find-the-value-of-integral-0-e-2-ia-2-t-2-dt-with-a-from-R-a-lt-1-

Question Number 35229 by abdo mathsup 649 cc last updated on 16/May/18

f i n d t h e v a l u e o f i n t e g r a l

\int_{0}^{\infty} e^{- {(2 + i a)}^{2} t^{2}} d t w i t h a f r o m R ∣ a ∣< 1 .

Commented by abdo mathsup 649 cc last updated on 17/May/18

R e m a r k w e h a v e A (α) = \int_{0}^{\infty} e^{- (4 + 4 i α - α^{2}) t^{2}} d t

= \int_{0}^{\infty} e^{- (4 - α^{2}) t^{2} - 4 i α t^{2}} d t

= \int_{0}^{\infty} e^{(α^{2} - 4) t^{2}} {c o s (4 α t^{2}) - i s i n (4 α t^{2})} d t =

\int_{0}^{\infty} e^{(α^{2} - 4) t^{2}} c o s (4 α t^{2}) - i \int_{0}^{\infty} e^{(α^{2} - 4) t^{2}} s i n (4 α t^{2}) d t

\Rightarrow \int_{0}^{\infty} e^{(α^{2} - 4) t^{2}} c o s (4 α t^{2}) d t = \frac{\sqrt{π}}{4 + α^{2}} a n d

\int_{0}^{\infty} e^{(α^{2} - 4) t^{2}} s i n (4 α t^{2}) d t = \frac{α \sqrt{π}}{8 + 2 α^{2}} .

Commented by abdo mathsup 649 cc last updated on 17/May/18

l e t p u t A (α) = \int_{0}^{\infty} e^{- {(2 + i α)}^{2} t^{2}} d t c h a n g e m e n t

(2 + i α) t = x g i v e

A (α) = \frac{1}{2 + i α} \int_{0}^{\infty} e^{- x^{2}} d x = \frac{\sqrt{π}}{2 (2 + i α)}

= \frac{\sqrt{π}}{2} \frac{2 - i α}{4 + α^{2}} = \frac{\sqrt{π}}{4 + α^{2}} - \frac{α \sqrt{π}}{2 (4 + α^{2})}

Answered by sma3l2996 last updated on 16/May/18

l e t u = (2 + i a) t \Rightarrow d t = \frac{d u}{2 + i a}

I = \int_{0}^{\infty} e^{- {(2 + i a)}^{2} t^{2}} d t = \frac{1}{2 + i a} \int_{0}^{\infty} e^{- u^{2}} d u

= \frac{\sqrt{π}}{2 (2 + i a)}