integration problem

∫[0~2^n] φ(u) /4^n du (u=2^n x)
= [∫[0~0.5] φ(u)du+∫[0.5~1.5] φ(u)du+...+
∫[2^n-1.5~2^n-0.5]φ(u)du+∫[2^n-0.5~2^n]φ(u)du] / 4^n
= [∫[0~0.5] udu+∫[0.5~1.5] |u-1|du+∫[1.5~2.5] |u-2|du+...+
∫[2^n-1.5~2^n-0.5] |u-2^n+1|du+∫[2^n-0.5~2^n] | u-2^n| du ]/4^n
= [1/8 + 2*(1/8)+2*(1/8)+...+2*(1/8)+ 1/8 ]/4^n
= (1/4)*(2^n) / 4^n=(1/4)/2^n

so, ∫[0~1] B(x)dx=1/2

註: By M-test, B(x)=Σ (0->∞) φ (2^n x)/2^n是 uniformly conv.
so, 可以逐項積分!

更正:∫[0~2^n] φ(u) /4^n du (u=2^n x)
= [∫[0~0.5] φ(u)du+∫[0.5~1.5] φ(u)du+...+
∫[2^n-1.5~2^n-0.5]φ(u)du+∫[2^n-0.5~2^n]φ(u)du] / 4^n
= [∫[0~0.5] udu+∫[0.5~1.5] |u-1|du+∫[1.5~2.5] |u-2|du+...+
∫[2^n-1.5~2^n-0.5] |u-2^n+1|du+∫[2^n-0.5~2^n] | u-2^n| du ]/4^n

2009-10-05 18:32:15 補充：
= [1/8 + 2*(1/8)+2*(1/8)+...+2*(1/8)+ 1/8 ]/4^n
= (1/4)*(2^n) / 4^n=(1/4)/2^n
so, ∫[0~1] B(x)dx=1/2

Sorry! 原作答誤為φ(x)=the nearest integer

∫[0~0.5] φ(u)du =∫[0~0.5] udu = 1/8
∫[0.5~1.5] φ(u)du = 2∫[0~0.5] φ(u)du = 1/4
...
∫[2^n-0.5~2^n]φ(u)du = ∫[0~0.5] φ(u)du = 1/8
所以，
∫[0~0.5] φ(u)du+∫[0.5~1.5] φ(u)du+...+
∫[2^n-1.5~2^n-0.5]φ(u)du+∫[2^n-0.5~2^n]φ(u)du
= (1/4)*2^n