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 34296 by math khazana by abdo last updated on 03/May/18

$$find{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{jx}^2}withj=e^{i\frac{2\pi }{3}}$$

Commented by candre last updated on 04/May/18

$$j=\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$$

Commented by abdo mathsup 649 cc last updated on 04/May/18

$$wehavej=cos\left(\frac{2\pi }{3}\right)+isin\left(\frac{2\pi }{3}\right)=-\frac{1}{2}+i\frac{\sqrt{3}}{2}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{jx}^2}dx={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{\left(\sqrt{j}x\right)}^2}dxch.\sqrt{j}x=tgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{jx}^2}dx={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-t^2}\frac{dt}{\sqrt{j}}$$$$=\frac{\sqrt{\pi }}{\sqrt{j}}=\frac{\sqrt{\pi }}{e^{i\frac{\pi }{3}}}=\sqrt{\pi }{e}^{-i\frac{\pi }{3}}=\sqrt{\pi }(cos\left(\frac{\pi }{3}\right)+isin\left(\frac{\pi }{3}\right))$$$$=\sqrt{\pi }(\frac{1}{2}+i\frac{\sqrt{3}}{2})\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{jx}^2}dx=\frac{\sqrt{\pi }}{2}+i\frac{\sqrt{3\pi }}{2}.$$

Commented by abdo mathsup 649 cc last updated on 04/May/18

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{jx}^2}dx=\sqrt{\pi }(cos\left(\frac{\pi }{3}\right)-isin\left(\frac{\pi }{3}\right))$$$$=\sqrt{\pi }(\frac{1}{2}-i\frac{\sqrt{3}}{2})\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{jx}^2}dx=\frac{\sqrt{\pi }}{2}-i\frac{\sqrt{3\pi }}{2}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com