^2-2(\mathrm{z}+\mathrm{i}){\mathrm{e}}^{\lambda \mathrm{cosz}}}{{(\mathrm{z}+\mathrm{i})}^4}$$ $$={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{-\lambda {\mathrm{sinze}}^{\lambda \mathrm{cosz}}(\mathrm{z}+\mathrm{i})-2{\mathrm{e}}^{\lambda \mathrm{cosz}}}{{(\mathrm{z}+\mathrm{i})}^3}$$ $$=\frac{-\lambda {\mathrm{sinie}}^{\lambda \cos \left(\mathrm{i}\right)}\left(2\mathrm{i}\right)-{2\mathrm{e}}^{\lambda \cos \left(\mathrm{i}\right)}}{-8\mathrm{i}}=\frac{\lambda \sin \left(\mathrm{i}\right){\mathrm{e}}^{\lambda \cos \left(\mathrm{i}\right)}+{\mathrm{e}}^{\lambda \cos \left(\mathrm{i}\right)}}{4\mathrm{i}}\Rightarrow$$ $${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \times \frac{(1+\lambda \sin \left(\mathrm{i}\right)){\mathrm{e}}^{\lambda \cos \left(\mathrm{i}\right)}}{4\mathrm{i}}$$ $$=\frac{\pi }{2}(1+\lambda \mathrm{sh}(-1)){\mathrm{e}}^{\lambda \mathrm{ch}(-1)}=\frac{\pi }{2}(1-\lambda \mathrm{sh}\left(1\right)){\mathrm{e}}^{\lambda \mathrm{ch}\left(1\right)}\Rightarrow$$ $$\mathrm{A}=\frac{\pi }{4}(1-\lambda \mathrm{sh}\left(1\right)){\mathrm{e}}^{\lambda \mathrm{ch}\left(1\right)}\mathrm{also}$$ $$\mathrm{B}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\lambda \mathrm{cosx}}}{{({\mathrm{x}}^2+1)}^2}=\frac{\pi }{4}(1+\lambda \mathrm{sh}\left(1\right)){\mathrm{e}}^{-\lambda \mathrm{ch}\left(1\right)}(\mathrm{change}\lambda \mathrm{by}-\lambda )\Rightarrow$$ $$\mathrm{I}=\frac{\pi \mathrm{e}}{8}(1-\lambda \mathrm{sh}\left(1\right)){\mathrm{e}}^{\lambda \mathrm{ch}\left(1\right)}+\frac{\pi {\mathrm{e}}^{-1}}{8}(1+\lambda \mathrm{sh}\left(1\right)){\mathrm{e}}^{-\lambda \mathrm{ch}\left(1\right)}$$ $$$$

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