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

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))

$${let}\:{r}\:=\sqrt{{p}^{\mathrm{2}} \:+{q}^{\mathrm{2}} }\:\:\:{p}\:{and}\:{q}\:{from}\:{R}\:\:{and}\:{p}>\mathrm{0}\:\:{q}>\mathrm{0} \\ $$ $$\left.\mathrm{1}\right){prove}\:{that}\:\:\int_{\mathrm{0}} ^{+\infty} \:\:{e}^{−{px}} \:\frac{{cos}\left({px}\right)}{\sqrt{{x}}}{dx}=\frac{\sqrt{\pi}}{{r}}\sqrt{\frac{{r}+{p}}{\mathrm{2}}} \\ $$ $$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{px}} \:\:\frac{{sin}\left({qx}\right)}{\sqrt{{x}}}{dx}\:=\frac{\sqrt{\pi}}{{r}}\:\sqrt{\frac{{r}−{p}}{\mathrm{2}}} \\ $$

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) )) .

$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{px}} \:\frac{{cos}\left({qx}\right)}{\sqrt{{x}}}\:{dx}\:{changement}\:\sqrt{{x}}={t} \\ $$ $${give}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{pt}^{\mathrm{2}} } \:\frac{{cos}\left({qt}^{\mathrm{2}} \right)}{{t}}\:\mathrm{2}{tdt} \\ $$ $$=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−{pt}^{\mathrm{2}} } {cos}\left({qt}^{\mathrm{2}} \right){dt}=\int_{−\infty} ^{+\infty} \:\:{e}^{−{pt}^{\mathrm{2}} } \:{cos}\left({qt}^{\mathrm{2}} \right){dt} \\ $$ $$={Re}\left(\:\int_{−\infty} ^{+\infty} \:{e}^{−{pt}^{\mathrm{2}} } \:{e}^{{iqt}^{\mathrm{2}} } {dt}\right)\:\:{but}\: \\ $$ $$\int_{−\infty} ^{+\infty} \:\:\:{e}^{−{pt}^{\mathrm{2}} \:+{iqt}^{\mathrm{2}} } {dt}\:\:=\:\int_{−\infty} ^{+\infty} \:\:{e}^{−\left({p}−{iq}\right){t}^{\mathrm{2}} } {dt} \\ $$ $$=_{\sqrt{{p}−{iq}}{t}={u}} \:\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{u}^{\mathrm{2}} } \:\:\frac{{du}}{\sqrt{{p}−{iq}}} \\ $$ $$=\:\frac{\mathrm{1}}{\sqrt{{p}−{iq}}}\sqrt{\pi}\:\:\:\:\:\:{but}\:\:{p}−{iq}\:=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }\left\{\frac{{p}}{\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}\:−{i}\frac{{q}}{\sqrt{{p}^{\mathrm{2}} \:+{q}^{\mathrm{2}} }}\right\} \\ $$ $$={r}\left\{\:\frac{{p}}{{r}}\:−{i}\frac{{q}}{{r}}\right\}={r}\:{e}^{{i}\theta} \:\Rightarrow\:{cos}\theta\:=\:\frac{{p}}{{r}}\:{and} \\ $$ $${sin}\theta\:=\frac{−{q}}{{r}}\:\Rightarrow\:{tan}\theta=\:−\frac{{q}}{{p}}\:\Rightarrow\theta\:=−{arctan}\left(\frac{{q}}{{p}}\right)\Rightarrow \\ $$ $${p}−{iq}\:={r}\:{e}^{−{iarctan}\left(\frac{{q}}{{p}}\right)} \:{and}\:\sqrt{{p}−{iq}}\:=\sqrt{{r}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{{q}}{\left.{p}\right)}\right)} \\ $$ $$\Rightarrow\:\int_{−\infty} ^{+\infty} \:\:{e}^{−\left({p}−{iq}\right)^{} {t}^{\mathrm{2}} } {dt}\:=\:\frac{\sqrt{\pi}}{\sqrt{{r}}}\:{e}^{\frac{{i}}{\mathrm{2}}\:{arctan}\left(\frac{{q}}{{p}}\right)} \\ $$ $$\Rightarrow\:{I}\:\:=\frac{\sqrt{\pi}}{\sqrt{{r}}}\:{cos}\left(\frac{{arctan}\left(\frac{{q}}{{p}}\right)}{\mathrm{2}}\right)\:{but} \\ $$ $${cos}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{{q}}{{p}}\right)\right)=\:\sqrt{\frac{\mathrm{1}+{cos}\left({arctan}\left(\frac{{q}}{{p}}\right)\right)}{\mathrm{2}}} \\ $$ $$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\sqrt{\:\mathrm{1}+\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\frac{{q}^{\mathrm{2}} }{{p}^{\mathrm{2}} }}}}\:\:=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\sqrt{\mathrm{1}+\frac{{p}}{\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }}} \\ $$ $$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\sqrt{\frac{{r}+{p}}{{r}}}\:\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\sqrt{{r}}}\sqrt{{r}+{p}}\:\:\Rightarrow \\ $$ $${I}\:\:=\frac{\sqrt{\pi}}{\sqrt{{r}}}\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\sqrt{{r}}}\:\sqrt{{r}+{p}}=\:\frac{\sqrt{\pi}}{{r}}\sqrt{\frac{{r}+{p}}{\mathrm{2}}\:}\:. \\ $$ $$ \\ $$

