# Simple unbounded integrand
# 
# 
# The integrand f in this example  is simple but unbounded and hence not Riemann integrable. For x = 1/2^n
# with `in`(n, `union`(nonnegint, {0}))we define f(x)=1/(n+1), otherwise f(x) = 0. A suitable gauge would be equal 
> epsilon/(2^n)(n+1)
# if x=1/2^n and 1otherwise. The definition of the Kurzweil inegral requires that  for
#  all tagged divisions which are gauge-fine
# 
# the Riemann sum is within  ϵ of the expected value of  the integral. Unfortunately and inevitably the computer will display only one sum. With the above choice our program will produce  0 for theRiemann sum,  a rather uninteresting result. Therefore we aim below  for the worst possible scenario for the Riemann sum. 
#  For a good display we well choose a fairly large ϵ. This might cause too large contribution from the subinterval which is tagged by 1.
#   In order to prevent this we make an adjustment in our definition of gauge. The choice of ϵ causes a fairly large error. However we are intersted in a good graphical display rather than in accuracy.
# 
> with(plots):
> read("khtools.m"):
> f:=proc(x)
> local n, f;
> if x=0 then f:=0;
> else n:=log[2](x): f:=(floor(n+1)-ceil(n))*(-n+1);
> end if:
> end:
> delta:=proc(x) 
> local n, delta;
>  if x = 0 then delta := eta 
> else n := evalf((log[2])(x));
>  delta := min(.1, floor(n+1)-ceil(n))*epsilon*x/(-n+1)+.9*(2^ceil(n)-x)
>  end if:
>  end: 
> epsilon:=0.9:eta:=0.05:
> khbox(f,0,1,delta);

The Riemann sum is   0.258750

> 
#