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Question Number 35228 by abdo mathsup 649 cc last updated on 16/May/18

find the value of integral ∫_0 ^∞ e^(−px) ((sin(qx))/(√x))dx with p>0 and q>0

$${find}\:{the}\:{value}\:{of}\:{integral} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{px}} \:\:\:\frac{{sin}\left({qx}\right)}{\sqrt{{x}}}{dx}\:\:{with}\:{p}>\mathrm{0}\:{and}\:{q}>\mathrm{0} \\ $$

Commented byabdo mathsup 649 cc last updated on 18/May/18

let put I =∫_0 ^∞ e^(−px) ((sin(qx))/(√x))dx chanbgement (√x) =t give I = ∫_0 ^∞ e^(−pt^2 ) ((sin(qt^2 ))/t) 2t dt I =2 ∫_0 ^∞ e^(−pt^2 ) sin(qt^2 )dt = ∫_(−∞) ^(+∞) e^(−pt^2 ) sin(qt^2 )dt = Im( ∫_(−∞) ^(+∞) e^(−pt^2 +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))) = ((√π)/(√(p−iq))) I = ((√π)/(√(p−iq)))

$${let}\:{put}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−{px}} \:\:\frac{{sin}\left({qx}\right)}{\sqrt{{x}}}{dx}\:\:{chanbgement} \\ $$ $$\sqrt{{x}}\:={t}\:{give} \\ $$ $${I}\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{pt}^{\mathrm{2}} } \:\:\:\:\:\frac{{sin}\left({qt}^{\mathrm{2}} \right)}{{t}}\:\mathrm{2}{t}\:{dt} \\ $$ $${I}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:{e}^{−{pt}^{\mathrm{2}} } \:{sin}\left({qt}^{\mathrm{2}} \right){dt} \\ $$ $$=\:\int_{−\infty} ^{+\infty} \:\:\:{e}^{−{pt}^{\mathrm{2}} } {sin}\left({qt}^{\mathrm{2}} \right){dt}\:=\:{Im}\left(\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{pt}^{\mathrm{2}} \:+{iqt}^{\mathrm{2}} } {dt}\right) \\ $$ $${but}\:\:\int_{−\infty} ^{+\infty} \:\:\:{e}^{−{pt}^{\mathrm{2}} \:+{iqt}^{\mathrm{2}} } {dt}=\:\int_{−\infty} ^{+\infty} \:\:{e}^{−\left({p}−{iq}\right){t}^{\mathrm{2}} } {dt} \\ $$ $$=_{\sqrt{{p}−{iq}}{t}=\:{u}} \:\:\:\int_{−\infty} ^{+\infty} \:\:\:{e}^{−{u}^{\mathrm{2}} } \:\:\:\frac{{du}}{\sqrt{{p}−{iq}}} \\ $$ $$=\:\frac{\sqrt{\pi}}{\sqrt{{p}−{iq}}} \\ $$ $${I}\:\:=\:\frac{\sqrt{\pi}}{\sqrt{{p}−{iq}}} \\ $$ $$ \\ $$

Commented byabdo mathsup 649 cc last updated on 18/May/18

I =Im( ((√π)/(√(p−iq)))) but p−iq =(√(p^2 +q^2 ))((p/(√(p^2 +q^2 ))) −i(q/(√(p^2 +q^2 ))))=r e^(iθ) ⇒ r=(√(p^2 +q^2 )) and tanθ=−(q/p) ⇒θ=−arctan((q/p)) (√(p−iq)) = (√r) e^(i(θ/2)) = (p^2 +q^2 )^(1/4) e^(−(i/2)arctan((q/p))) I =Im( (√π) (p^2 +q^2 )^(−(1/4)) e^((i/2)arctan((q/p))) ) I = (√π) (p^2 +q^2 )^(−(1/4)) sin((1/2)arctan((q/p)))

$${I}\:={Im}\left(\:\:\frac{\sqrt{\pi}}{\sqrt{{p}−{iq}}}\right)\:\:{but} \\ $$ $${p}−{iq}\:=\sqrt{{p}^{\mathrm{2}} \:+{q}^{\mathrm{2}} }\left(\frac{{p}}{\sqrt{{p}^{\mathrm{2}} \:+{q}^{\mathrm{2}} }}\:−{i}\frac{{q}}{\sqrt{{p}^{\mathrm{2}} \:+{q}^{\mathrm{2}} }}\right)={r}\:{e}^{{i}\theta} \:\Rightarrow \\ $$ $${r}=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }\:\:\:{and}\:\:\:{tan}\theta=−\frac{{q}}{{p}}\:\Rightarrow\theta=−{arctan}\left(\frac{{q}}{{p}}\right) \\ $$ $$\sqrt{{p}−{iq}}\:=\:\sqrt{{r}}\:\:{e}^{{i}\frac{\theta}{\mathrm{2}}} =\:\left({p}^{\mathrm{2}} \:+{q}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{{q}}{{p}}\right)} \\ $$ $${I}\:={Im}\left(\:\sqrt{\pi}\:\:\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{{q}}{{p}}\right)} \right) \\ $$ $${I}\:=\:\sqrt{\pi}\:\:\left({p}^{\mathrm{2}} \:+{q}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} \:\:{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{{q}}{{p}}\right)\right) \\ $$ $$ \\ $$

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