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Question Number 131050 by mathmax by abdo last updated on 31/Jan/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{tsin}\left(\mathrm{xt}\right)}{{({\mathrm{x}}^2+{\mathrm{t}}^2)}^2}\mathrm{dt}(\mathrm{x}>0)$$ $$\mathrm{calculate}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)$$

Answered by mathmax by abdo last updated on 02/Feb/21

$$\mathrm{changement}\mathrm{t}=\mathrm{xz}\mathrm{give}\mathrm{f}\left(\mathrm{x}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{xzsin}\left({\mathrm{x}}^2\mathrm{z}\right)}{{\mathrm{x}}^4{({\mathrm{z}}^2+1)}^2}\mathrm{xdz}$$ $$=\frac{1}{{\mathrm{x}}^2}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{zsin}\left({\mathrm{x}}^2\mathrm{z}\right)}{{({\mathrm{z}}^2+1)}^2}\mathrm{dz}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{{2\mathrm{x}}^2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{zsin}\left({\mathrm{x}}^2\mathrm{z}\right)}{{({\mathrm{z}}^2+1)}^2}\mathrm{dz}=\frac{1}{{2\mathrm{x}}^2}\mathrm{Im}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{ze}}^{{\mathrm{ix}}^2\mathrm{z}}}{{({\mathrm{z}}^2+1)}^2}\mathrm{dz}\right)$$ $$\Rightarrow \phi \left(\mathrm{z}\right)=\frac{\mathrm{z}{\mathrm{e}}^{{\mathrm{ix}}^2\mathrm{z}}}{{({\mathrm{z}}^2+1)}^2}\Rightarrow \phi \left(\mathrm{z}\right)=\frac{{\mathrm{ze}}^{{\mathrm{ix}}^2\mathrm{z}}}{{(\mathrm{z}-\mathrm{i})}^2{(\mathrm{z}+\mathrm{i})}^2}\mathrm{residus}\mathrm{theorem}\mathrm{give}$$ $${\int }_{\mathrm{R}}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,\mathrm{i})$$ $$\mathrm{Res}(\phi ,\mathrm{i})={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{1}{(2-1)!}{\left\{{(\mathrm{z}-\mathrm{i})}^2\phi \left(\mathrm{z}\right)\right\}}^{\left(1\right)}$$ $$={\lim }_{\mathrm{z}\to \mathrm{i}}{\left\{\frac{{\mathrm{ze}}^{{\mathrm{ix}}^2\mathrm{z}}}{{(\mathrm{z}+\mathrm{i})}^2}\right\}}^{\left(1\right)}={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{({\mathrm{e}}^{{\mathrm{ix}}^2\mathrm{z}}+{\mathrm{ix}}^2\mathrm{z}{\mathrm{e}}^{{\mathrm{ix}}^2\mathrm{z}}){(\mathrm{z}+\mathrm{i})}^2-2(\mathrm{z}+\mathrm{i}){\mathrm{ze}}^{{\mathrm{ix}}^2\mathrm{z}}}{{(\mathrm{z}+\mathrm{i})}^4}$$ $$={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{{\mathrm{e}}^{{\mathrm{ix}}^2\mathrm{z}}(1+{\mathrm{ix}}^2\mathrm{z})(\mathrm{z}+\mathrm{i})-{2\mathrm{ze}}^{{\mathrm{ix}}^2\mathrm{z}}}{{(\mathrm{z}+\mathrm{i})}^3}$$ $$=\frac{{\mathrm{e}}^{-{\mathrm{x}}^2}(1-{\mathrm{x}}^2)\left(2\mathrm{i}\right)-2\mathrm{i}{\mathrm{e}}^{-{\mathrm{x}}^2}}{{\left(2\mathrm{i}\right)}^3}=\frac{2\mathrm{i}\{(1-{\mathrm{x}}^2){\mathrm{e}}^{-{\mathrm{x}}^2}-{\mathrm{e}}^{-{\mathrm{x}}^2}\}}{-8\mathrm{i}}$$ $$=-\frac{1}{4}(-{\mathrm{x}}^2){\mathrm{e}}^{-{\mathrm{x}}^2}=\frac{{\mathrm{x}}^2}{4}{\mathrm{e}}^{-{\mathrm{x}}^2}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \times \frac{{\mathrm{x}}^2}{4}{\mathrm{e}}^{-{\mathrm{x}}^2}$$ $$=\frac{\mathrm{i}\pi {\mathrm{x}}^2}{2}{\mathrm{e}}^{-{\mathrm{x}}^2}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{{2\mathrm{x}}^2}.\frac{\pi }{2}{\mathrm{x}}^2{\mathrm{e}}^{-{\mathrm{x}}^2}=\frac{\pi }{4}{\mathrm{e}}^{-{\mathrm{x}}^2}\Rightarrow$$ $${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{\pi }{4}(-2\mathrm{x}){\mathrm{e}}^{-{\mathrm{x}}^2}=-\frac{\pi \mathrm{x}}{2}{\mathrm{e}}^{-{\mathrm{x}}^2}$$

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