G(x)=∫(0,x)[2f(t)-∫(t,t+2)f(s)ds]dt
證明:
因為f(x)是周期為2的連續(xù)函數(shù),f(x)=f(x+2)
又∫(t,t+2)f(s)ds=∫(t,2)f(s)ds+∫(2,t+2)f(s)ds
令s-2=v,ds=dv,則∫(2,t+2)f(s)ds=∫(0,t)f(v+2)dv=∫(0,t)f(v)dv=∫(0,t)f(s)ds
從而∫(t,t+2)f(s)ds=∫(0,t)f(s)ds+∫(t,2)f(s)ds=∫(0,2)f(s)ds=k（記）
那么G(x)=∫(0,x)[2f(t)-∫(t,t+2)f(s)ds]dt=∫(0,x)[2f(t)-k]dt
而G(x+2)=∫(0,x+2)[2f(t)-k]dt
令t-2=u,dt=du
則G(x+2)=∫(0,x+2)[2f(t)-k]dt=∫(0,x)[2f(u+2)-k]du+∫(x,x+2)[2f(u+2)-k]du
=∫(0,x)[2f(u)-k]du+∫(x,x+2)[2f(u)-k]du
注意到∫(x,x+2)[2f(u)-k]du=∫(x,2)[2f(u)-k]du+∫(2,x+2)[2f(u)-k]du
再令u-2=r,du=dr,∫(2,x+2)[2f(u)-k]du=∫(0,x)[2f(r+2)-k]dr=∫(0,x)[2f(u)-k]du
于是∫(x,x+2)[2f(u)-k]du=∫(0,2)[2f(u)-k]du=2∫(0,2)f(u)du-2k=2k-2k=0
因此G(x+2)=∫(0,x)[2f(u)-k]du=∫(0,x)[2f(t)-k]dt=G(x)
即G(x+2)=G(x)命題得證.