find the value of integral ∫_0 ^∞ e^(−(2+ia)^2 t^2 ) dt with a from R ∣a∣<1.


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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
$Remark we have A(α)= ∫_0 ^∞ e^(−(4 +4iα −α^2 )t^2 ) dt = ∫_0 ^∞ e^(−(4 −α^2 )t^2 −4iαt^2 ) dt = ∫_0 ^∞ e^((α^2 −4)t^2 ) {cos(4αt^2 ) −isin(4αt^2 )}dt = ∫_0 ^∞ e^((α^2 −4)t^2 ) cos(4αt^2 ) −i ∫_0 ^∞ e^((α^2 −4)t^2 ) sin(4αt^2 )dt ⇒ ∫_0 ^∞ e^((α^2 −4)t^2 ) cos(4αt^2 )dt = ((√π)/(4+α^2 )) and ∫_0 ^∞ e^((α^2 −4)t^2 ) sin(4αt^2 )dt = ((α(√π))/(8 +2α^2 )) .$
$R e m a r k w e h a v e A (α) = \int_{0}^{\infty} e^{- (4 + 4 i α - α^{2}) t^{2}} d t$ $= \int_{0}^{\infty} e^{- (4 - α^{2}) t^{2} - 4 i α t^{2}} d t$ $= \int_{0}^{\infty} e^{(α^{2} - 4) t^{2}} {c o s (4 α t^{2}) - i s i n (4 α t^{2})} d t =$ $\int_{0}^{\infty} e^{(α^{2} - 4) t^{2}} c o s (4 α t^{2}) - i \int_{0}^{\infty} e^{(α^{2} - 4) t^{2}} s i n (4 α t^{2}) d t$ $\Rightarrow \int_{0}^{\infty} e^{(α^{2} - 4) t^{2}} c o s (4 α t^{2}) d t = \frac{\sqrt{π}}{4 + α^{2}} a n d$ $\int_{0}^{\infty} e^{(α^{2} - 4) t^{2}} s i n (4 α t^{2}) d t = \frac{α \sqrt{π}}{8 + 2 α^{2}} .$
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

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