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Question Number 136950 by BHOOPENDRA last updated on 28/Mar/21

Answered by Olaf last updated on 28/Mar/21

f^∧ ^c (ν) = ∫_(−∞) ^(+∞) f(s)cos(2πνs)ds = Ref^∧ (ν)  F((1/s)) = −iπsign(ν)  F(g(s)sin(as)) = (1/2)[g^∧ (ν−(a/(2π)))+g^∧ (ν+(a/(2π)))]  F(((sin(as))/s)) = (1/2)[−iπsign(ν−(a/(2π)))−iπsign(ν+(a/(2π)))]  F(((sin(as))/s)) = −((iπ)/2)[sign(ν−(a/(2π)))+sign(ν+(a/(2π)))]  f^∧ ^c (ν) = Ref^∧ (ν) = 0 ????

fc(ν)=+f(s)cos(2πνs)ds=Ref(ν)F(1s)=iπsign(ν)F(g(s)sin(as))=12[g(νa2π)+g(ν+a2π)]F(sin(as)s)=12[iπsign(νa2π)iπsign(ν+a2π)]F(sin(as)s)=iπ2[sign(νa2π)+sign(ν+a2π)]fc(ν)=Ref(ν)=0????

Answered by mathmax by abdo last updated on 28/Mar/21

f^★ (x)=(1/( (√(2π))))∫_R f(t)e^(−ixt)  dt  =(1/( (√(2π))))∫_(−∞) ^∞  f(t)e^(−ixt)  dt  =(√(2/π))∫_0 ^∞ f(t)cos(xt)dt(if f is even) ⇒ F^∧ (((sin(ax))/x))=(√(2/π))∫_0 ^∞ ((sin(at))/t)cos(xt)dt  we have cosp sinq =cosp.cos((π/2)−q)  =(1/2){cos(p−q+(π/2))+cos(p+q−(π/2)))  =(1/2){−sin(p−q)+sin(p+q)} ⇒cos(xt)sin(at)  =(1/2){sin(x+a)t−sin(x−a)t} ⇒ case 1   x>a  ∫_0 ^∞  ((cos(xt)sin(at))/t)dt =(1/2)∫_0 ^∞  ((sin(x+a)t)/t)dt(→(x+a)t=z)−(1/2)∫_0 ^∞ ((sin(x−a)t)/t)dt(→(x−a)t=z)  =(1/2)∫_0 ^∞  ((sinz)/(z/(x+a)))(dz/(x+a))  −(1/2)∫_0 ^∞  ((sinz)/(z/(x−a)))(dz/(x−a))=0 ⇒f^∧ (x)=0  case 2 −a<x<a ⇒∫_0 ^∞  ((cos(xt)sin(at))/t)dt  =(1/2)∫_0 ^∞  ((sinz)/z)dz +(1/2)∫_0 ^∞  ((sin(a−x)t)/t)dt =∫_0 ^∞  ((sinz)/z)dz=(π/2)  ⇒f^∧ (x)=(√(2/π)).(π/2)=((π(√2))/(2(√π))) =(√(π/2))

f(x)=12πRf(t)eixtdt=12πf(t)eixtdt=2π0f(t)cos(xt)dt(iffiseven)F(sin(ax)x)=2π0sin(at)tcos(xt)dtwehavecospsinq=cosp.cos(π2q)=12{cos(pq+π2)+cos(p+qπ2))=12{sin(pq)+sin(p+q)}cos(xt)sin(at)=12{sin(x+a)tsin(xa)t}case1x>a0cos(xt)sin(at)tdt=120sin(x+a)ttdt((x+a)t=z)120sin(xa)ttdt((xa)t=z)=120sinzzx+adzx+a120sinzzxadzxa=0f(x)=0case2a<x<a0cos(xt)sin(at)tdt=120sinzzdz+120sin(ax)ttdt=0sinzzdz=π2f(x)=2π.π2=π22π=π2

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