When x>0 we see that x+|x|=2x.
Let k≤x<k+1,k=0,1,2,.........,n−1 ⇒[x]=k,therefore,x−[x]=x−k
So,x+|x|=2x>x−k=x−[x]⇒max{x+|x|,x−[x]}=2x
∫n0max{x+|x|,x−[x]}dx=2∫n0xdx=x2|n20=n2…(A)
When x<0 we see that x+|x|=x−x=0.
Let −(k+1)≤x<−k,k=0,1,2,.........,n−1 ⇒[x]=−(k+1),therefore,x−[x]=x+(k+1)
So,x+|x|=0<x+(k+1)=x−[x]⇒max{x+|x|,x−[x]}=x−[x]
∫−k−(k+1)max{x+|x|,x−[x]}dx=∫−k−(k+1)x+(k+1)dx
∫0−nmax{x+|x|,x−[x]}dx=
∫−(n−1)−nx+(k+1)dx+∫−(n−2)−(n−1)x+(k+1)dx+⋯+∫0−1x+(k+1)dx=∑n−1k=0∫−k−(k+1)x+(k+1)dx
=∫0−nxdx+∑n−1k=0∫−k−(k+1)(k+1)dx=−n22+∑n−1k=0(k+1)x|−k−(k+1)=−n22+∑n−1k=0(k+1)=−n22+n(n+1)2…(B)
Adding A and B we have, ∫0−nmax{x+|x|,x−[x]}dx+∫n0max{x+|x|,x−[x]}dx=
∫n−nmax{x+|x|,x−[x]}dx=−n22+n(n+1)2+n2=n22+n(n+1)2
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