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Question Number 37347 by math khazana by abdo last updated on 12/Jun/18

l e t r = \sqrt{p^{2} + q^{2}} p a n d q f r o m R a n d p > 0 q > 0

1) p r o v e t h a t \int_{0}^{+ \infty} e^{- p x} \frac{c o s (p x)}{\sqrt{x}} d x = \frac{\sqrt{π}}{r} \sqrt{\frac{r + p}{2}}

2) \int_{0}^{\infty} e^{- p x} \frac{s i n (q x)}{\sqrt{x}} d x = \frac{\sqrt{π}}{r} \sqrt{\frac{r - p}{2}}

Commented by math khazana by abdo last updated on 12/Jun/18

l e t I = \int_{0}^{\infty} e^{- p x} \frac{c o s (q x)}{\sqrt{x}} d x c h a n g e m e n t \sqrt{x} = t

g i v e I = \int_{0}^{\infty} e^{- {p t}^{2}} \frac{c o s ({q t}^{2})}{t} 2 t d t

= 2 \int_{0}^{\infty} e^{- {p t}^{2}} c o s ({q t}^{2}) d t = \int_{- \infty}^{+ \infty} e^{- {p t}^{2}} c o s ({q t}^{2}) d t

= R e (\int_{- \infty}^{+ \infty} e^{- {p t}^{2}} e^{{i q t}^{2}} d t) b u t

\int_{- \infty}^{+ \infty} e^{- {p t}^{2} + {i q t}^{2}} d t = \int_{- \infty}^{+ \infty} e^{- (p - i q) t^{2}} d t

=_{\sqrt{p - i q} t = u} \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{d u}{\sqrt{p - i q}}

= \frac{1}{\sqrt{p - i q}} \sqrt{π} b u t p - i q = \sqrt{p^{2} + q^{2}} {\frac{p}{\sqrt{p^{2} + q^{2}}} - i \frac{q}{\sqrt{p^{2} + q^{2}}}}

= r {\frac{p}{r} - i \frac{q}{r}} = r e^{i θ} \Rightarrow c o s θ = \frac{p}{r} a n d

s i n θ = \frac{- q}{r} \Rightarrow t a n θ = - \frac{q}{p} \Rightarrow θ = - a r c t a n (\frac{q}{p}) \Rightarrow

p - i q = r e^{- i a r c t a n (\frac{q}{p})} a n d \sqrt{p - i q} = \sqrt{r} e^{- \frac{i}{2} a r c t a n (\frac{q}{p)})}

\Rightarrow \int_{- \infty}^{+ \infty} e^{- {(p - i q)}^{} t^{2}} d t = \frac{\sqrt{π}}{\sqrt{r}} e^{\frac{i}{2} a r c t a n (\frac{q}{p})}

\Rightarrow I = \frac{\sqrt{π}}{\sqrt{r}} c o s (\frac{a r c t a n (\frac{q}{p})}{2}) b u t

c o s (\frac{1}{2} a r c t a n (\frac{q}{p})) = \sqrt{\frac{1 + c o s (a r c t a n (\frac{q}{p}))}{2}}

= \frac{1}{\sqrt{2}} \sqrt{1 + \frac{1}{\sqrt{1 + \frac{q^{2}}{p^{2}}}}} = \frac{1}{\sqrt{2}} \sqrt{1 + \frac{p}{\sqrt{p^{2} + q^{2}}}}

= \frac{1}{\sqrt{2}} \sqrt{\frac{r + p}{r}} = \frac{1}{\sqrt{2} \sqrt{r}} \sqrt{r + p} \Rightarrow

I = \frac{\sqrt{π}}{\sqrt{r}} \frac{1}{\sqrt{2} \sqrt{r}} \sqrt{r + p} = \frac{\sqrt{π}}{r} \sqrt{\frac{r + p}{2}} .

Commented by math khazana by abdo last updated on 12/Jun/18

1) t h e Q i s p r o v e t h a t

\int_{0}^{\infty} e^{- p x} \frac{c o s (q x)}{\sqrt{x}} d x = \frac{\sqrt{π}}{r} \sqrt{\frac{r + p}{2}} .

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jun/18

A = \int_{0}^{\infty} e^{- p x} . \frac{c o s q x}{\sqrt{x}} d x

B = \int_{0}^{\infty} e^{- p x} \frac{s i n q x}{\sqrt{x}} d x

A + i B = \int_{0}^{\infty} e^{- p x} . \frac{e^{i q x}}{\sqrt{x}} d x

= \int_{0}^{\infty} e^{- x (p - i q)} \times x^{- \frac{1}{2}} d x

t = (p - i q) x

d t = (p - i q) d x

= \frac{1}{p - i q} \int_{0}^{\infty} e^{- i t} \times {(\frac{t}{p - i q})}^{- \frac{1}{2}} d t

= \frac{1}{{(p - i q)}^{\frac{1}{2}}} \int_{0}^{\infty} e^{- i t} \times t^{- \frac{1}{2}} d t

= \frac{\sqrt{Π}}{{(p - i q)}^{\frac{1}{2}}}

p = r c o s θ q = r s i n θ

{(r c o s θ - r s i n θ)}^{\frac{1}{2}}

= \sqrt{r} \times e^{- i \frac{θ}{2}}

= \sqrt{r} \times (c o s \frac{θ}{2}) + i \sqrt{r} s i n \frac{θ}{2}

c o n t d