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find-the-value-of-integral-0-e-px-sin-qx-x-dx-with-p-gt-0-and-q-gt-0-




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
findthevalueofintegral0epxsin(qx)xdxwithp>0andq>0
Commented by abdo 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))))
letputI=0epxsin(qx)xdxchanbgementx=tgiveI=0ept2sin(qt2)t2tdtI=20ept2sin(qt2)dt=+ept2sin(qt2)dt=Im(+ept2+iqt2dt)but+ept2+iqt2dt=+e(piq)t2dt=piqt=u+eu2dupiq=πpiqI=πpiq
Commented by abdo 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(πpiq)butpiq=p2+q2(pp2+q2iqp2+q2)=reiθr=p2+q2andtanθ=qpθ=arctan(qp)piq=reiθ2=(p2+q2)14ei2arctan(qp)I=Im(π(p2+q2)14ei2arctan(qp))I=π(p2+q2)14sin(12arctan(qp))

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