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Tuesday, April 14, 2015

Indian Statistica Institute : Special Integration

Evaluate: nnmax{x+|x|,x[x]}dx where [x] is the floor function
When x>0 we see that x+|x|=2x.
Let kx<k+1,k=0,1,2,.........,n1 [x]=k,therefore,x[x]=xk
So,x+|x|=2x>xk=x[x]max{x+|x|,x[x]}=2x
n0max{x+|x|,x[x]}dx=2n0xdx=x2|n20=n2(A)
When x<0 we see that x+|x|=xx=0.
Let (k+1)x<k,k=0,1,2,.........,n1 [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
0nmax{x+|x|,x[x]}dx=
(n1)nx+(k+1)dx+(n2)(n1)x+(k+1)dx++01x+(k+1)dx=n1k=0k(k+1)x+(k+1)dx
=0nxdx+n1k=0k(k+1)(k+1)dx=n22+n1k=0(k+1)x|k(k+1)=n22+n1k=0(k+1)=n22+n(n+1)2(B)
Adding A and B we have, 0nmax{x+|x|,x[x]}dx+n0max{x+|x|,x[x]}dx=
nnmax{x+|x|,x[x]}dx=n22+n(n+1)2+n2=n22+n(n+1)2

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