求∫∞0sin(at)tdt及∫∞−∞sin(at)te−jωtdt令f(t)=sin(at)t⇒I(s)=L{f(t)}=∫∞0sin(at)te−stdt=∫∞sax2+a2dx=[tan−1xa]|∞s=π2−tan−1sa⇒∫∞0sin(at)tdt=I(0)=π2又f(−t)=−sin(at)−t=sin(at)t=f(t)⇒f is even function⇒∫∞−∞f(t)sin(ωt)dt=0⇒Fourier transform F(ω)=∫∞−∞f(t)e−jωtdt=∫∞−∞f(t)(cos(ωt)−isin(ωt))dt=∫∞−∞f(t)cos(ωt)dt=2∫∞0f(t)cos(ωt)dt=2∫∞0sin(at)tcos(ωt)dt=∫∞0sin(a+ω)t+sin(a−ω)tdt=π2+π2=π
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