Thread: help me again riemann integral

1. help me again riemann integral

sorry for posting another question but I have a class test and I'm getting full of doubts.. ok, the definition of integral by Riemann that I've studied is(I know you know it): let f bounded on (a;b), f is integrable on (a,b) if the upper and lower sum's limits converge to the same finite number.
then should it be true that
if f(x) is bounded on (a,b) then it is integrable according to Riemann?..
the book says that if F(x) is continuous it is integrablle on (a,b) according to Riemann.
thanks a lot

2. Originally Posted by 0123
sorry for posting another question but I have a class test and I'm getting full of doubts.. ok, the definition of integral by Riemann that I've studied is(I know you know it): let f bounded on (a;b), f is integrable on (a,b) if the upper and lower sum's limits converge to the same finite number.
then should it be true that
if f(x) is bounded on (a,b) then it is integrable according to Riemann?..
the book says that if F(x) is continuous it is integrablle on (a,b) according to Riemann.
thanks a lot
The answer is no. Consider, for example:
$\int_0^{\infty}dx sin(x)$

sin(x) is bounded on $[0, \infty)$ but as the function has no limiting value for large x (and does not blow up) we can't say anything about the value of this integral. (Though, obviously, we can do the indefinite integral of sin(x).)

-Dan

Edit: I'm rethinking this. I don't recall off the top of my head if an integral over $[0, \infty)$ is, by definition, Riemann...

3. Originally Posted by 0123
sorry for posting another question but I have a class test and I'm getting full of doubts.. ok, the definition of integral by Riemann that I've studied is(I know you know it): let f bounded on (a;b), f is integrable on (a,b) if the upper and lower sum's limits converge to the same finite number.
then should it be true that
if f(x) is bounded on (a,b) then it is integrable according to Riemann?..
The classical counter example to this is the function on an open interval which is 1 if its argument is rational and zero otherwise. (the limit of lower Riemann sums is 0, and of upper Riemann sums is (b-a))

RonL

4. Originally Posted by topsquark
The answer is no. Consider, for example:
$\int_0^{\infty}dx sin(x)$

sin(x) is bounded on $[0, \infty)$ but as the function has no limiting value for large x (and does not blow up) we can't say anything about the value of this integral. (Though, obviously, we can do the indefinite integral of sin(x).)

-Dan
This is an improper integral, any partition of the range into a finite number of sub-intervals must have a subinterval of infinite length, and so the Riemann sums are not properly defined.

This is usually handled by constructing the integral over a finite range and then taking the limit as one end goes to infinity.

RonL

5. Originally Posted by 0123
the book says that if F(x) is continuous it is integrablle on (a,b) according to Riemann.
Now let:

$F(x)=\frac{1}{(x-b)^4}$

This is continuous on the open interval $(a,b)$ but is not Riemann integrable over the interval. You may need continuity on the closed interval $[a,b]$ or uniform continuity on $(a,b)$ or boundedness (and continuity) for this to work

RonL

6. If a function is defined and bounded not necessary it is integrable. Look here. The Dirichlet function is not integrable.. The if and only if theorem says that if a function (defined on intervao) is Riemann integrable if and only if it is almost everywhere continous. The term "almost everywhere" is more set theortetic.