Re: [微積] 一題積分..
Method III
Sin[Nx] = Exp[iNx]-Exp[-iNx]/2i
Sin[x] = Exp[ix]-Exp[-ix]/2i
Integrand
=(Sin[Nx]/Sin[x])^2
=(Exp[iNx]-Exp[-iNx]/Exp[ix]-Exp[-ix])^2
=Σ(Exp[ix])^2r * (Exp[-ix])^2(N-r-1) r from 0~N-1
=ΣExp[i*2(2r-N+1)x]
when 2r-N+1≠0
∫Exp[i*2(2r-N+1)x]dx=0
when 2r-N+1=0
∫Exp[i*2(2r-N+1)x]dx=∫dx=π
Hence,
Σ∫Exp[i*2(2r-N+1)x]dx=Nπ
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