Does this sequence of Riemann sums converge to Int a^b f(x) dx?

Question

Suppose the real valued f is unbounded in a neighborhood of a but its improper Riemann integrable exists anyway over the compact interval [a, b]. (A typical example is f(x) =1/sqrt(x) in [0,1], no matter the definition of f at 0). Let P_n be a sequence of partitions of [a, b] such that the norm of P_n (length of its largest interval) goes to 0 as n --> oo. Let S_n be a sequence of Riemann sums corresponding to P_n. Then, is it true that lim S_n = Integral a^b f(x) dx.

If f were bounded the answer would sure be yes, but since we have improper integrals I'm not sure.

Thank you for any help.

Answer 1

No. The reason is that you can always take the first term in the Riemann sum to be arbitrarily large. Let me explain in more detail. Say f is positive, P_n is your sequence of partitions, and (P_n)_1 is the first interval in your partition. Then since f is unbounded near a you can find for every n a point x_n in (P_n)_1 so that:

f(x_n) > n/length((P_n)_1).

Now we take x_n to be the sample point in that first interval and so:

S_n = f(x_n)*length((P_n)_1) + other terms

Since f is positive the "other terms" are >0, and so:

S_n > n.

Thus S_n -> infinity as n ->infinity.

Edit: Here's an extension to the question. If f is a function as in the original question, does there exist a sequence of Riemann sums S_n as described so that S_n -> int_a^b f dx? Note that what I and the other answerer already said shows that not all Riemann sums converge, but what I'm asking here is whether you can find some Riemann sum that converges to the integral. A final question would be, if a Riemann sum converges, does it neccesarily equal the integral? Just some more food for thought : )

Answer 2

The answer is no since the Riemann sum will have one term that is infinite (since the function is unbounded in one of the intervals of P_n).