find-in-terms-of-0-e-t-sint-t-dt-with-gt-0-

Question Number 28756 by abdo imad last updated on 29/Jan/18

f i n d i n t e r m s o f λ \int_{0}^{\infty} e^{- λ t} \frac{s i n t}{\sqrt{t}} d t w i t h λ > 0

Commented by abdo imad last updated on 30/Jan/18

l e t p u t I = \int_{0}^{\infty} e^{- λ t} \frac{s i n t}{\sqrt{t}} d t w i t h λ > 0 t h e c h . \sqrt{t} = x g i v e

I = \int_{0}^{\infty} e^{- λ x^{2}} \frac{s i n (x^{2})}{x} 2 x d x = 2 \int_{0}^{\infty} e^{- λ x^{2}} s i n (x^{2}) d x

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

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

= \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{d u}{\sqrt{λ + i}} (c h . \sqrt{λ + i} x = u)

= {(λ + i)}^{- \frac{1}{2}} . \sqrt{π} w e h a v e

λ + i = \sqrt{1 + λ^{2}} (\frac{λ}{\sqrt{1 + λ^{2}}} + \frac{i}{\sqrt{1 + λ^{2}}}) = r e^{i θ} \Rightarrow r = \sqrt{1 + λ^{2}}

a n d c o s θ = \frac{λ}{\sqrt{1 + λ^{2}}} a n d s i n θ = \frac{1}{\sqrt{1 + λ^{2}}} \Rightarrow t a n θ = \frac{1}{λ}

\Rightarrow θ = a r c t a n (\frac{1}{λ}) s o λ + i = {(1 + λ^{2})}^{\frac{1}{2}} e^{i a r c t a n (\frac{1}{λ})}

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

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

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

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

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