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Question Number 106691 by Laplace last updated on 06/Aug/20

$$\mathit{d}\mathit{e}\mathit{t}\mathit{e}\mathit{r}\mathit{m}\mathit{i}\mathit{n}\mathit{e}\mathit{u}\mathit{s}\mathit{i}\mathit{n}\mathit{g}\mathit{l}\mathit{a}\mathit{p}\mathit{l}\mathit{c}\mathit{e}\mathit{t}\mathit{r}\mathit{a}\mathit{n}\mathit{s}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{t}\mathit{h}\mathit{i}\mathit{s}$$$$\mathit{i}\mathit{n}\mathit{t}\mathit{e}\mathit{g}\mathit{r}\mathit{a}\mathit{l}\mathit{e}$$$${\int }_0^{+\mathrm{\infty }}\frac{\mathit{t}\mathit{s}\mathit{i}\mathit{n}\left(\mathit{t}\mathit{x}\right)}{{\mathit{a}}^2+{\mathit{t}}^2}\mathit{d}\mathit{t}$$

Answered by mathmax by abdo last updated on 06/Aug/20

$${\mathrm{f}}_{\mathrm{a}}\left(\mathrm{x}\right)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{tsin}\left(\mathrm{tx}\right)}{{\mathrm{a}}^2+{\mathrm{t}}^2}\mathrm{dt}\Rightarrow {\mathrm{f}}_{\mathrm{a}}\left(\mathrm{x}\right)={\int }_0^{\mathrm{\infty }}\mathrm{tsin}\left(\mathrm{tx}\right){\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-({\mathrm{a}}^2+{\mathrm{t}}^2)\mathrm{x}}\mathrm{dx}\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}\left({\int }_0^{\mathrm{\infty }}\sin \left(\mathrm{tx}\right){\mathrm{e}}^{-({\mathrm{a}}^2+{\mathrm{t}}^2)\mathrm{x}}\mathrm{dx}\right)\mathrm{tdt}\left(\mathrm{by}\mathrm{use}\mathrm{of}\mathrm{fubini}\right)$$$${\int }_0^{\mathrm{\infty }}\sin \left(\mathrm{tx}\right){\mathrm{e}}^{-({\mathrm{a}}^2+{\mathrm{t}}^2)\mathrm{x}}\mathrm{dx}=\mathrm{Im}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{\mathrm{itx}-({\mathrm{a}}^2+{\mathrm{t}}^2)\mathrm{x}}\mathrm{dx}\right)\mathrm{and}$$$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{(\mathrm{it}-({\mathrm{a}}^2+{\mathrm{t}}^2))\mathrm{x}}\mathrm{dx}={\left[\frac{1}{\mathrm{it}-({\mathrm{a}}^2+{\mathrm{t}}^2)}{\mathrm{e}}^{(\mathrm{it}-({\mathrm{a}}^2+{\mathrm{t}}^2))\mathrm{x}}\right]}_0^{\mathrm{\infty }}$$$$=\frac{-1}{\mathrm{it}-({\mathrm{a}}^2+{\mathrm{t}}^2)}=\frac{1}{({\mathrm{a}}^2+{\mathrm{t}}^2)-\mathrm{it}}=\frac{{\mathrm{a}}^2+{\mathrm{t}}^2+\mathrm{it}}{{({\mathrm{a}}^2+{\mathrm{t}}^2)}^2+{\mathrm{t}}^2}\Rightarrow$$$$\mathrm{Im}(...)=\frac{\mathrm{t}}{{({\mathrm{a}}^2+{\mathrm{t}}^2)}^2+{\mathrm{t}}^2}\Rightarrow$$$${\mathrm{f}}_{\mathrm{a}}\left(\mathrm{x}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^2}{{({\mathrm{a}}^2+{\mathrm{t}}^2)}^2+{\mathrm{t}}^2}\mathrm{dt}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^2}{({\mathrm{a}}^4+{2\mathrm{a}}^2{\mathrm{t}}^2+{\mathrm{t}}^4+{\mathrm{t}}^2}\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^2}{{\mathrm{t}}^4+({2\mathrm{a}}^2+1){\mathrm{t}}^2+{\mathrm{a}}^4}\mathrm{dt}=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{t}}^2}{{\mathrm{t}}^4+({2\mathrm{a}}^2+1){\mathrm{t}}^2+{\mathrm{a}}^4}\mathrm{dt}$$$$\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{z}}^2}{{\mathrm{z}}^4+({2\mathrm{a}}^2+1){\mathrm{z}}^2+{\mathrm{a}}^4}\mathrm{poles}\mathrm{of}\phi ?$$$$\mathrm{\Delta }={({2\mathrm{a}}^2+1)}^2-{4\mathrm{a}}^4={4\mathrm{a}}^4+{4\mathrm{a}}^2+1-{4\mathrm{a}}^4={4\mathrm{a}}^2+1({\mathrm{z}}^2=\mathrm{u})$$$${\mathrm{u}}_1=\frac{-({2\mathrm{a}}^2+1)+\sqrt{{4\mathrm{a}}^2+1}}{2}\mathrm{and}{\mathrm{u}}_2=\frac{-({2\mathrm{a}}^2+1)-\sqrt{{4\mathrm{a}}^2+1}}{2}$$$$\Rightarrow \phi \left(\mathrm{z}\right)=\frac{{\mathrm{z}}^2}{({\mathrm{z}}^2-{\mathrm{u}}_1)({\mathrm{z}}^2-{\mathrm{u}}_2)}=\frac{{\mathrm{z}}^2}{(\mathrm{z}-\sqrt{{\mathrm{u}}_1})(\mathrm{z}+\sqrt{{\mathrm{u}}_1})(\mathrm{z}-\sqrt{{\mathrm{u}}_2})(\mathrm{z}+\sqrt{{\mathrm{u}}_2})}$$$$\sqrt{{\mathrm{u}}_1}=\mathrm{i}\sqrt{\frac{{2\mathrm{a}}^2+1-\sqrt{{4\mathrm{a}}^2+1}}{2}}\mathrm{and}\sqrt{{\mathrm{u}}_2}=\mathrm{i}\sqrt{\frac{{2\mathrm{a}}^2+1+\sqrt{{4\mathrm{a}}^2+1}}{2}}$$$$...\mathrm{be}\mathrm{continued}...$$

