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Is this Riemann Integrable?

  • Thread starter Lizzie11
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  • #1
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let f(x) = 1 when x in in [0,1)
f(x) = -1/2 when x is in [1,2)
f(x) = 1/3 when x is in [2, 3)

and so on, in othe words its the sequence (1/n)(-1)^n, whose series obviously converges to log 2. However is f(x) Riemann integrable and equal to this series?
If so, how to give an upper sum lower sum proof?, just choose a good partion?

thanks,

Lizzie
 

Answers and Replies

  • #2
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that function is riemann integrable but i'm not sure what it has to do with that sequence/series?
 
  • #3
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Ok, but how to prove it?

It has to do with that sequence because on each interval [n, n+1) it is equal to (1/n+1)(-1)^n.
 
  • #4
HallsofIvy
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Lizzie11 said:
Ok, but how to prove it?
It has to do with that sequence because on each interval [n, n+1) it is equal to (1/n+1)(-1)^n.

Perhaps it would help if you clarified what your function is. In your original post you had
"let f(x) = 1 when x in in [0,1)
f(x) = -1/2 when x is in [1,2)
f(x) = 1/3 when x is in [2, 3)"

Which is a piecewise constant function that has trivially integrable. Yes, it is equal to (-1)^n/(n+1) on the interval [n,n+1) but what does that have to do with the series 1- 1/2+ 1/3+....

Ah, wait- you aren't asking whether the function is Riemann integrable- you are asking whether the definite integral from 0 to infinity exists! I think you've answered your own question. The improper integral [tex]\int_0^\infty f(x)dx[/tex] exists if and only if the limit [tex]lim_{A->0}\int_0^A f(x) dx[/tex] exists.

With f as you give it, [tex]\int_0^A f(x)dx= \Sigma_{i=1}^{N}\frac{(-1)^i}{i+1}[/tex] where N is the largest integer less than or equal to A and, as you say, that series converges to log 2.
 
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