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

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 by Abdo msup. last updated on 06/Dec/18

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})}) \dots b e c o n t i n u e d \dots

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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