sintdt ≥0,∀k=0,1,2,... 从而 f(___) +∞ = ∫+∞ ___ F(t)sindtn == k=0 nF(nt) sint dt (2k+2)π nF(nt) sint dt 2kπ 2π ≥nF(nt)sintdt ∫π() =nF(nt)−F(2nπ−nt) ___() sintdt nF(nt)−F(2n...