Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 196013 by pticantor last updated on 15/Aug/23

$$quelestlatransformerdeFourierdelafonction$$$$suivante:$$$$f\left(x\right)={\mathit{e}}^{-\frac{{\mathit{x}}^2}{2}}$$$$\mathit{F}indtheFouriertransformofthe$$$$followingfonction.$$

Answered by witcher3 last updated on 16/Aug/23

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{ixt}}{\mathrm{e}}^{-\frac{{\mathrm{x}}^2}{2}}\mathrm{dx}$$$$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{-\frac{1}{2}({\mathrm{x}}^2+2\mathrm{ixt})}\mathrm{dx}$$$$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{-\frac{1}{2}{(\mathrm{x}+\mathrm{it})}^2-\frac{{\mathrm{t}}^2}{2}}\mathrm{dx}$$$$={\mathrm{e}}^{-\frac{{\mathrm{t}}^2}{2}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{-\frac{1}{2}{(\mathrm{x}+\mathrm{it})}^2}\mathrm{dx}$$$${\mathrm{e}}^{-\frac{{\mathrm{t}}^2}{2}}{\int }_{-\mathrm{\infty }+\mathrm{it}}^{\mathrm{\infty }+\mathrm{it}}{\mathrm{e}}^{-\frac{1}{2}{\mathrm{y}}^2}\mathrm{dy}$$$$\mathrm{D}=]-\mathrm{R},\mathrm{R}[\cup [\mathrm{R},\mathrm{R}+\mathrm{it}]\cup [\mathrm{R}+\mathrm{it},-\mathrm{R}+\mathrm{it}]\cup [-\mathrm{R}+\mathrm{it},-\mathrm{R}]$$$${\int }_{\mathrm{D}}{\mathrm{e}}^{-\frac{1}{2}{\mathrm{y}}^2}\mathrm{dy}=0$$$$\underset{\mathrm{R}\to \mathrm{\infty }}{\lim }\mid {\int }_{\underset{-}{+}\mathrm{R}}^{\underset{-}{+}\mathrm{R}+\mathrm{it}}{\mathrm{e}}^{-\frac{{\mathrm{y}}^2}{2}}\mathrm{dy}\mid ⩽\underset{\mathrm{R}\to \mathrm{\infty }}{\lim }\mid {\mathrm{te}}^{-\frac{1}{2}({\mathrm{R}}^2+{\mathrm{t}}^2)}\mid =0$$$$\Rightarrow {\int }_{-\mathrm{\infty }+\mathrm{it}}^{\mathrm{\infty }+\mathrm{it}}{\mathrm{e}}^{-\frac{{\mathrm{y}}^2}{2}}\mathrm{dy}={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{-\frac{{\mathrm{y}}^2}{2}}=$$$$2{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\frac{{\mathrm{y}}^2}{2}}\mathrm{dy}=\sqrt{2}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}.{\mathrm{x}}^{-\frac{1}{2}}\mathrm{dx}$$$$=\sqrt{2}.\mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{2\pi }$$$$\mathcal{F}\left({\mathrm{e}}^{-\frac{{\mathrm{x}}^2}{2}}\right)=\sqrt{2\pi }{\mathrm{e}}^{-\frac{{\mathrm{t}}^2}{2}}.$$

Commented by witcher3 last updated on 16/Aug/23

$$\mathrm{no}-\frac{{\mathrm{x}}^2}{2}-\mathrm{ixt}=-\frac{1}{2}({\mathrm{x}}^2+2\mathrm{ixt})$$

Commented by pticantor last updated on 16/Aug/23

$$inthe2ndlineyoumakesomemistakelookwell$$

Commented by pticantor last updated on 16/Aug/23

$$thefouriertranformationformuleis:$$$$F\left(f\right)\left(t\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-2\pi ixt}f\left(x\right)dx!!!$$

Commented by witcher3 last updated on 17/Aug/23

$$\mathrm{Ah}\mathrm{yes}\mathrm{forget}\mathrm{the}\mathrm{formula}$$$$\mathrm{i}\mathrm{thougth}\mathrm{it}{\mathrm{e}}^{-\mathrm{ixt}}\mathrm{sorry}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com