換元即可
∫[a,a+T] f(x) dx 令 u = x﹣a,du = dx
= ∫[0,T] f(u) du
= ∫[0,T] f(x) dx換元后被積函數(shù)應(yīng)該是f(u+a)哦，還沒可以變到f(u)對啊，暈了?！襕a,a+T] f(x) dx= ∫[a,0] f(x) dx + ∫[0,T] f(x) dx + ∫[T,a+T] f(x) dx 其中∫[a,0] f(x) dx + ∫[T,a+T] f(x) dx = ∫[a,0] f(x) dx +∫[0,a] f(u+T) du= ∫[a,0] f(x) dx + ∫[0,a] f(u) du= 0