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

1) the Q is prove that  ∫_0 ^∞    e^(−px)   ((cos(qx))/(√x)) dx = ((√π)/r)(√((r+p)/2))  .

$$\left.\mathrm{1}\right)\:{the}\:{Q}\:{is}\:{prove}\:{that} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−{px}} \:\:\frac{{cos}\left({qx}\right)}{\sqrt{{x}}}\:{dx}\:=\:\frac{\sqrt{\pi}}{{r}}\sqrt{\frac{{r}+{p}}{\mathrm{2}}}\:\:. \\ $$

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

A=∫_0 ^∞ e^(−px) .((cosqx)/(√x))dx  B=∫_0 ^∞ e^(−px) ((sinqx)/(√x))dx  A+iB=∫_0 ^∞ e^(−px) .(e^(iqx) /(√x))dx  =∫_0 ^∞ e^(−x(p−iq)) ×x^(−(1/2)) dx  t=(p−iq)x  dt=(p−iq)dx  =(1/(p−iq))∫_0 ^∞ e^(−it) ×((t/(p−iq)))^(−(1/2)) dt  =(1/((p−iq)^(1/2) ))∫_0 ^∞ e^(−it) ×t^(−(1/2)) dt  =((√Π)/((p−iq)^(1/2) ))  p=rcosθ  q=rsinθ  (rcosθ−rsinθ)^(1/2)   =(√r) ×e^(−i(θ/2))   =(√r) ×(cos(θ/2))+i(√r) sin(θ/2)  contd

$${A}=\int_{\mathrm{0}} ^{\infty} {e}^{−{px}} .\frac{{cosqx}}{\sqrt{{x}}}{dx} \\ $$ $${B}=\int_{\mathrm{0}} ^{\infty} {e}^{−{px}} \frac{{sinqx}}{\sqrt{{x}}}{dx} \\ $$ $${A}+{iB}=\int_{\mathrm{0}} ^{\infty} {e}^{−{px}} .\frac{{e}^{{iqx}} }{\sqrt{{x}}}{dx} \\ $$ $$=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}\left({p}−{iq}\right)} ×{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$ $${t}=\left({p}−{iq}\right){x} \\ $$ $${dt}=\left({p}−{iq}\right){dx} \\ $$ $$=\frac{\mathrm{1}}{{p}−{iq}}\int_{\mathrm{0}} ^{\infty} {e}^{−{it}} ×\left(\frac{{t}}{{p}−{iq}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dt} \\ $$ $$=\frac{\mathrm{1}}{\left({p}−{iq}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\int_{\mathrm{0}} ^{\infty} {e}^{−{it}} ×{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} {dt} \\ $$ $$=\frac{\sqrt{\Pi}}{\left({p}−{iq}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$ $${p}={rcos}\theta\:\:{q}={rsin}\theta \\ $$ $$\left({rcos}\theta−{rsin}\theta\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$ $$=\sqrt{{r}}\:×{e}^{−{i}\frac{\theta}{\mathrm{2}}} \\ $$ $$=\sqrt{{r}}\:×\left({cos}\frac{\theta}{\mathrm{2}}\right)+{i}\sqrt{{r}}\:{sin}\frac{\theta}{\mathrm{2}} \\ $$ $${contd} \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$

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