let r =(√(p^2 +q^2 )) p and q from R and p>0 q>0 1)prove that ∫_0 ^(+∞) e^(−px) ((cos(px))/(√x))dx=((√π)/r)(√((r+p)/2)) 2) ∫_0 ^∞ e^(−px) ((sin(qx))/(√x))dx =((√π)/r) (√((r−p)/2))


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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 bymath khazana by abdo last updated on 12/Jun/18
$let I = ∫_0 ^∞ e^(−px) ((cos(qx))/(√x)) dx changement (√x)=t give I = ∫_0 ^∞ e^(−pt^2 ) ((cos(qt^2 ))/t) 2tdt = 2 ∫_0 ^∞ e^(−pt^2 ) cos(qt^2 )dt=∫_(−∞) ^(+∞) e^(−pt^2 ) cos(qt^2 )dt =Re( ∫_(−∞) ^(+∞) e^(−pt^2 ) e^(iqt^2 ) dt) but ∫_(−∞) ^(+∞) e^(−pt^2 +iqt^2 ) dt = ∫_(−∞) ^(+∞) e^(−(p−iq)t^2 ) dt =_((√(p−iq))t=u) ∫_(−∞) ^(+∞) e^(−u^2 ) (du/(√(p−iq))) = (1/(√(p−iq)))(√π) but p−iq =(√(p^2 +q^2 )){(p/(√(p^2 +q^2 ))) −i(q/(√(p^2 +q^2 )))} =r{ (p/r) −i(q/r)}=r e^(iθ) ⇒ cosθ = (p/r) and sinθ =((−q)/r) ⇒ tanθ= −(q/p) ⇒θ =−arctan((q/p))⇒ p−iq =r e^(−iarctan((q/p))) and (√(p−iq)) =(√r)e^(−(i/2)arctan((q/(p))))) ⇒ ∫_(−∞) ^(+∞) e^(−(p−iq)^ t^2 ) dt = ((√π)/(√r)) e^((i/2) arctan((q/p))) ⇒ I =((√π)/(√r)) cos(((arctan((q/p)))/2)) but cos((1/2)arctan((q/p)))= (√((1+cos(arctan((q/p))))/2)) =(1/(√2))(√( 1+(1/(√(1+(q^2 /p^2 )))))) =(1/(√2))(√(1+(p/(√(p^2 +q^2 ))))) =(1/(√2))(√((r+p)/r)) = (1/((√2)(√r)))(√(r+p)) ⇒ I =((√π)/(√r)) (1/((√2)(√r))) (√(r+p))= ((√π)/r)(√(((r+p)/2) )) .$
$l e t I = \int_{0}^{\infty} e^{- p x} \frac{c o s (q x)}{\sqrt{x}} d x c h a n g e m e n t \sqrt{x} = t$ $g i v e I = \int_{0}^{\infty} e^{- {p t}^{2}} \frac{c o s ({q t}^{2})}{t} 2 t d t$ $= 2 \int_{0}^{\infty} e^{- {p t}^{2}} c o s ({q t}^{2}) d t = \int_{- \infty}^{+ \infty} e^{- {p t}^{2}} c o s ({q t}^{2}) d t$ $= R e (\int_{- \infty}^{+ \infty} e^{- {p t}^{2}} e^{{i q t}^{2}} d t) b u t$ $\int_{- \infty}^{+ \infty} e^{- {p t}^{2} + {i q t}^{2}} d t = \int_{- \infty}^{+ \infty} e^{- (p - i q) t^{2}} d t$ $=_{\sqrt{p - i q} t = u} \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{d u}{\sqrt{p - i q}}$ $= \frac{1}{\sqrt{p - i q}} \sqrt{π} b u t p - i q = \sqrt{p^{2} + q^{2}} {\frac{p}{\sqrt{p^{2} + q^{2}}} - i \frac{q}{\sqrt{p^{2} + q^{2}}}}$ $= r {\frac{p}{r} - i \frac{q}{r}} = r e^{i θ} \Rightarrow c o s θ = \frac{p}{r} a n d$ $s i n θ = \frac{- q}{r} \Rightarrow t a n θ = - \frac{q}{p} \Rightarrow θ = - a r c t a n (\frac{q}{p}) \Rightarrow$ $p - i q = r e^{- i a r c t a n (\frac{q}{p})} a n d \sqrt{p - i q} = \sqrt{r} e^{- \frac{i}{2} a r c t a n (\frac{q}{p)})}$ $\Rightarrow \int_{- \infty}^{+ \infty} e^{- {(p - i q)}^{} t^{2}} d t = \frac{\sqrt{π}}{\sqrt{r}} e^{\frac{i}{2} a r c t a n (\frac{q}{p})}$ $\Rightarrow I = \frac{\sqrt{π}}{\sqrt{r}} c o s (\frac{a r c t a n (\frac{q}{p})}{2}) b u t$ $c o s (\frac{1}{2} a r c t a n (\frac{q}{p})) = \sqrt{\frac{1 + c o s (a r c t a n (\frac{q}{p}))}{2}}$ $= \frac{1}{\sqrt{2}} \sqrt{1 + \frac{1}{\sqrt{1 + \frac{q^{2}}{p^{2}}}}} = \frac{1}{\sqrt{2}} \sqrt{1 + \frac{p}{\sqrt{p^{2} + q^{2}}}}$ $= \frac{1}{\sqrt{2}} \sqrt{\frac{r + p}{r}} = \frac{1}{\sqrt{2} \sqrt{r}} \sqrt{r + p} \Rightarrow$ $I = \frac{\sqrt{π}}{\sqrt{r}} \frac{1}{\sqrt{2} \sqrt{r}} \sqrt{r + p} = \frac{\sqrt{π}}{r} \sqrt{\frac{r + p}{2}} .$
Commented bymath 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

	$A = \int_{0}^{\infty} e^{- p x} . \frac{c o s q x}{\sqrt{x}} d x$ $B = \int_{0}^{\infty} e^{- p x} \frac{s i n q x}{\sqrt{x}} d x$ $A + i B = \int_{0}^{\infty} e^{- p x} . \frac{e^{i q x}}{\sqrt{x}} d x$ $= \int_{0}^{\infty} e^{- x (p - i q)} \times x^{- \frac{1}{2}} d x$ $t = (p - i q) x$ $d t = (p - i q) d x$ $= \frac{1}{p - i q} \int_{0}^{\infty} e^{- i t} \times {(\frac{t}{p - i q})}^{- \frac{1}{2}} d t$ $= \frac{1}{{(p - i q)}^{\frac{1}{2}}} \int_{0}^{\infty} e^{- i t} \times t^{- \frac{1}{2}} d t$ $= \frac{\sqrt{Π}}{{(p - i q)}^{\frac{1}{2}}}$ $p = r c o s θ q = r s i n θ$ ${(r c o s θ - r s i n θ)}^{\frac{1}{2}}$ $= \sqrt{r} \times e^{- i \frac{θ}{2}}$ $= \sqrt{r} \times (c o s \frac{θ}{2}) + i \sqrt{r} s i n \frac{θ}{2}$ $c o n t d$
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