find-e-jx-2-with-j-e-i-2pi-3-

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

f i n d \int_{- \infty}^{+ \infty} e^{- {j x}^{2}} w i t h j = e^{i \frac{2 π}{3}}

Commented by candre last updated on 04/May/18

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

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

w e h a v e j = c o s (\frac{2 π}{3}) + i s i n (\frac{2 π}{3}) = - \frac{1}{2} + i \frac{\sqrt{3}}{2} \Rightarrow

\int_{- \infty}^{+ \infty} e^{- {j x}^{2}} d x = \int_{- \infty}^{+ \infty} e^{- {(\sqrt{j} x)}^{2}} d x c h . \sqrt{j} x = t g i v e

\int_{- \infty}^{+ \infty} e^{- {j x}^{2}} d x = \int_{- \infty}^{+ \infty} e^{- t^{2}} \frac{d t}{\sqrt{j}}

= \frac{\sqrt{π}}{\sqrt{j}} = \frac{\sqrt{π}}{e^{i \frac{π}{3}}} = \sqrt{π} e^{- i \frac{π}{3}} = \sqrt{π} (c o s (\frac{π}{3}) + i s i n (\frac{π}{3}))

= \sqrt{π} (\frac{1}{2} + i \frac{\sqrt{3}}{2}) \Rightarrow

\int_{- \infty}^{+ \infty} e^{- {j x}^{2}} d x = \frac{\sqrt{π}}{2} + i \frac{\sqrt{3 π}}{2} .

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

\int_{- \infty}^{+ \infty} e^{- {j x}^{2}} d x = \sqrt{π} (c o s (\frac{π}{3}) - i s i n (\frac{π}{3}))

= \sqrt{π} (\frac{1}{2} - i \frac{\sqrt{3}}{2}) \Rightarrow

\int_{- \infty}^{+ \infty} e^{- {j x}^{2}} d x = \frac{\sqrt{π}}{2} - i \frac{\sqrt{3 π}}{2}