证明：
设 F(x) = [x,a]∫f(u)du ；[x,a]表示积分限下限为a,上限为x
F(y) = [y,a]∫dx*[x,a]∫xf(u)du = [y,a]∫x*F(x)*dx
首先证明F(x)的奇偶性
F(-x) = [-x,a]∫f(u)du
设 u =-v,则上式变为：
F(-x) = [x,-a]∫f(-v)d(-v),由于f(v)为奇函数,f(-v) = -f(v),因此
F(-x) = [x,-a]∫f(v)dv
= [x,a]∫f(v)dv + [a,-a]∫f(v)dv
= F(x) + 0 =F(x)
即：F(x) 为偶函数
同样：
F(-y) = [-y,a]∫x*F(x)*dx,令 x=-z,则有：
F(-y) = [y,-a]∫(-z)F(-z)*d(-z)
= [y,-a]∫zF(z)dz
= [y,a]∫zF(z)dz + [a,-a]∫zF(z)dz
= F(y) + 0 =F(y)
因此F(y)为偶函数,即[y,a]∫dx*[x,a]∫xf(u)du 为偶函数