换元即可
∫[a,a+T] f(x) dx 令 u = x﹣a,du = dx
= ∫[0,T] f(u) du
= ∫[0,T] f(x) dx换元后被积函数应该是f(u+a)哦，还没可以变到f(u)对啊，晕了。∫[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