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Question Number 42260 by math khazana by abdo last updated on 21/Aug/18

let f(a) = ∫_(−∞) ^(+∞)  cos(ax^2 )dx with a>0  1) calculate f(a) interms of a  ) calculate ∫_(−∞) ^(+∞)    cos(2x^2 )dx  3) find the value of  ∫_(−∞) ^(+∞)  cos(x^2  +x+1)dx .

letf(a)=+cos(ax2)dxwitha>01)calculatef(a)intermsofa)calculate+cos(2x2)dx3)findthevalueof+cos(x2+x+1)dx.

Commented by maxmathsup by imad last updated on 21/Aug/18

3) ∫_(−∞) ^(+∞)  cos(x^2  +x+1)dx =∫_(−∞) ^(+∞)  cos((x+(1/2))^2  +(3/4))dx  =_(x+(1/2)=((√3)/2)t)       ∫_(−∞) ^(+∞)   cos((3/4)(t^2  +1))((√3)/2) dt  =((√3)/2) ∫_(−∞) ^(+∞)   cos((3/4)t^2  +(3/4))dt  =((√3)/2)  ∫_(−∞) ^(+∞)   {cos((3/4)t^2 )cos((3/4))−sin((3/4)t^2 )sin((3/4))}dt  =((√3)/2) cos((3/4)) ∫_(−∞) ^(+∞)  cos((3/4)t^2 )dt −((√3)/2) sin((3/4))∫_(−∞) ^(+∞)  sin((3/4)t^2 )dt  =((√3)/2) cos((3/4))((√(2π))/(2(√(3/4))))  −((√3)/2) sin((3/4))((√(2π))/(2(√(3/4))))  =((√(2π))/2) cos((3/4))  −((√(2π))/2) sin((3/4)) ⇒  ∫_(−∞) ^(+∞)  cos(x^2  +x +1)dx =(√(π/2)){cos((3/4))−sin((3/4))} .

3)+cos(x2+x+1)dx=+cos((x+12)2+34)dx=x+12=32t+cos(34(t2+1))32dt=32+cos(34t2+34)dt=32+{cos(34t2)cos(34)sin(34t2)sin(34)}dt=32cos(34)+cos(34t2)dt32sin(34)+sin(34t2)dt=32cos(34)2π23432sin(34)2π234=2π2cos(34)2π2sin(34)+cos(x2+x+1)dx=π2{cos(34)sin(34)}.

Commented by maxmathsup by imad last updated on 21/Aug/18

we have ∫_(−∞) ^(+∞)  cos(ax^2 )dx−i ∫_(−∞) ^(+∞)  sin(ax^2 )dx  = ∫_(−∞) ^(+∞)   e^(−iax^2 ) dx  =∫_(−∞) ^(+∞)   e^(−((√(ia))x)^2 ) dx =_((√(ia))x =t)     ∫_(−∞) ^(+∞)   e^(−t^2 )  (dt/(√(ia)))  = (1/((√a)e^((iπ)/4) )) (√π)=((√π)/(√a)) e^(−((iπ)/4))   =((√π)/(√a)) {cos((π/4))−isin((π/4))} ⇒  f(a) =((√π)/(√a)) ((√2)/2)  =((√(2π))/(2(√a)))  . also we have ∫_(−∞) ^(+∞)   sin(ax^2 )dx =((√(2π))/(2(√a)))  2) ∫_(−∞) ^(+∞)   cos(2x^2 )dx =f(2) = ((√(2π))/(2(√2)))  =((√π)/2)  .

wehave+cos(ax2)dxi+sin(ax2)dx=+eiax2dx=+e(iax)2dx=iax=t+et2dtia=1aeiπ4π=πaeiπ4=πa{cos(π4)isin(π4)}f(a)=πa22=2π2a.alsowehave+sin(ax2)dx=2π2a2)+cos(2x2)dx=f(2)=2π22=π2.

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