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Question Number 131637 by rs4089 last updated on 07/Feb/21

Answered by Dwaipayan Shikari last updated on 07/Feb/21

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{sinx}{x(x^2+1)}dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{sinx}{x}-\frac{xsinx}{(x^2+1)}dx=\pi -\frac{\pi }{e}$$$$I\left(\alpha \right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{cos\alpha x}{(x^2+1)}=\pi e^{-\alpha }$$$$I^{\prime }\left(\alpha \right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{xsin\left(\alpha x\right)}{x^2+1}dx=\pi e^{-\alpha }$$$$I^{\prime }\left(1\right)=\frac{\pi }{e}$$

Commented by Lordose last updated on 07/Feb/21

$$\mathrm{why}{\mathrm{I}}^{\prime }\left(\alpha \right)=\pi {\mathrm{e}}^{-\alpha }\mathrm{and}\mathrm{not}-\alpha \pi {\mathrm{e}}^{-\alpha }$$

Commented by Dwaipayan Shikari last updated on 07/Feb/21

$$I\left(\alpha \right)=\pi e^{-\alpha }$$$$I^{\prime }\left(\alpha \right)=-\pi e^{-\alpha }$$$$Also{I}^{\prime }\left(\alpha \right)=-{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{xsin\left(\alpha x\right)}{x^2+1}$$$$So\pi e^{-\alpha }={\int }_0^{\mathrm{\infty }}\frac{xsin\left(\alpha x\right)}{x^2+1}dx$$

Answered by mathmax by abdo last updated on 07/Feb/21

$$\mathrm{\Phi }={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{sinx}}{{\mathrm{x}}^3+\mathrm{x}}\mathrm{dx}\Rightarrow \mathrm{\Phi }={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{sinx}}{\mathrm{x}({\mathrm{x}}^2+1)}\mathrm{dx}$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{sinx}(\frac{1}{\mathrm{x}}-\frac{\mathrm{x}}{{\mathrm{x}}^2+1})\mathrm{dx}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{sinx}}{\mathrm{x}}\mathrm{dx}-{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{xsinx}}{{\mathrm{x}}^2+1}\mathrm{dx}$$$$=\pi -{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{xsinx}}{{\mathrm{x}}^2+1}\mathrm{dx}=\pi -\mathrm{I}$$$$\mathrm{I}=\mathrm{Im}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{xe}}^{\mathrm{ix}}}{{\mathrm{x}}^2+1}\mathrm{dx}\right)\mathrm{but}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{xe}}^{\mathrm{ix}}}{{\mathrm{x}}^2+1}\mathrm{dx}$$$$=2\mathrm{i}\pi \mathrm{Res}(\mathrm{f},\mathrm{i})=2\mathrm{i}\pi .\frac{{\mathrm{ie}}^{-1}}{2\mathrm{i}}=\pi {\mathrm{ie}}^{-1}\Rightarrow \mathrm{I}=\frac{\pi }{\mathrm{e}}\Rightarrow \mathrm{\Phi }=\pi -\frac{\pi }{\mathrm{e}}$$

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