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Question Number 147166 by puissant last updated on 18/Jul/21

$${\int }_{\mathbb{R}}{\mathrm{e}}^{\mathrm{ixt}}{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt}..$$

Answered by mathmax by abdo last updated on 18/Jul/21

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{t}}^2}{\mathrm{e}}^{\mathrm{ixt}}\mathrm{dt}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\mathrm{e}}^{-({\mathbf{t}}^2-\mathbf{ixt})}\mathbf{dt}$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\mathrm{e}}^{-({\mathrm{t}}^2-2\frac{\mathrm{ix}}{2}\mathrm{t}-\frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{x}}^2}{4})}\mathrm{dt}$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\mathrm{e}}^{-{(\mathrm{t}-\frac{\mathrm{ix}}{2})}^2}{\mathrm{e}}^{-\frac{{\mathrm{x}}^2}{4}}\mathrm{dt}{=}_{\mathrm{t}-\frac{\mathrm{ix}}{2}=\mathrm{z}}{\mathrm{e}}^{-\frac{{\mathrm{x}}^2}{4}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{z}}^2}\mathrm{dz}$$$$=\sqrt{\pi }{\mathrm{e}}^{-\frac{{\mathrm{x}}^2}{4}}$$

Commented by puissant last updated on 18/Jul/21

$$\mathrm{thanks}$$

Commented by Kamel last updated on 18/Jul/21

$${it}^{\prime }swrongtowritet-\frac{ix}{2}\to +\mathrm{\infty },thisisacomplexnumber.$$

Commented by mathmax by abdo last updated on 18/Jul/21

$$\mathrm{study}\mathrm{first}\mathrm{when}\mathrm{a}\mathrm{complex}\mathrm{z}\mathrm{go}\mathrm{to}\mathrm{\infty }....$$

Answered by Kamel last updated on 18/Jul/21

$$$$$$\mathrm{\Omega }\left(x\right)={\int }_{\mathbb{R}}{e}^{ixt}{e}^{-t^2}dt=2{\int }_0^{+\mathrm{\infty }}cos\left(xt\right)e^{-t^2}dt$$$$\mid cos\left(xt\right)e^{-t^2}\mid ⩽e^{-t^2},\mid tsin\left(tx\right)e^{-t^2}\mid ⩽{te}^{-t^2},t\to e^{-t^2},t\to {te}^{-t^2}integrablesin[0;+\mathrm{\infty }[.$$$${\mathrm{\Omega }}^{\prime }\left(x\right)=-2{\int }_0^{+\mathrm{\infty }}tsin\left(xt\right)e^{-t^2}dt=-\frac{1}{2}x\mathrm{\Omega }\left(x\right)$$$$\therefore \frac{{\mathrm{\Omega }}^{\prime }\left(x\right)}{\mathrm{\Omega }\left(x\right)}=-\frac{1}{2}x\Rightarrow Ln\mid \mathrm{\Omega }\left(x\right)\mid =-\frac{x^2}{4}+c/c\in \mathbb{R}.$$$$\therefore \mathrm{\Omega }\left(x\right)={Ae}^{-\frac{x^2}{4}},\mathrm{\Omega }\left(0\right)=2{\int }_0^{+\mathrm{\infty }}{e}^{-t^2}dt=\sqrt{\pi }$$$$\therefore \mathrm{\Omega }\left(x\right)={\int }_{\mathbb{R}}{e}^{ixt}{e}^{-t^2}dt=\sqrt{\pi }{e}^{-\frac{x^2}{4}}$$

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