let α>0 calculate ∫_(−∞) ^(+∞) (1+αi)^(−x^2 ) dx .


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Question Number 49344 by maxmathsup by imad last updated on 05/Dec/18

$l e t α > 0 c a l c u l a t e \int_{- \infty}^{+ \infty} {(1 + α i)}^{- x^{2}} d x .$
Commented byAbdo msup. last updated on 06/Dec/18
$let A =∫_(−∞) ^(+∞) (1+αi)^(−x^2 ) dx ⇒A =∫_(−∞) ^(+∞) e^(−x^2 ln(1+αi)) dx =_(x(√(ln(1+αi)))=t) ∫_(−∞) ^(+∞) e^(−t^2 ) (dt/(√(ln(1+αi)))) =((√π)/(√(ln(1+αi)))) but 1+αi =(√(1+α^2 ))((1/(√(1+α^2 ))) +i(α/(√(1+α^2 )))) =r e^(iθ) ⇒r =(1+α^2 )^(1/2) and tanθ =α ⇒θ =arctan(α) ⇒ ln(1+αi)=ln(r)+iθ =(1/2)ln(1+α^2 )+iarctan(α) ⇒ A = ((√π)/(√((1/2)ln(1+α^2 )+i arctan(α)))) leyZ=(1/2)ln(1+α^2 )+i arctan(α) we have ∣Z∣=(√((1/4)ln^2 (1+α^2 )+arctan^2 (α))) ⇒ Z =∣Z∣{((ln(1+α^2 ))/(2(√((1/4)ln^2 (1+α^2 )+arctan^2 (α))))) +i((arctan(α))/(√((1/4)ln^2 (1+α^2 )+arctan^2 (α))))} =r e^(iθ) ⇒r =∣Z∣ and tanθ = ((2arctan(α))/(ln(1+α^2 ))) ⇒ θ =arctan(2 ((arctan(α))/(ln(1+α^2 ))))...be continued...$
$l e t A = \int_{- \infty}^{+ \infty} {(1 + α i)}^{- x^{2}} d x \Rightarrow A = \int_{- \infty}^{+ \infty} e^{- x^{2} l n (1 + α i)} d x$ $=_{x \sqrt{l n (1 + α i)} = t} \int_{- \infty}^{+ \infty} e^{- t^{2}} \frac{d t}{\sqrt{l n (1 + α i)}}$ $= \frac{\sqrt{π}}{\sqrt{l n (1 + α i)}} b u t 1 + α i = \sqrt{1 + α^{2}} (\frac{1}{\sqrt{1 + α^{2}}} + i \frac{α}{\sqrt{1 + α^{2}}})$ $= r e^{i θ} \Rightarrow r = {(1 + α^{2})}^{\frac{1}{2}} a n d t a n θ = α \Rightarrow θ = a r c t a n (α) \Rightarrow$ $l n (1 + α i) = l n (r) + i θ = \frac{1}{2} l n (1 + α^{2}) + i a r c t a n (α) \Rightarrow$ $A = \frac{\sqrt{π}}{\sqrt{\frac{1}{2} l n (1 + α^{2}) + i a r c t a n (α)}} l e y Z = \frac{1}{2} l n (1 + α^{2}) + i a r c t a n (α)$ $w e h a v e ∣ Z ∣= \sqrt{\frac{1}{4} {l n}^{2} (1 + α^{2}) + {a r c t a n}^{2} (α)} \Rightarrow$ $Z =∣ Z ∣ {\frac{l n (1 + α^{2})}{2 \sqrt{\frac{1}{4} {l n}^{2} (1 + α^{2}) + {a r c t a n}^{2} (α)}} + i \frac{a r c t a n (α)}{\sqrt{\frac{1}{4} {l n}^{2} (1 + α^{2}) + {a r c t a n}^{2} (α)}}}$ $= r e^{i θ} \Rightarrow r =∣ Z ∣ a n d t a n θ = \frac{2 a r c t a n (α)}{l n (1 + α^{2})} \Rightarrow$ $θ = a r c t a n (2 \frac{a r c t a n (α)}{l n (1 + α^{2})}) . . . b e c o n t i n u e d . . .$
	Answered by Smail last updated on 06/Dec/18

	${(1 + α i)}^{- x^{2}} = e^{- x^{2} l n (1 + α i)}$ $u = x \sqrt{l n (1 + α i)}$ $d x = \frac{d u}{\sqrt{l n (1 + α i)}}$ $\int_{- \infty}^{\infty} {(1 + α i)}^{- x^{2}} d x = \frac{1}{\sqrt{l n (1 + α i)}} \int_{- \infty}^{\infty} e^{- u^{2}} d u$ $= \frac{\sqrt{π}}{\sqrt{l n (1 + α i)}}$
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