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1-find-f-a-e-ax-2-cos-3-x-2-dx-with-a-gt-0-2-find-the-value-of-0-e-3x-2-cos-3-x-2-dx-




Question Number 65286 by mathmax by abdo last updated on 27/Jul/19
1)find   f(a)=∫_(−∞) ^(+∞)   e^(−ax^2 ) cos(3−x^2 )dx  with a>0  2) find the value of ∫_0 ^∞  e^(−3x^2 ) cos(3−x^2 )dx
1)findf(a)=+eax2cos(3x2)dxwitha>02)findthevalueof0e3x2cos(3x2)dx
Commented by mathmax by abdo last updated on 29/Jul/19
1) we have f(a) =∫_(−∞) ^(+∞)  e^(−ax^2 ) cos(3−x^2 )dx ⇒  f(a) =Re(∫_(−∞) ^(+∞)   e^(−ax^2 +i(3−x^2 )) dx)  I=∫_(−∞) ^(+∞)  e^(−ax^2  +i(3−x^2 )) dx =e^(3i)  ∫_(−∞) ^(+∞)  e^(−(a+i)x^2 ) dx  changement (√(a+i))x=t  give I =e^(3i)    ∫_(−∞) ^(+∞)   e^(−t^2 )  (dt/( (√(a+i)))) =(e^(3i) /( (√(a+i)))) (√π)  a+i =(√(1+a^2 )) {(a/( (√(1+a^2 )))) +(i/( (√(1+a^2 ))))} =r e^(iθ)  ⇒r =(√(1+a^2 ))  and θ =arctan((1/a))  (√(a+i))=(√r)e^((i/2)θ)  =(1+a^2 )^(1/4)  e^((i/2)arctan((1/a)))  =^4 (√(1+a^2 ))e^((i/2)((π/2)−arctana))   =^4 (√(1+a^2 )) e^(((iπ)/4)−(i/2)arctana)  ⇒  I =(√π)  (e^(3i) /((^4 (√(1+a^2 )))e^(((iπ)/4)−(i/2)arctan(a)) )) =((√π)/((^4 (√(1+a^2 ))))) e^(3i−((iπ)/4) +(i/2)arctan(a))   =((√π)/((^4 (√(1+a^2 ))))) e^(i(3−(π/4) +(1/2)arctan(a)))  ⇒  f(a) =((√π)/((^4 (√(1+a^2 ))))) cos(3−(π/4)+((arctana)/2)) .  2)∫_0 ^∞  e^(−3x^2 ) cos(3−x^2 )dx =(1/2)∫_(−∞) ^(+∞)  e^(−3x^2 ) cos(3−x^2 )dx  =(1/2)f(3) =((√π)/((^4 (√(10)))))cos(3−(π/4) +((arctan(3))/2))
1)wehavef(a)=+eax2cos(3x2)dxf(a)=Re(+eax2+i(3x2)dx)I=+eax2+i(3x2)dx=e3i+e(a+i)x2dxchangementa+ix=tgiveI=e3i+et2dta+i=e3ia+iπa+i=1+a2{a1+a2+i1+a2}=reiθr=1+a2andθ=arctan(1a)a+i=rei2θ=(1+a2)14ei2arctan(1a)=41+a2ei2(π2arctana)=41+a2eiπ4i2arctanaI=πe3i(41+a2)eiπ4i2arctan(a)=π(41+a2)e3iiπ4+i2arctan(a)=π(41+a2)ei(3π4+12arctan(a))f(a)=π(41+a2)cos(3π4+arctana2).2)0e3x2cos(3x2)dx=12+e3x2cos(3x2)dx=12f(3)=π(410)cos(3π4+arctan(3)2)
Commented by mathmax by abdo last updated on 29/Jul/19
∫_0 ^∞   e^(−3x^2 ) cos(3−x^2 )dx =((√π)/(2(^4 (√(10)))))cos(3−(π/4) +((arctan(3))/2))
0e3x2cos(3x2)dx=π2(410)cos(3π4+arctan(3)2)

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