The Logic Off Topic list

--- In logic-ot@y..., Hendrik Jan Veenstra <h@k...> wrote:
> Thoughts from the mind of yoonchinet@y..., 08-11-2001:
> 
> >--- In logic-ot@y..., GAmoore@a... wrote:
> >  >    let  f_n(x) = 1/n for 0 <= x <= n, and f_n(x) = 0 elsewhere.
> >>
> >>    Then the
> >>         integral(   f_n(x) , x = 0 to x = infinity)  = 1 for all n,
> >
> >I'm sorry but wouldn't the result of this be: limit( ((1/n)*x, x = 0 
> >to x = y), y -> infinity) which is not equal to 1 for all n?
> 
> The graph of the functions Greg describes looks like (in Courier please)
> 
> 
>       |
> 1/n -|=======================
>       |
>       |
>       |
>       +----------------------+=======================
>                              n
> 
> i.e. a horizontal line at height 1/n, from x=0 to x=n, and at height 
> 0 for x>n.  The integral is simply the surface between the function 
> and the x-axis, which is clearly n*1/n = 1.
> 
> Your
>      limit( ((1/n)*x, x = 0 to x = y), y -> infinity)
> is indeed not 1 for all n (in fact, it's infinity for all n>0). 
> However, this limit is not the same as the previous integral. 
> (1/n)*x is a primitive function for 1/n -- but 1/n is _not_ what's 
> being integrated.


You are right! My mistake; the integration should've been done piece-wise. Thanks for correcting me.

Yoonchi.