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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 by math 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 by math 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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