Commented by Ar Brandon last updated on 06/Aug/20

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Answered by abdomathmax last updated on 07/Aug/20

$$\mathrm{residus}\mathrm{method}\mathrm{let}\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{tsin}\left(\mathrm{tx}\right)}{{\mathrm{t}}^2+{\mathrm{a}}^2}\mathrm{dt}$$$$\Rightarrow \mathrm{I}{=}_{\mathrm{t}=\mid \mathrm{a}\mid \mathrm{u}}{\int }_0^{\mathrm{\infty }}\frac{\mid \mathrm{a}\mid \mathrm{usin}(\mid \mathrm{a}\mid \mathrm{ux})}{{\mathrm{a}}^2({\mathrm{u}}^2+1)}\mid \mathrm{a}\mid \mathrm{du}$$$$={\int }_0^{\mathrm{\infty }}\frac{\mathrm{u}\sin (\mathrm{x}\mid \mathrm{a}\mid \mathrm{u})}{{\mathrm{u}}^2+1}=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{u}\sin (\mathrm{x}\mid \mathrm{a}\mid \mathrm{u})}{{\mathrm{u}}^2+1}\mathrm{du}$$$$=\frac{1}{2}\mathrm{Im}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{u}{\mathrm{e}}^{\mathrm{ix}\mid \mathrm{a}\mid \mathrm{u}}}{{\mathrm{u}}^2+1}\mathrm{du}\right)\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{\mathrm{z}{\mathrm{e}}^{\mathrm{ix}\mid \mathrm{a}\mid \mathrm{z}}}{{\mathrm{z}}^2+1}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dx}=2\mathrm{i}\pi \mathrm{Res}(\phi ,\mathrm{i})=2\mathrm{i}\pi \times \frac{{\mathrm{ie}}^{-\mathrm{x}\mid \mathrm{a}\mid }}{2\mathrm{i}}$$$$=\mathrm{i}\pi {\mathrm{e}}^{-\mathrm{x}\mid \mathrm{a}\mid }\Rightarrow \mathrm{I}=\frac{\pi }{2}{\mathrm{e}}^{-\mathrm{x}\mid \mathrm{a}\mid }$$

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