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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(1s)=−iπsign(ν)F(g(s)sin(as))=12[g∧(ν−a2π)+g∧(ν+a2π)]F(sin(as)s)=12[−iπsign(ν−a2π)−iπsign(ν+a2π)]F(sin(as)s)=−iπ2[sign(ν−a2π)+sign(ν+a2π)]f∧c(ν)=Ref∧(ν)=0????
Answered by mathmax by abdo last updated on 28/Mar/21
f★(x)=12π∫Rf(t)e−ixtdt=12π∫−∞∞f(t)e−ixtdt=2π∫0∞f(t)cos(xt)dt(iffiseven)⇒F∧(sin(ax)x)=2π∫0∞sin(at)tcos(xt)dtwehavecospsinq=cosp.cos(π2−q)=12{cos(p−q+π2)+cos(p+q−π2))=12{−sin(p−q)+sin(p+q)}⇒cos(xt)sin(at)=12{sin(x+a)t−sin(x−a)t}⇒case1x>a∫0∞cos(xt)sin(at)tdt=12∫0∞sin(x+a)ttdt(→(x+a)t=z)−12∫0∞sin(x−a)ttdt(→(x−a)t=z)=12∫0∞sinzzx+adzx+a−12∫0∞sinzzx−adzx−a=0⇒f∧(x)=0case2−a<x<a⇒∫0∞cos(xt)sin(at)tdt=12∫0∞sinzzdz+12∫0∞sin(a−x)ttdt=∫0∞sinzzdz=π2⇒f∧(x)=2π.π2=π22π=π2
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