calculate-g-e-x-2-sin-sin-x-2-dx-

Question Number 37291 by math khazana by abdo last updated on 11/Jun/18

c a l c u l a t e g (θ) = \int_{- \infty}^{+ \infty} e^{- x^{2}} s i n (s i n θ x^{2}) d x .

Commented by math khazana by abdo last updated on 16/Jun/18

l e t λ = s i n θ \Rightarrow g (θ) = \int_{- \infty}^{+ \infty} e^{- x^{2}} s i n (λ x^{2}) d x

= I m {\int_{- \infty}^{+ \infty} e^{- x^{2} + i λ x^{2}} d x} b u t

\int_{- \infty}^{+ \infty} e^{- x^{2} + i λ x^{2}} d x = \int_{- \infty}^{+ \infty} e^{- (1 - i λ) x^{2}} d x

=_{\sqrt{1 - i λ} x = t} \int_{- \infty}^{+ \infty} e^{- t^{2}} \frac{d t}{\sqrt{1 - i λ}}

= \frac{1}{\sqrt{1 - i λ}} \sqrt{π} b u t

1 - i λ = \sqrt{1 + λ^{2}} {\frac{1}{\sqrt{1 + λ^{2}}} + i \frac{- λ}{\sqrt{1 + λ^{2}}}} = r e^{i φ} \Rightarrow

r = \sqrt{1 + λ^{2}} a n d c o s φ = \frac{1}{\sqrt{1 + λ^{2}}} a n d s i n φ = \frac{- λ}{\sqrt{1 + λ^{2}}}

\Rightarrow t a n φ = - λ \Rightarrow φ = - a r c t a n (λ)

1 - i λ = \sqrt{1 + λ^{2}} e^{- i a r t a n (λ)} \Rightarrow

\sqrt{1 - i λ_{}} = {(1 + λ^{2})}^{\frac{1}{4}} e^{- \frac{i}{2} a r c t a n (λ)} \Rightarrow

\int_{- \infty}^{+ \infty} e^{- x^{2} + i λ x^{2}} d x = \sqrt{π} {(1 + λ^{2})}^{- \frac{1}{4}} e^{\frac{i}{2} a r c t a n (λ)}

= \sqrt{π} {(1 + λ^{2})}^{- \frac{1}{4}} {c o s (\frac{a r c t a n (λ)}{2}) + i s i n (\frac{a r c t a n (λ)}{2}}

g (θ) = \sqrt{π} {(1 + {s i n}^{2} θ)}^{- \frac{1}{4}} s i n (\frac{a r c t a n (s i n θ}{2}) .