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Question Number 143190 by Ar Brandon last updated on 11/Jun/21

$${\int }_{\mathbb{R}}\frac{{\mathrm{e}}^{\mathrm{its}}}{\mathrm{s}+3}\mathrm{ds}$$

Commented by mohammad17 last updated on 11/Jun/21

$$w=\frac{1}{s+3}\Rightarrow s=\frac{1}{w}-3\Rightarrow ds=-\frac{1}{w^2}dw$$$$$$$$\int {we}^{it(\frac{1}{w}-3)}(-\frac{1}{w^2})dw=\frac{1}{e^{3it}}\int {we}^{\left(\frac{it}{w}\right)}(-\frac{1}{w^2})dw$$$$$$$$bybart::u=w\to du=dw,dv=e^{\left(\frac{it}{w}\right)}(-\frac{1}{w^2})\to v=\frac{1}{it}{e}^{\left(\frac{it}{w}\right)}$$$$$$$$=\frac{1}{{ite}^{3it}}{we}^{\left(\frac{it}{w}\right)}-\frac{1}{{ite}^{3it}}\int e^{\left(\frac{it}{w}\right)}dw$$$$$$$$=\frac{1}{{ite}^{3it}}{we}^{\left(\frac{it}{w}\right)}-\frac{1}{{ite}^{3it}}{we}^{\left(\frac{it}{w}\right)}+\frac{1}{e^{3it}}Ei\left(\frac{it}{w}\right)+C$$$$$$$${\int }_R\frac{e^{its}}{s+3}ds=\frac{1}{e^{3it}}Ei\left(it(s+3)\right)+C$$$$$$$$⟨mohammadaldolaimy⟩$$$$$$

Commented by Ar Brandon last updated on 11/Jun/21

$$\mathrm{Thanks}\mathrm{Sir}.\mathrm{Can}\mathrm{you}\mathrm{insert}$$$$\mathrm{the}\mathrm{limits}\mathrm{and}\mathrm{evaluate}?$$

Answered by mathmax by abdo last updated on 11/Jun/21

$$\mathrm{\Phi }={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{itx}}}{\mathrm{x}+3}\mathrm{dx}={\int }_{-\mathrm{\infty }}^0\frac{{\mathrm{e}}^{\mathrm{itx}}}{\mathrm{x}+3}\mathrm{dx}+{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{itx}}}{\mathrm{x}+3}\mathrm{dx}\mathrm{but}$$$${\int }_{-\mathrm{\infty }}^0\frac{{\mathrm{e}}^{\mathrm{itx}}}{\mathrm{x}+3}\mathrm{dx}{=}_{\mathrm{x}=-\mathrm{u}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{itu}}}{3-\mathrm{u}}\mathrm{du}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{itx}}}{3-\mathrm{x}}\mathrm{dx}\Rightarrow$$$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{itx}}}{\mathrm{x}+3}\mathrm{dx}+{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{itx}}}{3-\mathrm{x}}\mathrm{dx}[\mathrm{we}\mathrm{have}$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{itx}}}{\mathrm{x}+3}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{x}+3)\mathrm{z}}\mathrm{dz}\right){\mathrm{e}}^{\mathrm{itx}}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{(-\mathrm{z}+\mathrm{it})\mathrm{x}}\mathrm{dx}\right){\mathrm{e}}^{-3\mathrm{z}}\mathrm{dz}$$$$={\int }_0^{\mathrm{\infty }}\left({\left[\frac{1}{-\mathrm{z}+\mathrm{it}}{\mathrm{e}}^{(-\mathrm{z}+\mathrm{it})\mathrm{x}}\right]}_0^{\mathrm{\infty }}\right){\mathrm{e}}^{-3\mathrm{z}}\mathrm{dz}={\int }_0^{\mathrm{\infty }}\left(\frac{1}{-\mathrm{z}+\mathrm{it}}\right){\mathrm{e}}^{-3\mathrm{z}}\mathrm{dz}$$$$=-{\int }_0^{\mathrm{\infty }}\left(\frac{1}{\mathrm{z}-\mathrm{it}}\right){\mathrm{e}}^{-3\mathrm{z}}\mathrm{dz}=-{\int }_0^{\mathrm{\infty }}\frac{(\mathrm{z}+\mathrm{it}){\mathrm{e}}^{-3\mathrm{z}}}{{\mathrm{z}}^2+{\mathrm{t}}^2}\mathrm{dz}$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{itx}}}{3-\mathrm{x}}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(3-\mathrm{x})\mathrm{z}}\mathrm{dz}\right){\mathrm{e}}^{-\mathrm{itx}}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{(\mathrm{z}-\mathrm{it})\mathrm{x}}\mathrm{dx}\right){\mathrm{e}}^{-3\mathrm{z}}\mathrm{dz}$$$$={\int }_0^{\mathrm{\infty }}\left({\left[\frac{1}{\mathrm{z}-\mathrm{it}}{\mathrm{e}}^{(\mathrm{z}-\mathrm{it})\mathrm{x}}\right]}_0^{\mathrm{\infty }}\right){\mathrm{e}}^{-3\mathrm{z}}\mathrm{dz}$$$$={\int }_0^{\mathrm{\infty }}(-\frac{1}{\mathrm{z}-\mathrm{it}}){\mathrm{e}}^{-3\mathrm{z}}\mathrm{dz}=-{\int }_0^{\mathrm{\infty }}\frac{(\mathrm{z}+\mathrm{it}){\mathrm{e}}^{-3\mathrm{z}}}{{\mathrm{z}}^2+{\mathrm{t}}^2}\mathrm{dz}\Rightarrow$$$$\mathrm{\Phi }=-{\int }_0^{\mathrm{\infty }}\frac{(\mathrm{z}+\mathrm{it}){\mathrm{e}}^{-3\mathrm{z}}}{{\mathrm{z}}^2+{\mathrm{t}}^2}\mathrm{dz}-{\int }_0^{\mathrm{\infty }}\frac{(\mathrm{z}+\mathrm{it}){\mathrm{e}}^{-3\mathrm{z}}}{{\mathrm{z}}^2+{\mathrm{t}}^2}\mathrm{dz}$$$$=-2{\int }_0^{\mathrm{\infty }}\frac{(\mathrm{z}+\mathrm{it}){\mathrm{e}}^{-3\mathrm{z}}}{{\mathrm{z}}^2+{\mathrm{t}}^2}\mathrm{dz}....\mathrm{be}\mathrm{continued}....$$

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