Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 193316 by gatocomcirrose last updated on 10/Jun/23

$$$$$$\mathrm{IS}\mathrm{THIS}\mathrm{RIGHT}?$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{ix}}^2}\mathrm{dx}$$$${\mathrm{I}}^2={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{ix}}^2}\mathrm{dx}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{iy}}^2}\mathrm{dy}$$$$={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{iy}}^2}{\mathrm{dye}}^{-{\mathrm{ix}}^2}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{i}({\mathrm{x}}^2+{\mathrm{y}}^2)}\mathrm{dydx}$$$$\mathrm{dydx}=\mathrm{dA}=\mathrm{rdrd}\theta$$$${\mathrm{I}}^2={\int }_0^{\frac{\pi }{2}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{ir}}^2}\mathrm{rdrd}\theta$$$$$$$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{ir}}^2}\mathrm{rdr};\mathrm{u}=-{\mathrm{ir}}^2\Rightarrow \mathrm{du}=-2\mathrm{irdr}$$$$-\frac{\mathrm{i}}{2}{\int }_{-\mathrm{\infty }}^0{\mathrm{e}}^{\mathrm{u}}\mathrm{du}=-\frac{\mathrm{i}}{2}$$$$$$$${\mathrm{I}}^2={\int }_0^{\frac{\pi }{2}}-\frac{\mathrm{i}}{2}\mathrm{d}\theta =-\frac{\mathrm{i}\pi }{4}\Rightarrow \mathrm{I}=\sqrt{\frac{-\mathrm{i}\pi }{4}}$$$$\mathrm{I}=\frac{\mathrm{i}}{2}\sqrt{{\mathrm{e}}^{\mathrm{i}\pi /2}\pi }=\frac{{\mathrm{ie}}^{\mathrm{i}\pi /4}}{2}\sqrt{\pi }$$$$\mathrm{I}=\frac{\mathrm{i}\sqrt{\pi }}{2}\left(\frac{\sqrt{2}}{2}(1+\mathrm{i})\right)=\frac{\mathrm{i}\sqrt{2\pi }}{4}(1+\mathrm{i})$$

Commented by aba last updated on 10/Jun/23

$$\mathrm{wrong}\mathrm{I}=-\frac{\mathrm{i}\sqrt{2\pi }}{4}(\mathrm{i}+1)$$

Answered by aba last updated on 10/Jun/23

$$\mathrm{I}=\sqrt{\frac{-\mathrm{i}\pi }{4}}=\frac{1}{2}\sqrt{{\mathrm{e}}^{-\mathrm{i}\frac{\pi }{2}}\pi }=\frac{{\mathrm{e}}^{-\mathrm{i}\frac{\pi }{4}}}{2}\sqrt{\pi }=\frac{\sqrt{2\pi }}{4}(1-\mathrm{i})=-\frac{\mathrm{i}\sqrt{2\pi }}{4}(\mathrm{i}+1)=\sqrt{\frac{\pi }{2}}(\frac{1}{2}-\frac{\mathrm{i}}{2})✓$$

Answered by Mathspace last updated on 15/Jun/23

$$I={\int }_0^{\mathrm{\infty }}{e}^{-{ix}^2}dxwedo\sqrt{i}x=t\Rightarrow$$$$I={\int }_0^{\mathrm{\infty }}{e}^{-t^2}\frac{dt}{\sqrt{i}}ori=e^{\frac{i\pi }{2}}\Rightarrow$$$$\sqrt{i}=e^{\frac{i\pi }{4}}\Rightarrow I=e^{-\frac{i\pi }{4}}.{\int }_0^{\mathrm{\infty }}{e}^{-t^2}dt$$$$=(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)\times \frac{\sqrt{\pi }}{2}$$$$=\frac{\sqrt{\pi }}{2\sqrt{2}}(1-i)$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com