calculate-0-1-i-x-2-dx-

Question Number 36917 by maxmathsup by imad last updated on 07/Jun/18

c a l c u l a t e \int_{0}^{+ \infty} {(1 + i)}^{- x^{2}} d x

Commented by math khazana by abdo last updated on 09/Jun/18

w e h a v e (1 + i) = \sqrt{2} e^{i \frac{π}{4}} \Rightarrow

I = \int_{0}^{+ \infty} {(\sqrt{2} e^{i \frac{π}{4}})}^{- x^{2}} d x

= \int_{0}^{+ \infty} {e^{l n (\sqrt{2}) + i \frac{π}{4}}}^{- x^{2}} d x

= \int_{0}^{+ \infty} e^{- {l n (\sqrt{2}) + i \frac{π}{4}} x^{2}} d x c h a n g e m e n t

\sqrt{l n (\sqrt{2}) + i \frac{π}{4}} x = t g i v e

I = \frac{1}{\sqrt{l n (\sqrt{2}) + i \frac{π}{4}}} \int_{0}^{\infty} e^{- u^{2}} d u

I = \frac{\sqrt{π}}{2 {\sqrt{l n (\sqrt{2}) + i \frac{π}{4}}}} .